The equation $2 x^{2} - 3 x y - p y^{2} + x + q y - 1 = 0$ represents two mutually perpendicular lines if

p = 3, q = 2

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p = 2, q = 3

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p = - 2, q = 3

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p = 2, q = - \frac{9}{2}

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Solution

The correct option is B $p = 2, q = 3$
The angle between the pair of lines $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ is same as the angle between the lines $a x^{2} + 2 h x y + b y^{2} = 0$ .
Angle between lines is given by $tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}$
Given,Lines are mutually perpendicular to each other,then we get $a + b = 0$
which gives $p = 2 a n d q = 3$ .

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The equation 2x2−3xy−py2+x+qy−1=0 represents two mutually perpendicular lines if

The correct option is B p=2,q=3The angle between the pair of lines ax2+2hxy+by2+2gx+2fy+c=0 is same as the angle between the lines ax2+2hxy+by2=0 .Angle between lines is given by tanθ=2√h2−aba+b Given,Lines are mutually perpendicular to each other,then we get a+b=0which gives p=2 and q=3.

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