Balbharti 12th Maharashtra State Board Maths Solutions Book
Pdf Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 Questions and Answers.
I : Choose correct alternatives.
Question 1.
If the equation 4x2 + hxy + y2 = 0 represents two coincident lines, then h = _________.
(A) Β± 2
(B) Β± 3
(C) Β± 4
(D) Β± 5
Solution:
(C) Β± 4
Question 2.
If the lines represented by kx2 β 3xy + 6y2 = 0 are perpendicular to each other then _________.
(A) k = 6
(B) k = -6
(C) k = 3
(D) k = -3
Solution:
(B) k = -6
Question 3.
Auxiliary equation of 2x2 + 3xy β 9y2 = 0 is _________.
(A) 2m2 + 3m β 9 = 0
(B) 9m2 β 3m β 2 = 0
(C) 2m2 β 3m + 9 = 0
(D) -9m2 β 3m + 2 = 0
Solution:
(B) 9m2 β 3m β 2 = 0
Question 4.
The difference between the slopes of the lines represented by 3x2 β 4xy + y2 = 0 is _________.
(A) 2
(B) 1
(C) 3
(D) 4
Solution:
(A) 2
Question 5.
If the two lines ax2 +2hxy+ by2 = 0 make angles Ξ± and Ξ² with X-axis, then tan (Ξ± + Ξ²) = _____.
(A) \(\frac{h}{a+b}\)
(B) \(\frac{h}{a-b}\)
(C) \(\frac{2 h}{a+b}\)
(D) \(\frac{2 h}{a-b}\)
Solution:
(D) \(\frac{2 h}{a-b}\)
Question 6.
If the slope of one of the two lines \(\frac{x^{2}}{a}+\frac{2 x y}{h}+\frac{y^{2}}{b}\) = 0 is twice that of the other, then ab:h2 = ___.
(A) 1 : 2
(B) 2 : 1
(C) 8 : 9
(D) 9 : 8
Solution:
(D) 9 : 8
Question 7.
The joint equation of the lines through the origin and perpendicular to the pair of lines 3x2 + 4xy β 5y2 = 0 is _________.
(A) 5x2 + 4xy β 3y2 = 0
(B) 3x2 + 4xy β 5y2 = 0
(C) 3x2 β 4xy + 5y2 = 0
(D) 5x2 + 4xy + 3y2 = 0
Solution:
(A) 5x2 + 4xy β 3y2 = 0
Question 8.
If acute angle between lines ax2 + 2hxy + by2 = 0 is, \(\frac{\pi}{4}\) then 4h2 = _________.
(A) a2 + 4ab + b2
(B) a2 + 6ab + b2
(C) (a + 2b)(a + 3b)
(D) (a β 2b)(2a + b)
Solution:
(B) a2 + 6ab + b2
Question 9.
If the equation 3x2 β 8xy + qy2 + 2x + 14y + p = 1 represents a pair of perpendicular lines then
the values of p and q are respectively _________.
(A) -3 and -7
(B) -7 and -3
(C) 3 and 7
(D) -7 and 3
Solution:
(B) -7 and -3
Question 10.
The area of triangle formed by the lines x2 + 4xy + y2 = 0 and x β y β 4 = 0 is _________.
(A) \(\frac{4}{\sqrt{3}}\) Sq. units
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
(C) \(\frac{16}{\sqrt{3}}\) Sq. units
(D)\(\frac{15}{\sqrt{3}}\) Sq. units
Solution:
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
[Hint : Area = \(\frac{p^{2}}{\sqrt{3}}\), where p is the length of perpendicular from the origin to x β y β 4 = 0]
Question 11.
The combined equation of the co-ordinate axes is _________.
(A) x + y = 0
(B) x y = k
(C) xy = 0
(D) x β y = k
Solution:
(C) xy = 0
Question 12.
If h2 = ab, then slope of lines ax2 + 2hxy + by2 = 0 are in the ratio _________.
(A) 1 : 2
(B) 2 : 1
(C) 2 : 3
(D) 1 : 1
Solution:
(D) 1 : 1
[Hint: If h2 = ab, then lines are coincident. Therefore slopes of the lines are equal.]
Question 13.
If slope of one of the lines ax2 + 2hxy + by2 = 0 is 5 times the slope of the other, then 5h2 = _________.
(A) ab
(B) 2 ab
(C) 7 ab
(D) 9 ab
Solution:
(D) 9 ab
Question 14.
If distance between lines (x β 2y)2 + k(x β 2y) = 0 is 3 units, then k =
(A) Β± 3
(B) Β± 5\(\sqrt {5}\)
(C) 0
(D) Β± 3\(\sqrt {5}\)
Solution:
(D) Β± 3\(\sqrt {5}\)
[Hint: (x β 2y)2 + k(x β 2y) = 0
β΄ (x β 2y)(x β 2y + k) = 0
β΄ equations of the lines are x β 2y = 0 and x β 2y + k = 0 which are parallel to each other.
β΄ \(\left|\frac{k-0}{\sqrt{1+4}}\right|\) = 3
β΄ k = Β± 3\(\sqrt {5}\)
II. Solve the following.
Question 1.
Find the joint equation of lines:
(i) x β y = 0 and x + y = 0
Solution:
The joint equation of the lines x β y = 0 and
x + y = 0 is
(x β y)(x + y) = 0
β΄ x2 β y2 = 0.
(ii) x + y β 3 = 0 and 2x + y β 1 = 0
Solution:
The joint equation of the lines x + y β 3 = 0 and 2x + y β 1 = 0 is
(x + y β 3)(2x + y β 1) = 0
β΄ 2x2 + xy β x + 2xy + y2 β y β 6x β 3y + 3 = 0
β΄ 2x2 + 3xy + y2 β 7x β 4y + 3 = 0.
(iii) Passing through the origin and having slopes 2 and 3.
Solution:
We know that the equation of the line passing through the origin and having slope m is y = mx. Equations of the lines passing through the origin and having slopes 2 and 3 are y = 2x and y = 3x respectively.
i.e. their equations are
2x β y = 0 and 3x β y = 0 respectively.
β΄ their joint equation is (2x β y)(3x β y) = 0
β΄ 6x2 β 2xy β 3xy + y2 = 0
β΄ 6x2 β 5xy + y2 = 0.
(iv) Passing through the origin and having inclinations 60Β° and 120Β°.
Solution:
Slope of the line having inclination ΞΈ is tan ΞΈ .
Inclinations of the given lines are 60Β° and 120Β°
β΄ their slopes are m1 = tan60Β° = \(\sqrt {3}\) and
m2 = tan 120Β° = tan (180Β° β 60Β°)
= -tan 60Β° = β\(\sqrt {3}\)
Since the lines pass through the origin, their equa-tions are
y = \(\sqrt {3}\)x and y= β\(\sqrt {3}\)x
i.e., \(\sqrt {3}\)x β y = 0 and \(\sqrt {3}\)x + y = 0
β΄ the joint equation of these lines is
(\(\sqrt {3}\)x β y)(\(\sqrt {3}\)x + y) = 0
β΄ 3x2 β y2 = 0.
(v) Passing through (1, 2) amd parallel to the co-ordinate axes.
Solution:
Equations of the coordinate axes are x = 0 and y = 0
β΄ the equations of the lines passing through (1, 2) and parallel to the coordinate axes are x = 1 and y =1
i.e. x β 1 = 0 and y β 2 0
β΄ their combined equation is
(x β 1)(y β 2) = 0
β΄ x(y β 2) β 1(y β 2) = 0
β΄ xy β 2x β y + 2 = 0
(vi) Passing through (3, 2) and parallel to the line x = 2 and y = 3.
Solution:
Equations of the lines passing through (3, 2) and parallel to the lines x = 2 and y = 3 are x = 3 and y = 2.
i.e. x β 3 = 0 and y β 2 = 0
β΄ their joint equation is
(x β 3)(y β 2) = 0
β΄ xy β 2x β 3y + 6 = 0.
(vii) Passing through (-1, 2) and perpendicular to the lines x + 2y + 3 = 0 and 3x β 4y β 5 = 0.
Solution:
Let L1 and L2 be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x β 4y β 5 = 0 respectively.
Slopes of the lines x + 2y + 3 = 0 and 3x β 4y β 5 = 0 are \(-\frac{1}{2}\) and \(-\frac{3}{-4}=\frac{3}{4}\) respectively.
β΄ slopes of the lines L1and L2 are 2 and \(\frac{-4}{3}\) respectively.
Since the lines L1 and L2 pass through the point (-1, 2), their equations are
β΄ (y β y1) = m(x β x1)
β΄ (y β 2) = 2(x + 1)
β y β 1 = 2x + 2
β 2x β y + 4 = 0 and
β΄ (y β 2) = \(\left(\frac{-4}{3}\right)\)(x + 1)
β 3y β 6 = (-4)(x + 1)
β 3y β 6 = -4x + 4
β 4x + 3y β 6 + 4 = 0
β 4x + 3y β 2 = 0
their combined equation is
β΄ (2x β y + 4)(4x + 3y β 2) = 0
β΄ 8x2 + 6xy β 4x β 4xy β 3y2 + 2y + 16x + 12y β 8 = 0
β΄ 8x2 + 2xy + 12x β 3y2 + 14y β 8 = 0
(viii) Passing through the origin and having slopes 1 + \(\sqrt {3}\) and 1 β \(\sqrt {3}\)
Solution:
Let l1 and l2 be the two lines. Slopes of l1 is 1 + \(\sqrt {3}\) and that of l2 is 1 β \(\sqrt {3}\)
Therefore the equation of a line (l1) passing through the origin and having slope is
y = (1 + \(\sqrt {3}\))x
β΄ (1 + \(\sqrt {3}\))x β y = 0 ..(1)
Similarly, the equation of the line (l2) passing through the origin and having slope is
y = (1 β \(\sqrt {3}\))x
β΄ (1 β \(\sqrt {3}\))x β y = 0 β¦(2)
From (1) and (2) the required combined equation is
β΄ (1 β 3)x2 β 2xy + y2 = 0
β΄ -2x2 β 2xy + y2 = 0
β΄ 2x2 + 2xy β y2 = 0
This is the required combined equation.
(ix) Which are at a distance of 9 units from the Y β axis.
Solution:
Equations of the lines, which are parallel to the Y-axis and at a distance of 9 units from it, are x = 9 and x = -9
i.e. x β 9 = 0 and x + 9 = 0
β΄ their combined equation is
(x β 9)(x + 9) = 0
β΄ x2 β 81 = 0.
(x) Passing through the point (3, 2), one of which is parallel to the line x β 2y = 2 and other is perpendicular to the line y = 3.
Solution:
Let L1 be the line passes through (3, 2) and parallel to the line x β 2y = 2 whose slope is \(\frac{-1}{-2}=\frac{1}{2}\)
β΄ slope of the line L1 is \(\frac{1}{2}\).
β΄ equation of the line L1 is
y β 2 = \(\frac{1}{2}\)(x β 3)
β΄ 2y β 4 = x β 3 β΄ x β 2y + 1 = 0
Let L2 be the line passes through (3, 2) and perpendicular to the line y = 3.
β΄ equation of the line L2 is of the form x = a.
Since L2 passes through (3, 2), 3 = a
β΄ equation of the line L2 is x = 3, i.e. x β 3 = 0
Hence, the equations of the required lines are
x β 2y + 1 = 0 and x β 3 = 0
β΄ their joint equation is
(x β 2y + 1)(x β 3) = 0
β΄ x2 β 2xy + x β 3x + 6y β 3 = 0
β΄ x2 β 2xy β 2x + 6y β 3 = 0.
(xi) Passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18.
Solution:
Let L1 and L2 be the lines passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18 respectively.
Slopes of the lines x + 2y = 19 and 3x + y = 18 are \(-\frac{1}{2}\) and \(-\frac{3}{1}\) = -3 respectively.
Since the lines L1 and L2 pass through the origin, their equations are
y = 2x and y = \(\frac{1}{3}\)x
i.e. 2x β y = 0 and x β 3y = 0
β΄ their combined equation is
(2x β y)(x β 3y) = 0
β΄ 2x2 β 6xy β xy + 3y2 = 0
β΄ 2x2 β 7xy + 3y2 = 0.
Question 2.
Show that each of the following equation represents a pair of lines.
(i) x2 + 2xy β y2 = 0
Solution:
Comparing the equation x2 + 2xy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 2, i.e. h = 1 and b = -1
β΄ h2 β ab = (1)2 β 1(-1) = 1 + 1=2 > 0
Since the equation x2 + 2xy β y2 = 0 is a homogeneous equation of second degree and h2 β ab > 0, the given equation represents a pair of lines which are real and distinct.
(ii) 4x2 + 4xy + y2 = 0
Solution:
Comparing the equation 4x2 + 4xy + y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 4, 2h = 4, i.e. h = 2 and b = 1
β΄ h2 β ab = (2)2 β 4(1) = 4 β 4 = 0
Since the equation 4x2 + 4xy + y2 = 0 is a homogeneous equation of second degree and h2 β ab = 0, the given equation represents a pair of lines which are real and coincident.
(iii) x2 β y2 = 0
Solution:
Comparing the equation x2 β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 0, i.e. h = 0 and b = -1
β΄ h2 β ab = (0)2 β 1(-1) = 0 + 1 = 1 > 0
Since the equation x2 β y2 = 0 is a homogeneous equation of second degree and h2 β ab > 0, the given equation represents a pair of lines which are real and distinct.
(iv) x2 + 7xy β 2y2 = 0
Solution:
Comparing the equation x2 + 7xy β 2y2 = 0
a = 1, 2h = 7 i.e., h = \(\frac{7}{2}\) and b = -2
β΄ h2 β ab = \(\left(\frac{7}{2}\right)^{2}\) β 1(-2)
= \(\frac{49}{4}\) + 2
= \(\frac{57}{4}\) i.e. 14.25 = 14 > 0
Since the equation x2 + 7xy β 2y2 = 0 is a homogeneous equation of second degree and h2 β ab > 0, the given equation represents a pair of lines which are real and distinct.
(v) x2 β 2\(\sqrt {3}\) xy β y2 = 0
Solution:
Comparing the equation x2 β 2\(\sqrt {3}\) xy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h= -2\(\sqrt {3}\), i.e. h = β\(\sqrt {3}\) and b = 1
β΄ h2 β ab = (-\(\sqrt {3}\))2 β 1(1) = 3 β 1 = 2 > 0
Since the equation x2 β 2\(\sqrt {3}\)xy β y2 = 0 is a homoΒ¬geneous equation of second degree and h2 β ab > 0, the given equation represents a pair of lines which are real and distinct.
Question 3.
Find the separate equations of lines represented by the following equations:
(i) 6x2 β 5xy β 6y2 = 0
Solution:
6x2 β 5xy β 6y2 = 0
β΄ 6x2 β 9xy + 4xy β 6y2 = 0
β΄ 3x(2x β 3y) + 2y(2x β 3y) = 0
β΄ (2x β 3y)(3x + 2y) = 0
β΄ the separate equations of the lines are
2x β 3y = 0 and 3x + 2y = 0.
(ii) x2 β 4y2 = 0
Solution:
x2 β 4y2 = 0
β΄ x2 β (2y)2 = 0
β΄(x β 2y)(x + 2y) = 0
β΄ the separate equations of the lines are
x β 2y = 0 and x + 2y = 0.
(iii) 3x2 β y2 = 0
Solution:
3x2 β y2 = 0
β΄ (\(\sqrt {3}\) x)2 β y2 = 0
β΄ (\(\sqrt {3}\)x β y)(\(\sqrt {3}\)x + y) = 0
β΄ the separate equations of the lines are
\(\sqrt {3}\)x β y = 0 and \(\sqrt {3}\)x + y = 0.
(iv) 2x2 + 2xy β y2 = 0
Solution:
2x2 + 2xy β y2 = 0
β΄ The auxiliary equation is -m2 + 2m + 2 = 0
β΄ m2 β 2m β 2 = 0
m1 = 1 + \(\sqrt {3}\) and m2 = 1 β \(\sqrt {3}\) are the slopes of the lines.
β΄ their separate equations are
y = m1x and y = m2x
i.e. y = (1 + \(\sqrt {3}\))x and y = (1 β \(\sqrt {3}\))x
i.e. (\(\sqrt {3}\) + 1)x β y = 0 and (\(\sqrt {3}\) β 1)x + y = 0.
Question 4.
Find the joint equation of the pair of lines through the origin and perpendicular to the lines
given by :
(i) x2 + 4xy β 5y2 = 0
Solution:
Comparing the equation x2 + 4xy β 5y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 4, b= -5
Let m1 and m2 be the slopes of the lines represented by x2 + 4xy β 5y2 = 0.
Now, required lines are perpendicular to these lines
β΄ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m1y = -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
β΄ their combined equation is
(x + m1x + m2y) = 0
β΄ x2 + (m1 + m2)xy + m1m2y2 = 0
β΄ x2 + \(\frac{4}{5}\)xy β \(\frac{1}{5}\)y2 = 0 β¦[By (1)]
β΄ 5x2 + 4xy β y2 = 0
(ii) 2x2 β 3xy β 9y2 = 0
Solution:
Comparing the equation 2x2 β 3xy β 9y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 2, 2h = -3, b = -9
Let m1 and m2 be the slopes of the lines represented by 2x2 β 3xy β 9y2 = 0
β΄ m1 + m2 =\(\frac{-2 h}{b}=-\frac{3}{9}\) and m1m2 = \(\frac{a}{b}=-\frac{2}{9}\) β¦(1)
Now, required lines are perpendicular to these lines
β΄ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m1y = -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
β΄ their combined equation is
(x + m1y)(x + m2y) = 0
β΄ x2 + (m1 + m2)xy + m1m2y2 = 0
β΄ x2 + \(\left(-\frac{3}{9}\right)\)xy + \(\left(-\frac{2}{9}\right)\)y2 = 0 β¦[By (1)]
β΄ 9x2 β 3xy β 2y2 = 0
(iii) x2 + xy β y2 = 0
Solution:
Comparing the equation x2+ xy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 1, b = -1
Let m1 and m2 be the slopes of the lines represented by x2 + xy β y2 = 0
β΄ m1 + m2 = \(\frac{-2 h}{b}=\frac{-1}{-1}\) and m1m2 = \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{-1}\) = -1 ..(1)
Now, required lines are perpendicular to these lines
β΄ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m1y = -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
β΄ their combined equation is
(x + m1y)(x + m2y) = 0
β΄ x2 + (m1 + m2) + m1m2y2 = 0
β΄ x2 + 1xy + (-1)y2 = 0 β¦[By (1)]
β΄ x2 + xy β y2 = 0
Question 5.
Find k if
(i) The sum of the slopes of the lines given by 3x2 + kxy β y2 = 0 is zero.
Solution:
Comparing the equation 3x2 + kxy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 3, 2h = k, b = -1
Let m1 and m2 be the slopes of the lines represented by 3x2 + kxy β y2 = 0.
β΄ m1 + m2 = \(\frac{-2 h}{b}=\frac{-k}{-1}\) = k
Now, m1 + m2 = 0 β¦ (Given)
β΄ k = 0.
(ii) The sum of slopes of the lines given by 2x2 + kxy β 3y2 = 0 is equal to their product.
Question is modified.
The sum of slopes of the lines given by x2 + kxy β 3y2 = 0 is equal to their product.
Solution:
Comparing the equation x2 + kxy β 3y2 = 0, with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = k, b = -3
Let m1 and m2 be the slopes of the lines represented by x2 + kxy β 3y2 = 0.
β΄ m1 + m2 = \(-\frac{2 h}{b}=\frac{-k}{-3}=\frac{k}{3}\)
and m1m2 = \(\frac{a}{b}=\frac{1}{-3}=\frac{-1}{3}\)
Now, m1 + m2 = m1m2 β¦ (Given)
β΄ \(\frac{k}{3}=\frac{-1}{3}\)
β΄ k = -1.
(iii) The slope of one of the lines given by 3x2 β 4xy + ky2 = 0 is 1.
Solution:
The auxiliary equation of the lines given by 3x2 β 4xy + ky2 = 0 is km2 β 4m + 3 = 0.
Given, slope of one of the lines is 1.
β΄ m = 1 is the root of the auxiliary equation km2 β 4m + 3 = 0.
β΄ k(1)2 β 4(1) + 3 = 0
β΄ k β 4 + 3 = 0
β΄ k = 1.
(iv) One of the lines given by 3x2 β kxy + 5y2 = 0 is perpendicular to the 5x + 3y = 0.
Solution:
The auxiliary equation of the lines represented by 3x2 β kxy + 5y2 = 0 is 5m2 β km + 3 = 0.
Now, one line is perpendicular to the line 5x + 3y = 0, whose slope is \(-\frac{5}{3}\).
β΄ slope of that line = m = \(\frac{3}{5}\)
β΄ m = \(\frac{3}{5}\) is the root of the auxiliary equation 5
5m2 β km + 3 = 0.
β΄ 5\(\left(\frac{3}{5}\right)^{2}\) β k\(\left(\frac{3}{5}\right)\) + 3 = 0
β΄ \(\frac{9}{5}-\frac{3 k}{5}\) + 3 = 0
β΄ 9 β 3k + 15 = 0
β΄ 3k = 24
β΄ k = 8.
(v) The slope of one of the lines given by 3x2 + 4xy + ky2 = 0 is three times the other.
Solution:
3x2 + 4xy + ky2 = 0
β΄ divide by x2
β΄ y = mx
β΄ \(\frac{\mathrm{y}}{\mathrm{x}}\) = m
put \(\frac{\mathrm{y}}{\mathrm{x}}\) = m in equation (1)
Comparing the equation km2 + 4m + 3 = 0 with ax2 + 2hxy+ by2 = 0, we get,
a = k, 2h = 4, b = 3
m1 = 3m2 ..(given condition)
m1 + m2 = \(\frac{-2 h}{k}=-\frac{4}{k}\)
m1m2 = \(\frac{a}{b}=\frac{3}{k}\)
m1 + m2 = \(-\frac{4}{\mathrm{k}}\)
4m2 = \(-\frac{4}{\mathrm{k}}\) β¦(m1 = 3m2)
m2 = \(-\frac{1}{\mathrm{k}}\)
m1m2 = \(\frac{3}{k}\)
\(3 \mathrm{~m}_{2}^{2}=\frac{3}{\mathrm{k}}\) β¦(m1 = 3m2)
\(3\left(-\frac{1}{\mathrm{k}}\right)^{2}=\frac{3}{\mathrm{k}}\) β¦(m2 = \(-\frac{1}{k}\))
\(\frac{1}{k^{2}}=\frac{1}{k}\)
k2 = k
k = 1 or k = 0
(vi) The slopes of lines given by kx2 + 5xy + y2 = 0 differ by 1.
Solution:
Comparing the equation kx2 + 5xy +y2 = 0 with ax2 + 2hxy + by2
a = k, 2h = 5 i.e. h = \(\frac{5}{2}\)
m1 + m2 = \(\frac{-2 h}{b}=-\frac{5}{1}\) = -5
and m1m2 = \(\frac{a}{b}=\frac{k}{1}\) = k
the slope of the line differ by (m1 β m2) = 1 β¦(1)
β΄ (m1 β m2)2 = (m1 + m2)2 β 4m1m2
(m1 β m2)2 = (-5)2 β 4(k)
(m1 β m2)2 = 25 β 4k
1 = 25 β 4k ..[By (1)]
4k = 24
k = 6
(vii) One of the lines given by 6x2 + kxy + y2 = 0 is 2x + y = 0.
Solution:
The auxiliary equation of the lines represented by 6x2 + kxy + y2 = 0 is
m2 + km + 6 = 0.
Since one of the line is 2x + y = 0 whose slope is m = -2.
β΄ m = -2 is the root of the auxiliary equation m2 + km + 6 = 0.
β΄ (-2)2 + k(-2) + 6 = 0
β΄ 4 β 2k + 6 = 0
β΄ 2k = 10 β΄ k = 5
Question 6.
Find the joint equation of the pair of lines which bisect angle between the lines given by x2 + 3xy + 2y2 = 0
Solution:
x2 + 3xy + 2y2 = 0
β΄ x2 + 2xy + xy + 2y2 = 0
β΄ x(x + 2y) + y(x + 2y) = 0
β΄ (x + 2y)(x + y) = 0
β΄ separate equations of the lines represented by x2 + 3xy + 2y2 = 0 are x + 2y = 0 and x + y = 0.
Let P (x, y) be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines,
the distance of P (x, y) from the line x + 2y = 0
= the distance of P(x, y) from the line x + y = 0
β΄ 2(x + 2y)2 = 5(x + y)2
β΄ 2(x2 + 4xy + 4y2) = 5(x2 + 2xy + y2)
β΄ 2x2 + 8xy + 8y2 = 5x2 + 10xy + 5y2
β΄ 3x2 + 2xy β 3y2 = 0.
This is the required joint equation of the lines which bisect the angles between the lines represented by x2 + 3xy + 2y2 = 0.
Question 7.
Find the joint equation of the pair of lies through the origin and making equilateral triangle with the line x = 3.
Solution:
Let OA and OB be the lines through the origin making an angle of 60Β° with the line x = 3.
β΄ OA and OB make an angle of 30Β° and 150Β° with the positive direction of X-axis
β΄ slope of OA = tan 30Β° = 1/\(\sqrt {3}\)
β΄ equation of the line OA is y = \(\frac{1}{\sqrt{3}}\)x
β΄ \(\sqrt {3}\)y = x β΄ x β \(\sqrt {3}\)y = 0
Slope of OB = tan 150Β° = tan (180Β° β 30Β°)
= β tan 30Β°= -1/\(\sqrt {3}\)
β΄ equation of the line OB is y = \(\frac{-1}{\sqrt{3}}\)x
β΄ \(\sqrt {3}\)y = -x β΄ x + \(\sqrt {3}\)y = 0
β΄ required combined equation of the lines is
(x β \(\sqrt {3}\)y) (x + \(\sqrt {3}\)y) = 0
i.e. x2 β 3y2 = 0.
Question 8.
Show that the lines x2 β 4xy + y2 = 0 and x + y = 10 contain the sides of an equilateral triangle. Find the area of the triangle.
Solution:
We find the joint equation of the pair of lines OA and OB through origin, each making an angle of 60Β° with x + y = 10 whose slope is -1.
Let OA (or OB) has slope m.
β΄ its equation is y = mx β¦ (1)
Also, tan 60Β° = \(\left|\frac{m-(-1)}{1+m(-1)}\right|\)
β΄ \(\sqrt {3}\) = \(\left|\frac{m+1}{1-m}\right|\)
Squaring both sides, we get,
3 = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
β΄ 3(1 β 2m + m2) = m2 + 2m + 1
β΄ 3 β 6m + 3m2 = m2 + 2m + 1
β΄ 2m2 β 8m + 2 = 0
β΄ m2 β 4m + 1 = 0
β΄ \(\left(\frac{y}{x}\right)^{2}\) β 4\(\left(\frac{y}{x}\right)\) + 1 = 0 β¦[By (1)]
β΄ y2 β 4xy + x2 = 0
β΄ x2 β 4xy + y\left(\frac{y}{x}\right) = 0 is the joint equation of the two lines through the origin each making an angle of 60Β° with x + y = 10
β΄ x2 β 4xy + y2 = 0 and x + y = 10 form a triangle OAB which is equilateral.
Let seg OM β₯r line AB whose question is x + y = 10
Question 9.
If the slope of one of the lines represented by ax2 + 2hxy + by2 = 0 is three times the other then prove that 3h2 = 4ab.
Solution:
Let m1 and m2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0.
β΄ m1 + m2 = \(-\frac{2 h}{b}\) and m1m2 = \(\frac{a}{b}\)
We are given that m2 = 3m1
β΄ m1 + 3m1 = \(-\frac{2 h}{b}\) 4m1 = \(-\frac{2 h}{b}\)
β΄ m1 = \(-\frac{h}{2 b}\) β¦(1)
Also, m1(3m1) = \(\frac{a}{b}\) β΄ 3m12 = \(\frac{a}{b}\)
β΄ 3\(\left(-\frac{h}{2 b}\right)^{2}\) = \(\frac{a}{b}\) β¦.[By (1)]
β΄ \(\frac{3 h^{2}}{4 b^{2}}=\frac{a}{b}\)
β΄ 3h2 = 4ab, as b β 0.
Question 10.
Find the combined equation of the bisectors of the angles between the lines represented by 5x2 + 6xy β y2 = 0.
Solution:
Comparing the equation 5x2 + 6xy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 5, 2h = 6, b = -1
Let m1 and m2 be the slopes of the lines represented by 5x2 + 6xy β y2 = 0.
The separate equations of the lines are
y = m1x and y = m2x, where m1 β m2
i.e. m1x β y = 0 and m1x β y = 0.
Let P (x, y) be any point on one of the bisector of the angles between the lines.
β΄ the distance of P from the line m1x β y = 0 is equal to the distance of P from the line m2x β y = 0.
β΄ (m22 + 1)(m1x β y)2 = (m12 + 1)(m2x β y)2
β΄ (m22 + 1)(m12x2 β 2m1xy + y2) = (m12 + 1)(m22x2 β 2m2xy + y2)
β΄ m12m22x2 β 2m1m12y2xy + m22y2 + m12x2 β 2m12xy + y2
= m12m22x2 β 2m12m2xy + m12y2 + m22x2 β 2m2xy + y2
β΄ (m12 β m22)x2 + 2m1m2(m1 β m2)xy β 2(m1 β m2)xy β (m12 β m22)y2 = 0
Dividing throughout by m1 β m2 (β 0), we get,
(m1 + m2)x2 + 2m1m2xy β 2xy β (m1 + m2)y2 = 0
β΄ 6x2 β 10xy β 2xy β 6y2 = 0 β¦[By (1)]
β΄ 6x2 β 12xy β 6y2 = 0
β΄ x2 β 2xy β y2 = 0
This is the joint equation of the bisectors of the angles between the lines represented by 5x2 + 6xy β y2 = 0.
Question 11.
Find a, if the sum of the slopes of the lines represented by ax2 + 8xy + 5y2 = 0 is twice their product.
Solution :
Comparing the equation ax2 + 8xy + 5y2 = 0 with ax2 + 2hxy + by2 = 0,
we get, a = a, 2h = 8, b = 5
Let m1 and m2 be the slopes of the lines represented by ax2 + 8xy + 5y2 = 0.
β΄ m1 + m2 = \(\frac{-2 h}{b}=-\frac{8}{5}\)
and m1m2 = \(\frac{a}{b}=\frac{a}{5}\)
Now, (m1 + m2) = 2(m1m2)
\(-\frac{8}{5}\) = \(2\left(\frac{a}{5}\right)\)
a = -4
Question 12.
If the line 4x β 5y = 0 coincides with one of the lines given by ax2 + 2hxy + by2 = 0, then show that 25a + 40h +16b = 0.
Solution :
The auxiliary equation of the lines represented by ax2 + 2hxy + by2 = 0 is bm2 + 2hm + a = 0
Given that 4x β 5y = 0 is one of the lines represented by ax2 + 2hxy + by2 = 0.
The slope of the line 4x β 5y = 0 is \(\frac{-4}{-5}=\frac{4}{5}\)
β΄ m = \(\frac{4}{5}\) is a root of the auxiliary equation bm2 + 2hm + a = 0.
β΄ b\(\left(\frac{4}{5}\right)^{2}\) + 2h\(\left(\frac{4}{5}\right)\) + a = 0
β΄ \(\frac{16 b}{25}+\frac{8 h}{5}\) + a = 0
β΄ 16b + 40h + 25a = 0 i.e.
β΄ 25a + 40h + 16b = 0
Question 13.
Show that the following equations represent a pair of lines. Find the acute angle between them :
(i) 9x2 β 6xy + y2 + 18x β 6y + 8 = 0
Solution:
Comparing this equation with
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get,
a = 9, h = -3, b = 1, g = 9, f = -3 and c = 8.
β΄ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
9 & -3 & 9 \\
-3 & 1 & -3 \\
9 & -3 & 8
\end{array}\right|\)
= 9(8 β 9) + 3(-24 + 27) + 9(9 β 9)
= 9(-1) + 3(3) + 9(0)
= -9 + 9 + 0 = 0
and h2 β ab = (-3)2 β 9(1) = 9 β 9 = 0
β΄ the given equation represents a pair of lines.
Let ΞΈ be the acute angle between the lines.
β΄ tan ΞΈ = tan0Β°
β΄ ΞΈ = 0Β°.
(ii) 2x2 + xy β y2 + x + 4y β 3 = 0
Solution:
Comparing this equation with
ax2 + 2hxy + by2 + 2gx + 2fy+ c = 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2 and c = -3
= -2 + 1 + 1
= -2 + 2= 0
β΄ the given equation represents a pair of lines.
Let ΞΈ be the acute angle between the lines.
β΄ tan ΞΈ = tan 3
β΄ ΞΈ = tan-1(3)
(iii) (x β 3)2 + (x β 3)(y β 4) β 2(y β 4)2 = 0.
Solution :
Put x β 3 = X and y β 4 = Y in the given equation, we get,
X2 + XY β 2Y2 = 0
Comparing this equation with ax2 + 2hxy + by2 = 0, we get,
a = 1, h = \(\frac{1}{2}\), b = -2
This is the homogeneous equation of second degreeand h2 β ab = \(\left(\frac{1}{2}\right)^{2}\) β 1(-2)
= \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0
Hence, it represents a pair of lines passing through the new origin (3, 4).
Let ΞΈ be the acute angle between the lines.
β΄ tanΞΈ = 3 β΄ ΞΈ = tan-1(3)
Question 14.
Find the combined equation of pair of lines through the origin each of which makes angle of 60Β° with the Y-axis.
Solution:
Let OA and OB be the lines through the origin making an angle of 60Β° with the Y-axis.
Then OA and OB make an angle of 30Β° and 150Β° with the positive direction of X-axis.
β΄ slope of OA = tan 30Β° = \(\frac{1}{\sqrt{3}}\)
β΄ equation of the line OA is
y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x β \(\sqrt {3}\)y = 0
Slope of OB = tan 150Β° = tan (180Β° β 30Β°)
= tan 30Β° = \(-\frac{1}{\sqrt{3}}\)
β΄ equation of the line OB is
y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0
β΄ required combined equation is
(x β \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0
i.e. x2 β 3y2 = 0.
Question 15.
If lines representedby ax2 + 2hxy + by2 = 0 make angles of equal measures with the co-ordinate
axes then show that a = Β± b.
OR
Show that, one of the lines represented by ax2 + 2hxy + by2 = 0 will make an angle of the same measure with the X-axis as the other makes with the Y-axis, if a = Β± b.
Solution:
Let OA and OB be the two lines through the origin represented by ax2 + 2hxy + by2 = 0.
Since these lines make angles of equal measure with the coordinate axes, they make angles β and \(\frac{\pi}{2}\) β β with the positive direction of X-axis or β and \(\frac{\pi}{2}\) + β with thepositive direction of X-axis.
β΄ slope of the line OA = m1 = tan β
and slope of the line OB = m2
= tan(\(\frac{\pi}{2}\) β β) or tan(\(\frac{\pi}{2}\) + β)
i.e. m2 = cot β or m2 = -cot β
β΄ m1m2 β tan β x cot β = 1
OR m1m2 = tan β (-cot β) = -1
i.e. m1m2 = Β± 1
But m1m2 = \(\frac{a}{b}\)
β΄ \(\frac{a}{b}\)= Β±1 β΄ a = Β±b
This is the required condition.
Question 16.
Show that the combined equation of a pair of lines through the origin and each making an angle of β with the line x + y = 0 is x2 + 2(sec 2β) xy + y2 = 0.
Solution:
Let OA and OB be the required lines.
Let OA (or OB) has slope m.
β΄ its equation is y = mx β¦ (1)
It makes an angle β with x + y = 0 whose slope is -1. m +1
β΄ tan β = \(\left|\frac{m+1}{1+m(-1)}\right|\)
Squaring both sides, we get,
tan2β = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
β΄ tan2β(1 β 2m + m2) = m2 + 2m + 1
β΄ tan2β β 2m tan2β + m2tan2β = m2 + 2m + 1
β΄ (tan2β β 1)m2 β 2(1 + tan2β)m + (tan2β β 1) = 0
β΄ y2 + 2xysec2β + x2 = 0
β΄ x2 + 2(sec2β)xy + y2 = 0 is the required equation.
Question 17.
Show that the line 3x + 4y+ 5 = 0 and the lines (3x + 4y)2 β 3(4x β 3y)2 =0 form an equilateral triangle.
Solution:
The slope of the line 3x + 4y + 5 = 0 is \(\frac{-3}{4}\)
Let m be the slope of one of the line making an angle of 60Β° with the line 3x + 4y + 5 = 0. The angle between the lines having slope m and m1 is 60Β°.
On squaring both sides, we get,
3 = \(\frac{(4 m+3)^{2}}{(4-3 m)^{2}}\)
β΄ 3 (4 β 3m)2 = (4m + 3)2
β΄ 3(16 β 24m + 9m2) = 16m2 + 24m + 9
β΄ 48 β 72m + 27m2 = 16m2 + 24m + 9
β΄ 11m2 β 96m + 39 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).
β΄ the combined equation of the two lines is
11\(\left(\frac{y}{x}\right)^{2}\) β 96\(\left(\frac{y}{x}\right)\) + 39 = 0
β΄ \(\frac{11 y^{2}}{x^{2}}-\frac{96 y}{x}\) + 39 = 0
β΄ 11y2 β 96xy + 39x2 = 0
β΄ 39x2 β 96xy + 11y2 = 0.
β΄ 39x2 β 96xy + 11y2 = 0 is the joint equation of the two lines through the origin each making an angle of 60Β° with the line 3x + 4y + 5 = 0.
The equation 39x2 β 96xy + 11y2 = 0 can be written as :
-39x2 + 96xy β 11y2 = 0
i.e., (9x2 β 48x2) + (24xy + 72xy) + (16y2 β 27y2) = 0
i.e. (9x2 + 24xy + 16y2) β (48x2 β 72xy + 27y2) = 0
i.e. (9x2 + 24xy + 16y2) β 3(16x2 β 24xy + 9y2) = 0
i.e. (3x + 4y)2 β 3(4x β 3y)2 = 0
Hence, the line 3x + 4y + 5 = 0 and the lines
(3x + 4y)2 β 3(4x β 3y)2 form the sides of an equilateral triangle.
Question 18.
Show that lines x2 β 4xy + y2 = 0 and x + y = \(\sqrt {6}\) form an equilateral triangle. Find its area and perimeter.
Solution:
x2 β 4xy + y2 = 0 and x + y = \(\sqrt {6}\) form a triangle OAB which is equilateral.
Let OM be the perpendicular from the origin O to AB whose equation is x + y = \(\sqrt {6}\)
In right angled triangle OAM,
sin 60Β° = \(\frac{\mathrm{OM}}{\mathrm{OA}}\) β΄ \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{\mathrm{OA}}\)
β΄ OA = 2
β΄ length of the each side of the equilateral triangle OAB = 2 units.
β΄ perimeter of β OAB = 3 Γ length of each side
= 3 Γ 2 = 6 units.
Question 19.
If the slope of one of the lines given by ax2 + 2hxy + by2 = 0 is square of the other then show that a2b + ab2 + 8h3 = 6abh.
Solution:
Let m be the slope of one of the lines given by ax2 + 2hxy + by2 = 0.
Then the other line has slope m2
Multiplying by b3, we get,
-8h3 = ab2 + a2b β 6abh
β΄ a2b + ab2 + 8h3 = 6abh
This is the required condition.
Question 20.
Prove that the product of lengths of perpendiculars drawn from P (x1, y1) to the lines repersented by ax2 + 2hxy + by2 = 0 is \(\left|\frac{a x_{1}^{2}+2 h x_{1} y_{1}+b y_{1}^{2}}{\sqrt{(a-b)^{2}+4 h^{2}}}\right|\)
Solution:
Let m1 and m2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0.
β΄ m1 + m2 = \(-\frac{2 h}{b}\) and m1m2 = \(\frac{a}{b}\) β¦(1)
The separate equations of the lines represented by
ax2 + 2hxy + by2 = 0 are
y = m1x and y = m2x
i.e. m1x β y = 0 and m2x β y = 0
Length of perpendicular from P(x1, 1) on
Question 21.
Show that the difference between the slopes of lines given by (tan2ΞΈ + cos2ΞΈ )x2 β 2xytanΞΈ + (sin2ΞΈ )y2 = 0 is two.
Solution:
Comparing the equation (tan2ΞΈ + cos2ΞΈ)x2 β 2xy tan ΞΈ + (sin2ΞΈ) y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = tan2ΞΈ + cos2ΞΈ, 2h = -2 tan ΞΈ and b = sin2ΞΈ
Let m1 and m2 be the slopes of the lines represented by the given equation.
Question 22.
Find the condition that the equation ay2 + bxy + ex + dy = 0 may represent a pair of lines.
Solution:
Comparing the equation
ay2 + bxy + ex + dy = 0 with
Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0, we get,
A = 0, H = \(\frac{b}{2}\), B = a,G = \(\frac{e}{2}\), F = \(\frac{d}{2}\), C = 0
The given equation represents a pair of lines,
i.e. if bed β ae2 = 0
i.e. if e(bd β ae) = 0
i.e. e = 0 or bd β ae = 0
i.e. e = 0 or bd = ae
This is the required condition.
Question 23.
If the lines given by ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1 then show that (3a + b)(a + 3b) = 4h2.
Solution:
Since the lines ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1, the angle between the lines ax2 + 2hxy + by2 = 0 is 60Β°.
β΄ 3(a + b)2 = 4(h2 β ab)
β΄ 3(a2 + 2ab + b2) = 4h2 β 4ab
β΄ 3a2 + 6ab + 3b2 + 4ab = 4h2
β΄ 3a2 + 10ab + 3b2 = 4h2
β΄ 3a2 + 9ab + ab + 3b2 = 4h2
β΄ 3a(a + 3b) + b(a + 3b) = 4h2
β΄ (3a + b)(a + 3b) = 4h2
This is the required condition.
Question 24.
If line x + 2 = 0 coincides with one of the lines represented by the equation x2 + 2xy + 4y + k = 0 then show that k = -4.
Solution:
One of the lines represented by
x2 + 2xy + 4y + k = 0 β¦ (1)
is x + 2 = 0.
Let the other line represented by (1) be ax + by + c = 0.
β΄ their combined equation is (x + 2)(ax + by + c) = 0
β΄ ax2 + bxy + cx + 2ax + 2by + 2c = 0
β΄ ax2 + bxy + (2a + c)x + 2by + 2c β 0 β¦ (2)
As the equations (1) and (2) are the combined equations of the same two lines, they are identical.
β΄ by comparing their corresponding coefficients, we get,
β΄ 1 = \(\frac{-4}{k}\)
β΄ k = -4.
Question 25.
Prove that the combined equation of the pair of lines passing through the origin and perpendicular to the lines represented by ax2 + 2hxy + by2 = 0 is bx2 β 2hxy + ay2 = 0
Solution:
Let m1 and m2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0.
Now, required lines are perpendicular to these lines.
β΄ their slopes are and \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(-\frac{1}{m_{1}}\)x and y = \(-\frac{1}{m_{2}}\)x
i.e. m1y= -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
β΄ their combined equation is
(x + m1y)(x + m2y) = 0
β΄ x2 + (m1 + m2)xy + m1m2y2 = 0
β΄ x2\(\frac{-2 h}{b}\)x + \(\frac{a}{b}\)y2 = 0
β΄ bx2 β 2hxy + ay2 = 0.
Question 26.
If equation ax2 β y2 + 2y + c = 1 represents a pair of perpendicular lines then find a and c.
Solution:
The given equation represents a pair of lines perpendicular to each other.
β΄ coefficient of x2 + coefficient of y2 = 0
β΄ a β 1 = 0 β΄ a = 1
With this value of a, the given equation is
x2 β y2 + 2y + c β 1 = 0
Comparing this equation with
Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0, we get,
A = 1, H = 0, B = -1, G = 0, F = 1, C = c β 1
Since the given equation represents a pair of lines,
D = \(\left|\begin{array}{ccc}
A & H & G \\
H & B & F \\
G & F & C
\end{array}\right|\) = 0
β΄ \(\left|\begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 1 \\
0 & 1 & c-1
\end{array}\right|\) = 0
β΄ 1(-c + 1 β 1) β 0 + 0 = 0
β΄ -c = 0
β΄ c = 0.
Hence, a = 1, c = 0.
Pdf Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 Questions and Answers.
I : Choose correct alternatives.
Question 1.
If the equation 4x2 + hxy + y2 = 0 represents two coincident lines, then h = _________.
(A) Β± 2
(B) Β± 3
(C) Β± 4
(D) Β± 5
Solution:
(C) Β± 4
Question 2.
If the lines represented by kx2 β 3xy + 6y2 = 0 are perpendicular to each other then _________.
(A) k = 6
(B) k = -6
(C) k = 3
(D) k = -3
Solution:
(B) k = -6
Question 3.
Auxiliary equation of 2x2 + 3xy β 9y2 = 0 is _________.
(A) 2m2 + 3m β 9 = 0
(B) 9m2 β 3m β 2 = 0
(C) 2m2 β 3m + 9 = 0
(D) -9m2 β 3m + 2 = 0
Solution:
(B) 9m2 β 3m β 2 = 0
Question 4.
The difference between the slopes of the lines represented by 3x2 β 4xy + y2 = 0 is _________.
(A) 2
(B) 1
(C) 3
(D) 4
Solution:
(A) 2
Question 5.
If the two lines ax2 +2hxy+ by2 = 0 make angles Ξ± and Ξ² with X-axis, then tan (Ξ± + Ξ²) = _____.
(A) \(\frac{h}{a+b}\)
(B) \(\frac{h}{a-b}\)
(C) \(\frac{2 h}{a+b}\)
(D) \(\frac{2 h}{a-b}\)
Solution:
(D) \(\frac{2 h}{a-b}\)
Question 6.
If the slope of one of the two lines \(\frac{x^{2}}{a}+\frac{2 x y}{h}+\frac{y^{2}}{b}\) = 0 is twice that of the other, then ab:h2 = ___.
(A) 1 : 2
(B) 2 : 1
(C) 8 : 9
(D) 9 : 8
Solution:
(D) 9 : 8
Question 7.
The joint equation of the lines through the origin and perpendicular to the pair of lines 3x2 + 4xy β 5y2 = 0 is _________.
(A) 5x2 + 4xy β 3y2 = 0
(B) 3x2 + 4xy β 5y2 = 0
(C) 3x2 β 4xy + 5y2 = 0
(D) 5x2 + 4xy + 3y2 = 0
Solution:
(A) 5x2 + 4xy β 3y2 = 0
Question 8.
If acute angle between lines ax2 + 2hxy + by2 = 0 is, \(\frac{\pi}{4}\) then 4h2 = _________.
(A) a2 + 4ab + b2
(B) a2 + 6ab + b2
(C) (a + 2b)(a + 3b)
(D) (a β 2b)(2a + b)
Solution:
(B) a2 + 6ab + b2
Question 9.
If the equation 3x2 β 8xy + qy2 + 2x + 14y + p = 1 represents a pair of perpendicular lines then
the values of p and q are respectively _________.
(A) -3 and -7
(B) -7 and -3
(C) 3 and 7
(D) -7 and 3
Solution:
(B) -7 and -3
Question 10.
The area of triangle formed by the lines x2 + 4xy + y2 = 0 and x β y β 4 = 0 is _________.
(A) \(\frac{4}{\sqrt{3}}\) Sq. units
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
(C) \(\frac{16}{\sqrt{3}}\) Sq. units
(D)\(\frac{15}{\sqrt{3}}\) Sq. units
Solution:
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
[Hint : Area = \(\frac{p^{2}}{\sqrt{3}}\), where p is the length of perpendicular from the origin to x β y β 4 = 0]
Question 11.
The combined equation of the co-ordinate axes is _________.
(A) x + y = 0
(B) x y = k
(C) xy = 0
(D) x β y = k
Solution:
(C) xy = 0
Question 12.
If h2 = ab, then slope of lines ax2 + 2hxy + by2 = 0 are in the ratio _________.
(A) 1 : 2
(B) 2 : 1
(C) 2 : 3
(D) 1 : 1
Solution:
(D) 1 : 1
[Hint: If h2 = ab, then lines are coincident. Therefore slopes of the lines are equal.]
Question 13.
If slope of one of the lines ax2 + 2hxy + by2 = 0 is 5 times the slope of the other, then 5h2 = _________.
(A) ab
(B) 2 ab
(C) 7 ab
(D) 9 ab
Solution:
(D) 9 ab
Question 14.
If distance between lines (x β 2y)2 + k(x β 2y) = 0 is 3 units, then k =
(A) Β± 3
(B) Β± 5\(\sqrt {5}\)
(C) 0
(D) Β± 3\(\sqrt {5}\)
Solution:
(D) Β± 3\(\sqrt {5}\)
[Hint: (x β 2y)2 + k(x β 2y) = 0
β΄ (x β 2y)(x β 2y + k) = 0
β΄ equations of the lines are x β 2y = 0 and x β 2y + k = 0 which are parallel to each other.
β΄ \(\left|\frac{k-0}{\sqrt{1+4}}\right|\) = 3
β΄ k = Β± 3\(\sqrt {5}\)
II. Solve the following.
Question 1.
Find the joint equation of lines:
(i) x β y = 0 and x + y = 0
Solution:
The joint equation of the lines x β y = 0 and
x + y = 0 is
(x β y)(x + y) = 0
β΄ x2 β y2 = 0.
(ii) x + y β 3 = 0 and 2x + y β 1 = 0
Solution:
The joint equation of the lines x + y β 3 = 0 and 2x + y β 1 = 0 is
(x + y β 3)(2x + y β 1) = 0
β΄ 2x2 + xy β x + 2xy + y2 β y β 6x β 3y + 3 = 0
β΄ 2x2 + 3xy + y2 β 7x β 4y + 3 = 0.
(iii) Passing through the origin and having slopes 2 and 3.
Solution:
We know that the equation of the line passing through the origin and having slope m is y = mx. Equations of the lines passing through the origin and having slopes 2 and 3 are y = 2x and y = 3x respectively.
i.e. their equations are
2x β y = 0 and 3x β y = 0 respectively.
β΄ their joint equation is (2x β y)(3x β y) = 0
β΄ 6x2 β 2xy β 3xy + y2 = 0
β΄ 6x2 β 5xy + y2 = 0.
(iv) Passing through the origin and having inclinations 60Β° and 120Β°.
Solution:
Slope of the line having inclination ΞΈ is tan ΞΈ .
Inclinations of the given lines are 60Β° and 120Β°
β΄ their slopes are m1 = tan60Β° = \(\sqrt {3}\) and
m2 = tan 120Β° = tan (180Β° β 60Β°)
= -tan 60Β° = β\(\sqrt {3}\)
Since the lines pass through the origin, their equa-tions are
y = \(\sqrt {3}\)x and y= β\(\sqrt {3}\)x
i.e., \(\sqrt {3}\)x β y = 0 and \(\sqrt {3}\)x + y = 0
β΄ the joint equation of these lines is
(\(\sqrt {3}\)x β y)(\(\sqrt {3}\)x + y) = 0
β΄ 3x2 β y2 = 0.
(v) Passing through (1, 2) amd parallel to the co-ordinate axes.
Solution:
Equations of the coordinate axes are x = 0 and y = 0
β΄ the equations of the lines passing through (1, 2) and parallel to the coordinate axes are x = 1 and y =1
i.e. x β 1 = 0 and y β 2 0
β΄ their combined equation is
(x β 1)(y β 2) = 0
β΄ x(y β 2) β 1(y β 2) = 0
β΄ xy β 2x β y + 2 = 0
(vi) Passing through (3, 2) and parallel to the line x = 2 and y = 3.
Solution:
Equations of the lines passing through (3, 2) and parallel to the lines x = 2 and y = 3 are x = 3 and y = 2.
i.e. x β 3 = 0 and y β 2 = 0
β΄ their joint equation is
(x β 3)(y β 2) = 0
β΄ xy β 2x β 3y + 6 = 0.
(vii) Passing through (-1, 2) and perpendicular to the lines x + 2y + 3 = 0 and 3x β 4y β 5 = 0.
Solution:
Let L1 and L2 be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x β 4y β 5 = 0 respectively.
Slopes of the lines x + 2y + 3 = 0 and 3x β 4y β 5 = 0 are \(-\frac{1}{2}\) and \(-\frac{3}{-4}=\frac{3}{4}\) respectively.
β΄ slopes of the lines L1and L2 are 2 and \(\frac{-4}{3}\) respectively.
Since the lines L1 and L2 pass through the point (-1, 2), their equations are
β΄ (y β y1) = m(x β x1)
β΄ (y β 2) = 2(x + 1)
β y β 1 = 2x + 2
β 2x β y + 4 = 0 and
β΄ (y β 2) = \(\left(\frac{-4}{3}\right)\)(x + 1)
β 3y β 6 = (-4)(x + 1)
β 3y β 6 = -4x + 4
β 4x + 3y β 6 + 4 = 0
β 4x + 3y β 2 = 0
their combined equation is
β΄ (2x β y + 4)(4x + 3y β 2) = 0
β΄ 8x2 + 6xy β 4x β 4xy β 3y2 + 2y + 16x + 12y β 8 = 0
β΄ 8x2 + 2xy + 12x β 3y2 + 14y β 8 = 0
(viii) Passing through the origin and having slopes 1 + \(\sqrt {3}\) and 1 β \(\sqrt {3}\)
Solution:
Let l1 and l2 be the two lines. Slopes of l1 is 1 + \(\sqrt {3}\) and that of l2 is 1 β \(\sqrt {3}\)
Therefore the equation of a line (l1) passing through the origin and having slope is
y = (1 + \(\sqrt {3}\))x
β΄ (1 + \(\sqrt {3}\))x β y = 0 ..(1)
Similarly, the equation of the line (l2) passing through the origin and having slope is
y = (1 β \(\sqrt {3}\))x
β΄ (1 β \(\sqrt {3}\))x β y = 0 β¦(2)
From (1) and (2) the required combined equation is
β΄ (1 β 3)x2 β 2xy + y2 = 0
β΄ -2x2 β 2xy + y2 = 0
β΄ 2x2 + 2xy β y2 = 0
This is the required combined equation.
(ix) Which are at a distance of 9 units from the Y β axis.
Solution:
Equations of the lines, which are parallel to the Y-axis and at a distance of 9 units from it, are x = 9 and x = -9
i.e. x β 9 = 0 and x + 9 = 0
β΄ their combined equation is
(x β 9)(x + 9) = 0
β΄ x2 β 81 = 0.
(x) Passing through the point (3, 2), one of which is parallel to the line x β 2y = 2 and other is perpendicular to the line y = 3.
Solution:
Let L1 be the line passes through (3, 2) and parallel to the line x β 2y = 2 whose slope is \(\frac{-1}{-2}=\frac{1}{2}\)
β΄ slope of the line L1 is \(\frac{1}{2}\).
β΄ equation of the line L1 is
y β 2 = \(\frac{1}{2}\)(x β 3)
β΄ 2y β 4 = x β 3 β΄ x β 2y + 1 = 0
Let L2 be the line passes through (3, 2) and perpendicular to the line y = 3.
β΄ equation of the line L2 is of the form x = a.
Since L2 passes through (3, 2), 3 = a
β΄ equation of the line L2 is x = 3, i.e. x β 3 = 0
Hence, the equations of the required lines are
x β 2y + 1 = 0 and x β 3 = 0
β΄ their joint equation is
(x β 2y + 1)(x β 3) = 0
β΄ x2 β 2xy + x β 3x + 6y β 3 = 0
β΄ x2 β 2xy β 2x + 6y β 3 = 0.
(xi) Passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18.
Solution:
Let L1 and L2 be the lines passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18 respectively.
Slopes of the lines x + 2y = 19 and 3x + y = 18 are \(-\frac{1}{2}\) and \(-\frac{3}{1}\) = -3 respectively.
Since the lines L1 and L2 pass through the origin, their equations are
y = 2x and y = \(\frac{1}{3}\)x
i.e. 2x β y = 0 and x β 3y = 0
β΄ their combined equation is
(2x β y)(x β 3y) = 0
β΄ 2x2 β 6xy β xy + 3y2 = 0
β΄ 2x2 β 7xy + 3y2 = 0.
Question 2.
Show that each of the following equation represents a pair of lines.
(i) x2 + 2xy β y2 = 0
Solution:
Comparing the equation x2 + 2xy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 2, i.e. h = 1 and b = -1
β΄ h2 β ab = (1)2 β 1(-1) = 1 + 1=2 > 0
Since the equation x2 + 2xy β y2 = 0 is a homogeneous equation of second degree and h2 β ab > 0, the given equation represents a pair of lines which are real and distinct.
(ii) 4x2 + 4xy + y2 = 0
Solution:
Comparing the equation 4x2 + 4xy + y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 4, 2h = 4, i.e. h = 2 and b = 1
β΄ h2 β ab = (2)2 β 4(1) = 4 β 4 = 0
Since the equation 4x2 + 4xy + y2 = 0 is a homogeneous equation of second degree and h2 β ab = 0, the given equation represents a pair of lines which are real and coincident.
(iii) x2 β y2 = 0
Solution:
Comparing the equation x2 β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 0, i.e. h = 0 and b = -1
β΄ h2 β ab = (0)2 β 1(-1) = 0 + 1 = 1 > 0
Since the equation x2 β y2 = 0 is a homogeneous equation of second degree and h2 β ab > 0, the given equation represents a pair of lines which are real and distinct.
(iv) x2 + 7xy β 2y2 = 0
Solution:
Comparing the equation x2 + 7xy β 2y2 = 0
a = 1, 2h = 7 i.e., h = \(\frac{7}{2}\) and b = -2
β΄ h2 β ab = \(\left(\frac{7}{2}\right)^{2}\) β 1(-2)
= \(\frac{49}{4}\) + 2
= \(\frac{57}{4}\) i.e. 14.25 = 14 > 0
Since the equation x2 + 7xy β 2y2 = 0 is a homogeneous equation of second degree and h2 β ab > 0, the given equation represents a pair of lines which are real and distinct.
(v) x2 β 2\(\sqrt {3}\) xy β y2 = 0
Solution:
Comparing the equation x2 β 2\(\sqrt {3}\) xy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h= -2\(\sqrt {3}\), i.e. h = β\(\sqrt {3}\) and b = 1
β΄ h2 β ab = (-\(\sqrt {3}\))2 β 1(1) = 3 β 1 = 2 > 0
Since the equation x2 β 2\(\sqrt {3}\)xy β y2 = 0 is a homoΒ¬geneous equation of second degree and h2 β ab > 0, the given equation represents a pair of lines which are real and distinct.
Question 3.
Find the separate equations of lines represented by the following equations:
(i) 6x2 β 5xy β 6y2 = 0
Solution:
6x2 β 5xy β 6y2 = 0
β΄ 6x2 β 9xy + 4xy β 6y2 = 0
β΄ 3x(2x β 3y) + 2y(2x β 3y) = 0
β΄ (2x β 3y)(3x + 2y) = 0
β΄ the separate equations of the lines are
2x β 3y = 0 and 3x + 2y = 0.
(ii) x2 β 4y2 = 0
Solution:
x2 β 4y2 = 0
β΄ x2 β (2y)2 = 0
β΄(x β 2y)(x + 2y) = 0
β΄ the separate equations of the lines are
x β 2y = 0 and x + 2y = 0.
(iii) 3x2 β y2 = 0
Solution:
3x2 β y2 = 0
β΄ (\(\sqrt {3}\) x)2 β y2 = 0
β΄ (\(\sqrt {3}\)x β y)(\(\sqrt {3}\)x + y) = 0
β΄ the separate equations of the lines are
\(\sqrt {3}\)x β y = 0 and \(\sqrt {3}\)x + y = 0.
(iv) 2x2 + 2xy β y2 = 0
Solution:
2x2 + 2xy β y2 = 0
β΄ The auxiliary equation is -m2 + 2m + 2 = 0
β΄ m2 β 2m β 2 = 0
m1 = 1 + \(\sqrt {3}\) and m2 = 1 β \(\sqrt {3}\) are the slopes of the lines.
β΄ their separate equations are
y = m1x and y = m2x
i.e. y = (1 + \(\sqrt {3}\))x and y = (1 β \(\sqrt {3}\))x
i.e. (\(\sqrt {3}\) + 1)x β y = 0 and (\(\sqrt {3}\) β 1)x + y = 0.
Question 4.
Find the joint equation of the pair of lines through the origin and perpendicular to the lines
given by :
(i) x2 + 4xy β 5y2 = 0
Solution:
Comparing the equation x2 + 4xy β 5y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 4, b= -5
Let m1 and m2 be the slopes of the lines represented by x2 + 4xy β 5y2 = 0.
Now, required lines are perpendicular to these lines
β΄ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m1y = -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
β΄ their combined equation is
(x + m1x + m2y) = 0
β΄ x2 + (m1 + m2)xy + m1m2y2 = 0
β΄ x2 + \(\frac{4}{5}\)xy β \(\frac{1}{5}\)y2 = 0 β¦[By (1)]
β΄ 5x2 + 4xy β y2 = 0
(ii) 2x2 β 3xy β 9y2 = 0
Solution:
Comparing the equation 2x2 β 3xy β 9y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 2, 2h = -3, b = -9
Let m1 and m2 be the slopes of the lines represented by 2x2 β 3xy β 9y2 = 0
β΄ m1 + m2 =\(\frac{-2 h}{b}=-\frac{3}{9}\) and m1m2 = \(\frac{a}{b}=-\frac{2}{9}\) β¦(1)
Now, required lines are perpendicular to these lines
β΄ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m1y = -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
β΄ their combined equation is
(x + m1y)(x + m2y) = 0
β΄ x2 + (m1 + m2)xy + m1m2y2 = 0
β΄ x2 + \(\left(-\frac{3}{9}\right)\)xy + \(\left(-\frac{2}{9}\right)\)y2 = 0 β¦[By (1)]
β΄ 9x2 β 3xy β 2y2 = 0
(iii) x2 + xy β y2 = 0
Solution:
Comparing the equation x2+ xy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 1, b = -1
Let m1 and m2 be the slopes of the lines represented by x2 + xy β y2 = 0
β΄ m1 + m2 = \(\frac{-2 h}{b}=\frac{-1}{-1}\) and m1m2 = \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{-1}\) = -1 ..(1)
Now, required lines are perpendicular to these lines
β΄ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m1y = -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
β΄ their combined equation is
(x + m1y)(x + m2y) = 0
β΄ x2 + (m1 + m2) + m1m2y2 = 0
β΄ x2 + 1xy + (-1)y2 = 0 β¦[By (1)]
β΄ x2 + xy β y2 = 0
Question 5.
Find k if
(i) The sum of the slopes of the lines given by 3x2 + kxy β y2 = 0 is zero.
Solution:
Comparing the equation 3x2 + kxy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 3, 2h = k, b = -1
Let m1 and m2 be the slopes of the lines represented by 3x2 + kxy β y2 = 0.
β΄ m1 + m2 = \(\frac{-2 h}{b}=\frac{-k}{-1}\) = k
Now, m1 + m2 = 0 β¦ (Given)
β΄ k = 0.
(ii) The sum of slopes of the lines given by 2x2 + kxy β 3y2 = 0 is equal to their product.
Question is modified.
The sum of slopes of the lines given by x2 + kxy β 3y2 = 0 is equal to their product.
Solution:
Comparing the equation x2 + kxy β 3y2 = 0, with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = k, b = -3
Let m1 and m2 be the slopes of the lines represented by x2 + kxy β 3y2 = 0.
β΄ m1 + m2 = \(-\frac{2 h}{b}=\frac{-k}{-3}=\frac{k}{3}\)
and m1m2 = \(\frac{a}{b}=\frac{1}{-3}=\frac{-1}{3}\)
Now, m1 + m2 = m1m2 β¦ (Given)
β΄ \(\frac{k}{3}=\frac{-1}{3}\)
β΄ k = -1.
(iii) The slope of one of the lines given by 3x2 β 4xy + ky2 = 0 is 1.
Solution:
The auxiliary equation of the lines given by 3x2 β 4xy + ky2 = 0 is km2 β 4m + 3 = 0.
Given, slope of one of the lines is 1.
β΄ m = 1 is the root of the auxiliary equation km2 β 4m + 3 = 0.
β΄ k(1)2 β 4(1) + 3 = 0
β΄ k β 4 + 3 = 0
β΄ k = 1.
(iv) One of the lines given by 3x2 β kxy + 5y2 = 0 is perpendicular to the 5x + 3y = 0.
Solution:
The auxiliary equation of the lines represented by 3x2 β kxy + 5y2 = 0 is 5m2 β km + 3 = 0.
Now, one line is perpendicular to the line 5x + 3y = 0, whose slope is \(-\frac{5}{3}\).
β΄ slope of that line = m = \(\frac{3}{5}\)
β΄ m = \(\frac{3}{5}\) is the root of the auxiliary equation 5
5m2 β km + 3 = 0.
β΄ 5\(\left(\frac{3}{5}\right)^{2}\) β k\(\left(\frac{3}{5}\right)\) + 3 = 0
β΄ \(\frac{9}{5}-\frac{3 k}{5}\) + 3 = 0
β΄ 9 β 3k + 15 = 0
β΄ 3k = 24
β΄ k = 8.
(v) The slope of one of the lines given by 3x2 + 4xy + ky2 = 0 is three times the other.
Solution:
3x2 + 4xy + ky2 = 0
β΄ divide by x2
β΄ y = mx
β΄ \(\frac{\mathrm{y}}{\mathrm{x}}\) = m
put \(\frac{\mathrm{y}}{\mathrm{x}}\) = m in equation (1)
Comparing the equation km2 + 4m + 3 = 0 with ax2 + 2hxy+ by2 = 0, we get,
a = k, 2h = 4, b = 3
m1 = 3m2 ..(given condition)
m1 + m2 = \(\frac{-2 h}{k}=-\frac{4}{k}\)
m1m2 = \(\frac{a}{b}=\frac{3}{k}\)
m1 + m2 = \(-\frac{4}{\mathrm{k}}\)
4m2 = \(-\frac{4}{\mathrm{k}}\) β¦(m1 = 3m2)
m2 = \(-\frac{1}{\mathrm{k}}\)
m1m2 = \(\frac{3}{k}\)
\(3 \mathrm{~m}_{2}^{2}=\frac{3}{\mathrm{k}}\) β¦(m1 = 3m2)
\(3\left(-\frac{1}{\mathrm{k}}\right)^{2}=\frac{3}{\mathrm{k}}\) β¦(m2 = \(-\frac{1}{k}\))
\(\frac{1}{k^{2}}=\frac{1}{k}\)
k2 = k
k = 1 or k = 0
(vi) The slopes of lines given by kx2 + 5xy + y2 = 0 differ by 1.
Solution:
Comparing the equation kx2 + 5xy +y2 = 0 with ax2 + 2hxy + by2
a = k, 2h = 5 i.e. h = \(\frac{5}{2}\)
m1 + m2 = \(\frac{-2 h}{b}=-\frac{5}{1}\) = -5
and m1m2 = \(\frac{a}{b}=\frac{k}{1}\) = k
the slope of the line differ by (m1 β m2) = 1 β¦(1)
β΄ (m1 β m2)2 = (m1 + m2)2 β 4m1m2
(m1 β m2)2 = (-5)2 β 4(k)
(m1 β m2)2 = 25 β 4k
1 = 25 β 4k ..[By (1)]
4k = 24
k = 6
(vii) One of the lines given by 6x2 + kxy + y2 = 0 is 2x + y = 0.
Solution:
The auxiliary equation of the lines represented by 6x2 + kxy + y2 = 0 is
m2 + km + 6 = 0.
Since one of the line is 2x + y = 0 whose slope is m = -2.
β΄ m = -2 is the root of the auxiliary equation m2 + km + 6 = 0.
β΄ (-2)2 + k(-2) + 6 = 0
β΄ 4 β 2k + 6 = 0
β΄ 2k = 10 β΄ k = 5
Question 6.
Find the joint equation of the pair of lines which bisect angle between the lines given by x2 + 3xy + 2y2 = 0
Solution:
x2 + 3xy + 2y2 = 0
β΄ x2 + 2xy + xy + 2y2 = 0
β΄ x(x + 2y) + y(x + 2y) = 0
β΄ (x + 2y)(x + y) = 0
β΄ separate equations of the lines represented by x2 + 3xy + 2y2 = 0 are x + 2y = 0 and x + y = 0.
Let P (x, y) be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines,
the distance of P (x, y) from the line x + 2y = 0
= the distance of P(x, y) from the line x + y = 0
β΄ 2(x + 2y)2 = 5(x + y)2
β΄ 2(x2 + 4xy + 4y2) = 5(x2 + 2xy + y2)
β΄ 2x2 + 8xy + 8y2 = 5x2 + 10xy + 5y2
β΄ 3x2 + 2xy β 3y2 = 0.
This is the required joint equation of the lines which bisect the angles between the lines represented by x2 + 3xy + 2y2 = 0.
Question 7.
Find the joint equation of the pair of lies through the origin and making equilateral triangle with the line x = 3.
Solution:
Let OA and OB be the lines through the origin making an angle of 60Β° with the line x = 3.
β΄ OA and OB make an angle of 30Β° and 150Β° with the positive direction of X-axis
β΄ slope of OA = tan 30Β° = 1/\(\sqrt {3}\)
β΄ equation of the line OA is y = \(\frac{1}{\sqrt{3}}\)x
β΄ \(\sqrt {3}\)y = x β΄ x β \(\sqrt {3}\)y = 0
Slope of OB = tan 150Β° = tan (180Β° β 30Β°)
= β tan 30Β°= -1/\(\sqrt {3}\)
β΄ equation of the line OB is y = \(\frac{-1}{\sqrt{3}}\)x
β΄ \(\sqrt {3}\)y = -x β΄ x + \(\sqrt {3}\)y = 0
β΄ required combined equation of the lines is
(x β \(\sqrt {3}\)y) (x + \(\sqrt {3}\)y) = 0
i.e. x2 β 3y2 = 0.
Question 8.
Show that the lines x2 β 4xy + y2 = 0 and x + y = 10 contain the sides of an equilateral triangle. Find the area of the triangle.
Solution:
We find the joint equation of the pair of lines OA and OB through origin, each making an angle of 60Β° with x + y = 10 whose slope is -1.
Let OA (or OB) has slope m.
β΄ its equation is y = mx β¦ (1)
Also, tan 60Β° = \(\left|\frac{m-(-1)}{1+m(-1)}\right|\)
β΄ \(\sqrt {3}\) = \(\left|\frac{m+1}{1-m}\right|\)
Squaring both sides, we get,
3 = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
β΄ 3(1 β 2m + m2) = m2 + 2m + 1
β΄ 3 β 6m + 3m2 = m2 + 2m + 1
β΄ 2m2 β 8m + 2 = 0
β΄ m2 β 4m + 1 = 0
β΄ \(\left(\frac{y}{x}\right)^{2}\) β 4\(\left(\frac{y}{x}\right)\) + 1 = 0 β¦[By (1)]
β΄ y2 β 4xy + x2 = 0
β΄ x2 β 4xy + y\left(\frac{y}{x}\right) = 0 is the joint equation of the two lines through the origin each making an angle of 60Β° with x + y = 10
β΄ x2 β 4xy + y2 = 0 and x + y = 10 form a triangle OAB which is equilateral.
Let seg OM β₯r line AB whose question is x + y = 10
Question 9.
If the slope of one of the lines represented by ax2 + 2hxy + by2 = 0 is three times the other then prove that 3h2 = 4ab.
Solution:
Let m1 and m2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0.
β΄ m1 + m2 = \(-\frac{2 h}{b}\) and m1m2 = \(\frac{a}{b}\)
We are given that m2 = 3m1
β΄ m1 + 3m1 = \(-\frac{2 h}{b}\) 4m1 = \(-\frac{2 h}{b}\)
β΄ m1 = \(-\frac{h}{2 b}\) β¦(1)
Also, m1(3m1) = \(\frac{a}{b}\) β΄ 3m12 = \(\frac{a}{b}\)
β΄ 3\(\left(-\frac{h}{2 b}\right)^{2}\) = \(\frac{a}{b}\) β¦.[By (1)]
β΄ \(\frac{3 h^{2}}{4 b^{2}}=\frac{a}{b}\)
β΄ 3h2 = 4ab, as b β 0.
Question 10.
Find the combined equation of the bisectors of the angles between the lines represented by 5x2 + 6xy β y2 = 0.
Solution:
Comparing the equation 5x2 + 6xy β y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 5, 2h = 6, b = -1
Let m1 and m2 be the slopes of the lines represented by 5x2 + 6xy β y2 = 0.
The separate equations of the lines are
y = m1x and y = m2x, where m1 β m2
i.e. m1x β y = 0 and m1x β y = 0.
Let P (x, y) be any point on one of the bisector of the angles between the lines.
β΄ the distance of P from the line m1x β y = 0 is equal to the distance of P from the line m2x β y = 0.
β΄ (m22 + 1)(m1x β y)2 = (m12 + 1)(m2x β y)2
β΄ (m22 + 1)(m12x2 β 2m1xy + y2) = (m12 + 1)(m22x2 β 2m2xy + y2)
β΄ m12m22x2 β 2m1m12y2xy + m22y2 + m12x2 β 2m12xy + y2
= m12m22x2 β 2m12m2xy + m12y2 + m22x2 β 2m2xy + y2
β΄ (m12 β m22)x2 + 2m1m2(m1 β m2)xy β 2(m1 β m2)xy β (m12 β m22)y2 = 0
Dividing throughout by m1 β m2 (β 0), we get,
(m1 + m2)x2 + 2m1m2xy β 2xy β (m1 + m2)y2 = 0
β΄ 6x2 β 10xy β 2xy β 6y2 = 0 β¦[By (1)]
β΄ 6x2 β 12xy β 6y2 = 0
β΄ x2 β 2xy β y2 = 0
This is the joint equation of the bisectors of the angles between the lines represented by 5x2 + 6xy β y2 = 0.
Question 11.
Find a, if the sum of the slopes of the lines represented by ax2 + 8xy + 5y2 = 0 is twice their product.
Solution :
Comparing the equation ax2 + 8xy + 5y2 = 0 with ax2 + 2hxy + by2 = 0,
we get, a = a, 2h = 8, b = 5
Let m1 and m2 be the slopes of the lines represented by ax2 + 8xy + 5y2 = 0.
β΄ m1 + m2 = \(\frac{-2 h}{b}=-\frac{8}{5}\)
and m1m2 = \(\frac{a}{b}=\frac{a}{5}\)
Now, (m1 + m2) = 2(m1m2)
\(-\frac{8}{5}\) = \(2\left(\frac{a}{5}\right)\)
a = -4
Question 12.
If the line 4x β 5y = 0 coincides with one of the lines given by ax2 + 2hxy + by2 = 0, then show that 25a + 40h +16b = 0.
Solution :
The auxiliary equation of the lines represented by ax2 + 2hxy + by2 = 0 is bm2 + 2hm + a = 0
Given that 4x β 5y = 0 is one of the lines represented by ax2 + 2hxy + by2 = 0.
The slope of the line 4x β 5y = 0 is \(\frac{-4}{-5}=\frac{4}{5}\)
β΄ m = \(\frac{4}{5}\) is a root of the auxiliary equation bm2 + 2hm + a = 0.
β΄ b\(\left(\frac{4}{5}\right)^{2}\) + 2h\(\left(\frac{4}{5}\right)\) + a = 0
β΄ \(\frac{16 b}{25}+\frac{8 h}{5}\) + a = 0
β΄ 16b + 40h + 25a = 0 i.e.
β΄ 25a + 40h + 16b = 0
Question 13.
Show that the following equations represent a pair of lines. Find the acute angle between them :
(i) 9x2 β 6xy + y2 + 18x β 6y + 8 = 0
Solution:
Comparing this equation with
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get,
a = 9, h = -3, b = 1, g = 9, f = -3 and c = 8.
β΄ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
9 & -3 & 9 \\
-3 & 1 & -3 \\
9 & -3 & 8
\end{array}\right|\)
= 9(8 β 9) + 3(-24 + 27) + 9(9 β 9)
= 9(-1) + 3(3) + 9(0)
= -9 + 9 + 0 = 0
and h2 β ab = (-3)2 β 9(1) = 9 β 9 = 0
β΄ the given equation represents a pair of lines.
Let ΞΈ be the acute angle between the lines.
β΄ tan ΞΈ = tan0Β°
β΄ ΞΈ = 0Β°.
(ii) 2x2 + xy β y2 + x + 4y β 3 = 0
Solution:
Comparing this equation with
ax2 + 2hxy + by2 + 2gx + 2fy+ c = 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2 and c = -3
= -2 + 1 + 1
= -2 + 2= 0
β΄ the given equation represents a pair of lines.
Let ΞΈ be the acute angle between the lines.
β΄ tan ΞΈ = tan 3
β΄ ΞΈ = tan-1(3)
(iii) (x β 3)2 + (x β 3)(y β 4) β 2(y β 4)2 = 0.
Solution :
Put x β 3 = X and y β 4 = Y in the given equation, we get,
X2 + XY β 2Y2 = 0
Comparing this equation with ax2 + 2hxy + by2 = 0, we get,
a = 1, h = \(\frac{1}{2}\), b = -2
This is the homogeneous equation of second degreeand h2 β ab = \(\left(\frac{1}{2}\right)^{2}\) β 1(-2)
= \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0
Hence, it represents a pair of lines passing through the new origin (3, 4).
Let ΞΈ be the acute angle between the lines.
β΄ tanΞΈ = 3 β΄ ΞΈ = tan-1(3)
Question 14.
Find the combined equation of pair of lines through the origin each of which makes angle of 60Β° with the Y-axis.
Solution:
Let OA and OB be the lines through the origin making an angle of 60Β° with the Y-axis.
Then OA and OB make an angle of 30Β° and 150Β° with the positive direction of X-axis.
β΄ slope of OA = tan 30Β° = \(\frac{1}{\sqrt{3}}\)
β΄ equation of the line OA is
y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x β \(\sqrt {3}\)y = 0
Slope of OB = tan 150Β° = tan (180Β° β 30Β°)
= tan 30Β° = \(-\frac{1}{\sqrt{3}}\)
β΄ equation of the line OB is
y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0
β΄ required combined equation is
(x β \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0
i.e. x2 β 3y2 = 0.
Question 15.
If lines representedby ax2 + 2hxy + by2 = 0 make angles of equal measures with the co-ordinate
axes then show that a = Β± b.
OR
Show that, one of the lines represented by ax2 + 2hxy + by2 = 0 will make an angle of the same measure with the X-axis as the other makes with the Y-axis, if a = Β± b.
Solution:
Let OA and OB be the two lines through the origin represented by ax2 + 2hxy + by2 = 0.
Since these lines make angles of equal measure with the coordinate axes, they make angles β and \(\frac{\pi}{2}\) β β with the positive direction of X-axis or β and \(\frac{\pi}{2}\) + β with thepositive direction of X-axis.
β΄ slope of the line OA = m1 = tan β
and slope of the line OB = m2
= tan(\(\frac{\pi}{2}\) β β) or tan(\(\frac{\pi}{2}\) + β)
i.e. m2 = cot β or m2 = -cot β
β΄ m1m2 β tan β x cot β = 1
OR m1m2 = tan β (-cot β) = -1
i.e. m1m2 = Β± 1
But m1m2 = \(\frac{a}{b}\)
β΄ \(\frac{a}{b}\)= Β±1 β΄ a = Β±b
This is the required condition.
Question 16.
Show that the combined equation of a pair of lines through the origin and each making an angle of β with the line x + y = 0 is x2 + 2(sec 2β) xy + y2 = 0.
Solution:
Let OA and OB be the required lines.
Let OA (or OB) has slope m.
β΄ its equation is y = mx β¦ (1)
It makes an angle β with x + y = 0 whose slope is -1. m +1
β΄ tan β = \(\left|\frac{m+1}{1+m(-1)}\right|\)
Squaring both sides, we get,
tan2β = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
β΄ tan2β(1 β 2m + m2) = m2 + 2m + 1
β΄ tan2β β 2m tan2β + m2tan2β = m2 + 2m + 1
β΄ (tan2β β 1)m2 β 2(1 + tan2β)m + (tan2β β 1) = 0
β΄ y2 + 2xysec2β + x2 = 0
β΄ x2 + 2(sec2β)xy + y2 = 0 is the required equation.
Question 17.
Show that the line 3x + 4y+ 5 = 0 and the lines (3x + 4y)2 β 3(4x β 3y)2 =0 form an equilateral triangle.
Solution:
The slope of the line 3x + 4y + 5 = 0 is \(\frac{-3}{4}\)
Let m be the slope of one of the line making an angle of 60Β° with the line 3x + 4y + 5 = 0. The angle between the lines having slope m and m1 is 60Β°.
On squaring both sides, we get,
3 = \(\frac{(4 m+3)^{2}}{(4-3 m)^{2}}\)
β΄ 3 (4 β 3m)2 = (4m + 3)2
β΄ 3(16 β 24m + 9m2) = 16m2 + 24m + 9
β΄ 48 β 72m + 27m2 = 16m2 + 24m + 9
β΄ 11m2 β 96m + 39 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).
β΄ the combined equation of the two lines is
11\(\left(\frac{y}{x}\right)^{2}\) β 96\(\left(\frac{y}{x}\right)\) + 39 = 0
β΄ \(\frac{11 y^{2}}{x^{2}}-\frac{96 y}{x}\) + 39 = 0
β΄ 11y2 β 96xy + 39x2 = 0
β΄ 39x2 β 96xy + 11y2 = 0.
β΄ 39x2 β 96xy + 11y2 = 0 is the joint equation of the two lines through the origin each making an angle of 60Β° with the line 3x + 4y + 5 = 0.
The equation 39x2 β 96xy + 11y2 = 0 can be written as :
-39x2 + 96xy β 11y2 = 0
i.e., (9x2 β 48x2) + (24xy + 72xy) + (16y2 β 27y2) = 0
i.e. (9x2 + 24xy + 16y2) β (48x2 β 72xy + 27y2) = 0
i.e. (9x2 + 24xy + 16y2) β 3(16x2 β 24xy + 9y2) = 0
i.e. (3x + 4y)2 β 3(4x β 3y)2 = 0
Hence, the line 3x + 4y + 5 = 0 and the lines
(3x + 4y)2 β 3(4x β 3y)2 form the sides of an equilateral triangle.
Question 18.
Show that lines x2 β 4xy + y2 = 0 and x + y = \(\sqrt {6}\) form an equilateral triangle. Find its area and perimeter.
Solution:
x2 β 4xy + y2 = 0 and x + y = \(\sqrt {6}\) form a triangle OAB which is equilateral.
Let OM be the perpendicular from the origin O to AB whose equation is x + y = \(\sqrt {6}\)
In right angled triangle OAM,
sin 60Β° = \(\frac{\mathrm{OM}}{\mathrm{OA}}\) β΄ \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{\mathrm{OA}}\)
β΄ OA = 2
β΄ length of the each side of the equilateral triangle OAB = 2 units.
β΄ perimeter of β OAB = 3 Γ length of each side
= 3 Γ 2 = 6 units.
Question 19.
If the slope of one of the lines given by ax2 + 2hxy + by2 = 0 is square of the other then show that a2b + ab2 + 8h3 = 6abh.
Solution:
Let m be the slope of one of the lines given by ax2 + 2hxy + by2 = 0.
Then the other line has slope m2
Multiplying by b3, we get,
-8h3 = ab2 + a2b β 6abh
β΄ a2b + ab2 + 8h3 = 6abh
This is the required condition.
Question 20.
Prove that the product of lengths of perpendiculars drawn from P (x1, y1) to the lines repersented by ax2 + 2hxy + by2 = 0 is \(\left|\frac{a x_{1}^{2}+2 h x_{1} y_{1}+b y_{1}^{2}}{\sqrt{(a-b)^{2}+4 h^{2}}}\right|\)
Solution:
Let m1 and m2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0.
β΄ m1 + m2 = \(-\frac{2 h}{b}\) and m1m2 = \(\frac{a}{b}\) β¦(1)
The separate equations of the lines represented by
ax2 + 2hxy + by2 = 0 are
y = m1x and y = m2x
i.e. m1x β y = 0 and m2x β y = 0
Length of perpendicular from P(x1, 1) on
Question 21.
Show that the difference between the slopes of lines given by (tan2ΞΈ + cos2ΞΈ )x2 β 2xytanΞΈ + (sin2ΞΈ )y2 = 0 is two.
Solution:
Comparing the equation (tan2ΞΈ + cos2ΞΈ)x2 β 2xy tan ΞΈ + (sin2ΞΈ) y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = tan2ΞΈ + cos2ΞΈ, 2h = -2 tan ΞΈ and b = sin2ΞΈ
Let m1 and m2 be the slopes of the lines represented by the given equation.
Question 22.
Find the condition that the equation ay2 + bxy + ex + dy = 0 may represent a pair of lines.
Solution:
Comparing the equation
ay2 + bxy + ex + dy = 0 with
Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0, we get,
A = 0, H = \(\frac{b}{2}\), B = a,G = \(\frac{e}{2}\), F = \(\frac{d}{2}\), C = 0
The given equation represents a pair of lines,
i.e. if bed β ae2 = 0
i.e. if e(bd β ae) = 0
i.e. e = 0 or bd β ae = 0
i.e. e = 0 or bd = ae
This is the required condition.
Question 23.
If the lines given by ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1 then show that (3a + b)(a + 3b) = 4h2.
Solution:
Since the lines ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1, the angle between the lines ax2 + 2hxy + by2 = 0 is 60Β°.
β΄ 3(a + b)2 = 4(h2 β ab)
β΄ 3(a2 + 2ab + b2) = 4h2 β 4ab
β΄ 3a2 + 6ab + 3b2 + 4ab = 4h2
β΄ 3a2 + 10ab + 3b2 = 4h2
β΄ 3a2 + 9ab + ab + 3b2 = 4h2
β΄ 3a(a + 3b) + b(a + 3b) = 4h2
β΄ (3a + b)(a + 3b) = 4h2
This is the required condition.
Question 24.
If line x + 2 = 0 coincides with one of the lines represented by the equation x2 + 2xy + 4y + k = 0 then show that k = -4.
Solution:
One of the lines represented by
x2 + 2xy + 4y + k = 0 β¦ (1)
is x + 2 = 0.
Let the other line represented by (1) be ax + by + c = 0.
β΄ their combined equation is (x + 2)(ax + by + c) = 0
β΄ ax2 + bxy + cx + 2ax + 2by + 2c = 0
β΄ ax2 + bxy + (2a + c)x + 2by + 2c β 0 β¦ (2)
As the equations (1) and (2) are the combined equations of the same two lines, they are identical.
β΄ by comparing their corresponding coefficients, we get,
β΄ 1 = \(\frac{-4}{k}\)
β΄ k = -4.
Question 25.
Prove that the combined equation of the pair of lines passing through the origin and perpendicular to the lines represented by ax2 + 2hxy + by2 = 0 is bx2 β 2hxy + ay2 = 0
Solution:
Let m1 and m2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0.
Now, required lines are perpendicular to these lines.
β΄ their slopes are and \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(-\frac{1}{m_{1}}\)x and y = \(-\frac{1}{m_{2}}\)x
i.e. m1y= -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
β΄ their combined equation is
(x + m1y)(x + m2y) = 0
β΄ x2 + (m1 + m2)xy + m1m2y2 = 0
β΄ x2\(\frac{-2 h}{b}\)x + \(\frac{a}{b}\)y2 = 0
β΄ bx2 β 2hxy + ay2 = 0.
Question 26.
If equation ax2 β y2 + 2y + c = 1 represents a pair of perpendicular lines then find a and c.
Solution:
The given equation represents a pair of lines perpendicular to each other.
β΄ coefficient of x2 + coefficient of y2 = 0
β΄ a β 1 = 0 β΄ a = 1
With this value of a, the given equation is
x2 β y2 + 2y + c β 1 = 0
Comparing this equation with
Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0, we get,
A = 1, H = 0, B = -1, G = 0, F = 1, C = c β 1
Since the given equation represents a pair of lines,
D = \(\left|\begin{array}{ccc}
A & H & G \\
H & B & F \\
G & F & C
\end{array}\right|\) = 0
β΄ \(\left|\begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 1 \\
0 & 1 & c-1
\end{array}\right|\) = 0
β΄ 1(-c + 1 β 1) β 0 + 0 = 0
β΄ -c = 0
β΄ c = 0.
Hence, a = 1, c = 0.