Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 4 Pair of Straight Lines Ex 4.2 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.2
Solution & Step-by-Step Answer:
Comparing the equation 3x2 – 4 xy – 3y2 = 0 with ax2 + 2hxy + by2 = 0, we get, a = 3, 2h = -4, b = -3 Since a + b = 3 + (-3) = 0, the lines represented by 3x2 – 4xy – 3y2 = 0 are perpendicular to each other.
Solution & Step-by-Step Answer:
Comparing the equation x2 + 6xy + 9y2 = 0 with ax2 + 2hxy + by2 = 0, we get, a = 1, 2h = 6, i.e. h = 3 and b = 9 Since h2 – ab = (3)2 – 1(9) = 9 – 9 = 0,. the lines represented by x2 + 6xy + 9y2 = 0 are coincident.
Solution & Step-by-Step Answer:
Comparing the equation kx2 + 4xy – 4y2 = 0 with ax2 + 2hxy + by2 = 0, we get, a = k, 2h = 4, b = -4 Since lines represented by kx2 + 4xy – 4y2 = 0 are perpendicular to each other, a + b = 0 ∴ k – 4 = 0 ∴ k = 4.
Solution & Step-by-Step Answer:
Comparing the equation 3x2 – 4xy + 3y2 = 0 with ax2 + 2hxy + by2 = 0, we get, a = 3, 2h = -4, i.e. h = -24 and b = 3 Let θ be the acute angle between the lines. ∴ θ = 30°.
(ii) 4x2+ 5xy + y2= 0
Solution:
Comparing the equation 4x2+ 5xy + y2= 0 with ax2+ 2hxy + by2= 0, we get,
a = 4, 2h = 5, i.e. h = and b = 1.
Let θ be the acute angle between the lines.
(iii) 2x2+ 7xy + 3y2= 0
Solution:
Comparing the equation
2x2+ 7xy + 3y2= 0 with
ax2+ 2hxy + by2= 0, we get,
a = 2, 2h = 7 i.e. h = and b = 3
Let θ be the acute angle between the lines.
tanθ = 1
∴ θ = tan 1 = 45°
∴ θ = 45°
(iv) (a2– 3b2)x2+ 8abxy + (b2– 3a2)y2= 0
Solution:
Comparing the equation
(a2– 3b2)x2+ 8abxy + (b2– 3a2)y2= 0, with
Ax2+ 2Hxy + By2= 0, we have,
A = a2– 3b2, H = 4ab, B = b2– 3a2.
∴ H2– AB = 16a2b2– (a2– 3b2)(b2– 3a2)
= 16a2b2+ (a2– 3b2)(3a2– b2)
= 16a2b2+ 3a4– 10a2b2+ 3b4
= 3a4+ 6a2b2+ 3b4
= 3(a4+ 2a2b2+ b4)
= 3 (a2+ b2)2
∴ = (a2+ b2)
Also, A + B = (a2– 3b2) + (b2– 3a2)
= -2 (a2+ b2)
If θ is the acute angle between the lines, then
tan θ =
= = tan 60°
∴ θ = 60°
Solution & Step-by-Step Answer:
The slope of the line 3x + 2y – 11 = 0 is m1 = . Let m be the slope of one of the lines making an angle of 30° with the line 3x + 2y – 11 = 0. The angle between the lines having slopes m and m1 is 30°. On squaring both sides, we get, ∴ (2 – 3m)2 = 3 (2m + 3)2 ∴ 4 – 12m + 9m2 = 3(4m2 + 12m + 9) ∴ 4 – 12m + 9m2 = 12m2 + 36m + 27 3m2 + 48m + 23 = 0 This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = . ∴ the combined equation of the two lines is 3 + 48 + 23 = 0 ∴ + 23 = 0 ∴ 3y2 + 48xy + 23x2 = 0 ∴ 23x2 + 48xy + 3y2 = 0.
Solution & Step-by-Step Answer:
The acute angle θ between the lines ax2 + 2hxy + by2 = 0 is given by tan θ = ..(1) Comparing the equation 2x2 – 5xy + 3y2 = 0 with ax2 + 2hxy + by2 = 0, we get, a = 2, 2h= -5, i.e. h = and b = 3 Let ∝ be the acute angle between the lines 2x2 – 5xy + 3y2 = 0. This is the required condition.
Solution & Step-by-Step Answer:
Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis. Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis. ∴ slope of OA = tan 30° = ∴ equation of the line OA is y = = x, i.e. x – y = 0 Slope of OB = tan 150° = tan (180° – 30°) = tan 30° = ∴ equation of the line OB is y = x, i.e. x + y = 0 ∴ required combined equation is (x – y)(x + y) = 0 i.e. x2 – 3y2 = 0.
