Question #155246

Prove that the equation – x2 − xy + 2y2+2x + y − 1 = 0 represents a pair of


straight lines. Also find the point of intersection and angle between them.


1
Expert's answer
2021-01-14T12:46:24-0500

The general equation of second-degree ax² + 28xy + by² + 2gx + 2fy + c = 0 represents a pair of straight lines if Δ=abc + 2fgh − af² − bg² − ch² =0

 For – x2 − xy + 2y2+2x + y − 1 = 0,

a = -1, h =-0.5, b = 2 , g = 1, f = 0.5, c = -1

∆ = (-1 x 2 x -1) + (2 x 0.5 x 1 x -0.5) - (-1 x 0.5²) - (2 x 1²) - (-1 x (-0.5)²) =0

The equation represents a pair of straight lines.

the point of intersection is found by partially differentiating the equation first with respect to x and then with respect to y and solving both the equations.

Dy/dx (– x2 − xy + 2y2+2x + y − 1 = 0)

-2x - y +0 +2 +0-0 =0

-2x -y +2 =0-------------(i)


0-x +4y +2 +0-0 =0

-x +4y +2 =0------------(ii)


Solving both equations simultaneously, x = 10/9 and y = -2/9

The point of intersection is (10/9, -2/9)

Angle between a pair of straight lines is determined by

tan(θ)=2(h2ab)/(a+b)tan( \theta) = |2\sqrt{(h² - ab)} / (a + b)|

tan(θ)=2(0.5)2(12)/(1+2)tan(\theta) = |2\sqrt{(-0.5)² -(-1 * 2)} / (-1+2)|

θ=\theta= 71.57°


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