Maharashtra State Board Class 12 (HSC Board) Mathematics & Statistics Solutions Chapter 4 Pair of Straight Lines Ex 4.1 - Digest Answers | Maharashtra Board Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 4 Pair of Straight Lines Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.1

Question 1 Maharashtra Board Solution

Find the combined equation of the following pairs of lines: (i) 2x + y = 0 and 3x – y = 0

Solution & Step-by-Step Answer:

The combined equation of the lines 2x + y = 0 and 3x – y = 0 is (2x + y)( 3x – y) = 0 ∴ 6x2 – 2xy + 3xy – y2 = 0 ∴ 6x2 – xy – y2 = 0.

(ii) x + 2y – 1 = 0 and x – 3y + 2 = 0
Solution:
The combined equation of the lines x + 2y – 1 = 0 and x – 3y + 2 = 0 is
(x + 2y – 1)(x – 3y + 2) = 0
∴ x²– 3xy + 2x + 2xy – 6y²+ 4y – x + 3y – 2 = 0
∴ x²– xy – 6y²+ x + 7y – 2 = 0.

(iii) Passing through (2, 3) and parallel to the co-ordinate axes.
Solution:
Equations of the coordinate axes are x = 0 and y = 0.
∴ the equations of the lines passing through (2, 3) and parallel to the coordinate axes are x = 2 and
i.e. x – 2 = 0 and y – 3 = 0.
∴ their combined equation is
(x – 2)(y – 3) = 0.
∴ xy – 3x – 2y + 6 = 0.

(iv) Passing through (2, 3) and perpendicular to lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0
Solution:
Let L₁and L₂be the lines passing through the point (2, 3) and perpendicular to the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 respectively.
Slopes of the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 are and respectively.
∴ slopes of the lines L₁and L₂are and -3 respectively.
Since the lines L₁and L₂pass through the point (2, 3), their equations are
y – 3 = (x – 2) and y – 3 = -3 (x – 2)
∴ 3y – 9 = 2x – 4 and y – 3= -3x + 6
∴ 2x – 3y + 5 = 0 and 3x – y – 9 = 0
∴ their combined equation is
(2x – 3y + 5)(3x + y – 9) = 0
∴ 6x²+ 2xy – 18x – 9xy – 3y²+ 27y + 15x + 5y – 45 = 0
∴ 6x²– 7xy – 3y²– 3x + 32y – 45 = 0.

(v) Passsing through (-1, 2),one is parallel to x + 3y – 1 = 0 and the other is perpendicular to 2x – 3y – 1 = 0.
Solution:
Let L₁be the line passing through (-1, 2) and parallel to the line x + 3y – 1 = 0 whose slope is –.
∴ slope of the line L₁is –
∴ equation of the line L₁is
y – 2 = –(x + 1)
∴ 3y – 6 = -x – 1
∴ x + 3y – 5 = 0
Let L₂be the line passing through (-1, 2) and perpendicular to the line 2x – 3y – 1 = 0
whose slope is .
∴ slope of the line L₂is –
∴ equation of the line L₂is
y – 2= –(x + 1)
∴ 2y – 4 = -3x – 3
∴ 3x + 2y – 1 = 0
Hence, the equations of the required lines are
x + 3y – 5 = 0 and 3x + 2y – 1 = 0
∴ their combined equation is
(x + 3y – 5)(3x + 2y – 1) = 0
∴ 3x²+ 2xy – x + 9xy + 6y²– 3y – 15x – 10y + 5 = 0
∴ 3x²+ 11xy + 6y²– 16x – 13y + 5 = 0

Question 2 Maharashtra Board Solution

Find the separate equations of the lines represented by following equations: (i) 3y2 + 7xy = 0

Solution & Step-by-Step Answer:

3y2 + 7xy = 0 ∴ y(3y + 7x) = 0 ∴ the separate equations of the lines are y = 0 and 7x + 3y = 0.

(ii) 5x²– 9y²= 0
Solution:
5x²– 9y²= 0
∴ ( x)²– (3y)²= 0
∴ (x + 3y)(x – 3y) = 0
∴ the separate equations of the lines are
x + 3y = 0 and x – 3y = 0.

(iii) x²– 4xy = 0
Solution:
x²– 4xy = 0
∴ x(x – 4y) = 0
∴ the separate equations of the lines are x = 0 and x – 4y = 0

(iv) 3x²– 10xy – 8y²= 0
Solution:
3x²– 10xy – 8y²= 0
∴ 3x²– 12xy + 2xy – 8y²= 0
∴ 3x(x – 4y) + 2y(x – 4y) = 0
∴ (x – 4y)(3x +2y) = 0
∴ the separate equations of the lines are x – 4y = 0 and 3x + 2y = 0.

(v) 3x²– xy – 3y²= 0
Solution:
3x²– 2xy – 3y2 = 0
∴ 3x²– 3xy + xy – 3y²= 0
∴ 3x(x – y) + y(x – y) = 0
∴ (x – y)(3x + y) = 0
∴ the separate equations of the lines are
∴ x – y = 0 and 3x + y = 0.

(vi) x²+ 2(cosec ∝)xy + y²= 0
Solution:
x²+ 2 (cosec ∝)xy – y²= 0
i.e. y²+ 2(cosec∝)xy + x²= 0
Dividing by x², we get,

∴ the separate equations of the lines are
(cosec ∝ – cot ∝)x + y = 0 and (cosec ∝ + cot ∝)x + y = 0.

(vii) x²+ 2xy tan ∝ – y²= 0
Solution:
x²+ 2xy tan ∝ – y²= 0
Dividind by y²

The separate equations of the lines are
(sec∝ – tan ∝)x + y = 0 and (sec ∝ + tan ∝)x – y = 0

Question 3 Maharashtra Board Solution

Find the combined equation of a pair of lines passing through the origin and perpendicular to the lines represented by following equations : (i) 5x2 – 8xy + 3y2 = 0

Solution & Step-by-Step Answer:

Comparing the equation 5x2 – 8xy + 3y2 = 0 with ax2 + 2hxy + by2 = 0, we get, a = 5, 2h = -8, b = 3 Let m1 and m2 be the slopes of the lines represented by 5x2 – 8xy + 3y2 = 0. ∴ m1 + m2 = amd m1m2 = …(1) Now required lines are perpendicular to these lines ∴ their slopes are -1 /m1 and -1/m2 Since these lines are passing through the origin, their separate equations are y = x and y = x i.e. m1y = -x and m2y = -x i.e. x + m1y = 0 and x + m2y = 0 ∴ their combined equation is (x + m1y) (x + m2y) = 0 ∴ x2 + (m1 + m2)xy + m1m2y2 = 0 ∴ x2 + xy + y2 = 0 … [By (1)] ∴ x2 + 8xy + 5y = 0

(ii) 5x²+ 2xy – 3y²= 0
Solution:
Comparing the equation 5x²+ 2xy – 3y²= 0 with ax²+ 2hxy + by²= 0, we get,
a = 5, 2h = 2, b = -3
Let m₁and m₂be the slopes of the lines represented by 5x²+ 2xy – 3y²= 0
∴ m₁+ m₂= and m₁m₂= ..(1)
Now required lines are perpendicular to these lines
∴ their slopes are and
Since these lines are passing through the origin, their separate equations are
y = x and y = x
i.e. m₁y = -x amd m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
∴ (x + m₁y)(x + m₂y) = 0
x²+ (m₁+ m₂)xy + m₁m₂y²= 0
∴ x²+ xy – y = 0 …[By (1)]
∴ 3x²+ 2xy – 5y²= 0

(iii) xy + y²= 0
Solution:
Comparing the equation xy + y²= 0 with ax²+ 2hxy + by²= 0, we get,
a = 0, 2h = 1, b = 1
Let m₁and m₂be the slopes of the lines represented by xy + y²= 0

Now required lines are perpendicular to these lines
∴ their slopes are and .
Since these lines are passing through the origin, their separate equations are
y = x and y = x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y) (x + m₂y) = 0
∴ x²+ (m₁+ m₂)xy + m₁m₂y²= 0
∴ x²– xy = 0.y²= 0 … [By (1)]
∴ x²– xy = 0.
Alternative Method :
Consider xy + y²= 0
∴ y(x + y) = 0
∴ separate equations of the lines are y = 0 and
3x²+ 8xy + 5y²= 0.
x + y = 0.
Let m₁and m₂be the slopes of these lines.
Then m₁= 0 and m₂= -1
Now, required lines are perpendicular to these lines.
∴ their slopes are and
Since, m₁= 0, does not exist.
Also, m₂= -1, = 1
Since these lines are passing through the origin, their separate equations are x = 0 and y = x,
i.e. x – y = 0
∴ their combined equation is
x(x – y) = 0
x²– xy = 0.

(iv) 3x²– 4xy = 0
Solution:
Consider 3x²– 4xy = 0
∴ x(3x – 4y) = 0
∴ separate equations of the lines are x = 0 and 3x – 4y = 0.
Let m₁and m₂be the slopes of these lines.
Then m₁does not exist and and m₁= .
Now, required lines are perpendicular to these lines.
∴ their slopes are and .
Since m₁does not exist, = 0
Also m₂=
Since these lines are passing through the origin, their separate equations are y = 0 and y = x,
i.e. 4x + 3y = 0
∴ their combined equation is
y(4x + 3y) = 0
∴ 4xy + 3y²= 0.

Question 4 Maharashtra Board Solution

Find k if, (i) the sum of the slopes of the lines represented by x2 + kxy – 3y2 = 0 is twice their product.

Solution & Step-by-Step Answer:

Comparing the equation x2 + kxy – 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 – 3y2 = 0. ∴ m1 + m2 = and m1m2 = Now, m1 + m2 = 2(m1m2)..(Given) ∴ ∴ k = -2

(ii) slopes of lines represent by 3x²+ kxy – y²= 0 differ by 4.
Solution:
(ii) Comparing the equation 3x²+ kxy – y²= 0 with ax²+ 2hxy + by²= 0, we get, a = 3, 2h = k, b = -1.
Let m₁and m₂be the slopes of the lines represented by 3x²+ kxy – y²= 0.
∴ m₁+ m₂= = k
and m₁₂= = -3
∴ (m₁– m₂)²= (m₁+ m₂)²– 4m₁m₂
= k²– 4 (-3)
= k²+ 12 … (1)
But |m₁– m₂| =4
∴ (m₁– m₂)²= 16 … (2)
∴ from (1) and (2), k²+ 12 = 16
∴ k²= 4 ∴ k= ±2.

(iii) slope of one of the lines given by kx²+ 4xy – y²= 0 exceeds the slope of the other by 8.
Solution:
Comparing the equation kx²+ 4xy – y²= 0 with²+ 2hxy + by²= 0, we get, a = k, 2h = 4, b = -1. Let m₁and m₂be the slopes of the lines represented by kx²+ 4xy – y²= 0.
∴ m₁+ m₂= = 4
and m₁m₂= = -k
We are given that m₂= m₁+ 8
m₁+ m₁+ 8 = 4
∴ 2m₁= -4 ∴ m₁= -2 … (1)
Also, m₁(m₁+ 8) = -k
(-2)(-2 + 8) = -k … [By(1)]
∴ (-2)(6) = -k
∴ -12= -k ∴ k = 12.

Question 5 Maharashtra Board Solution

Find the condition that : (i) the line 4x + 5y = 0 coincides with one of the lines given by ax2 + 2hxy + by2 = 0.

Solution & Step-by-Step Answer:

The auxiliary equation of the lines represented by ax2 + 2hxy + by2 = 0 is bm2 + 2hm + a = 0. Given that 4x + 5y = 0 is one of the lines represented by ax2 + 2hxy + by2 = 0. The slope of the line 4x + 5y = 0 is . ∴ m = is a root of the auxiliary equation bm2 + 2hm + a = 0. ∴ b + 2h + a = 0 ∴ + a = 0 ∴ 16b – 40h + 25a = 0 ∴ 25a + 16b = 40k. This is the required condition.

(ii) the line 3x + y = 0 may be perpendicular to one of the lines given by ax²+ 2hxy + by²= 0.
Solution:
The auxiliary equation of the lines represented by ax²+ 2hxy + by²= 0 is bm²+ 2hm + a = 0.
Since one line is perpendicular to the line 3x + y = 0
whose slope is = -3
∴ slope of that line = m =
∴ m = is the root of the auxiliary equation bm²+ 2hm + a = 0.
∴ b + 2h + a = 0
∴ + a = 0
∴ b + 6h + 9a = 0
∴ 9a + b + 6h = 0
This is the required condition.

Question 6 Maharashtra Board Solution

If one of the lines given by ax2 + 2hxy + by2 = 0 is perpendicular to px + qy = 0 then show that ap2 + 2hpq + bq2 = 0.

Solution & Step-by-Step Answer:

To prove ap2 + 2hpq + bq2 = 0. Let the slope of the pair of straight lines ax2 + 2hxy + by2 = 0 be m1 and m2 Then, m1 + m2 = and m1m2 = Slope of the line px + qy = 0 is But one of the lines of ax2 + 2hxy + by2 = 0 is perpendicular to px + qy = 0 ⇒ bq2 + ap2 = -2hpq ⇒ ap2 + 2hpq + bq2 = 0

Question 7 Maharashtra Board Solution

Find the combined equation of the pair of lines passing through the origin and making an equilateral triangle with the line y = 3.

Solution & Step-by-Step Answer:

Let OA and OB be the lines through the origin making.an angle of 60° with the line y = 3. ∴ OA and OB make an angle of 60° and 120° with the positive direction of X-axis. ∴ slope of OA = tan60° = ∴ equation of the line OA is y = x, i.e. x – y = 0 Slope of OB = tan 120° = tan (180° – 60°) = -tan 60°= – ∴ equation of the line OB is y = – x, i.e. x + y = 0 ∴ required joint equation of the lines is ( x – y)( x + y) = 0 i.e. 3x2 – y2 = 0.

Question 8 Maharashtra Board Solution

If slope of one of the lines given by ax2 + 2hxy + by2 = 0 is four times the other then show that 16h2 = 25ab.

Solution & Step-by-Step Answer:

Let m1 and m2 be the slopes of the lines given by ax2 + 2hxy + by2 = 0. ∴ m1 + m2 = and m1m2 = We are given that m2 = 4m1 ∴ 16h2 = 25ab This is the required condition.

Question 9 Maharashtra Board Solution

If one of the lines given by ax2 + 2hxy + by2 = 0 bisects an angle between co-ordinate axes then show that (a + b) 2 = 4h2.

Solution & Step-by-Step Answer:

The auxiliary equation of the lines given by ax2 + 2hxy + by2 = 0 is bm2 + 2hm + a = 0. Since one of the line bisects an angle between the coordinate axes, that line makes an angle of 45° or 135° with the positive direction of X-axis. ∴ slope of that line = tan45° or tan 135° ∴ m = tan45° = 1 or m = tan 135° = tan (180° – 45°) = -tan 45°= -1 ∴ m = ±1 are the roots of the auxiliary equation bm2 + 2hm + a = 0. ∴ b(±1)2 + 2h(±1) + a = 0 ∴ b ± 2h + a = 0 ∴ a + b = ±2h ∴ (a + b)2 = 4h2 This is the required condition.

Chapter 4 Pair of Straight Lines Ex 4.1 Solutions