Let $P Q R$ be a right angled isoceles triangle, right angled at $P (2, 1)$ . If the equation of the line QR is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is -

3 x^{2} - 3 y^{2} + 8 x y + 20 x + 10 y + 25 = 0

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3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0

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3 x^{2} - 3 y^{2} + 8 x y + 10 x + 15 y + 20 = 0

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3 x^{2} - 3 y^{2} - 8 x y - 10 x - 15 y - 20 = 0

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Solution

The correct option is A $3 x^{2} - 3 y^{2} + 8 x y + 20 x + 10 y + 25 = 0$
Given , equation of QR
$2 x + y = 3 \Rightarrow y = - 2 x + 3 ∴ m_{1} = - 2$ [y = mx + c]
Now, we know
$tan α = ∣ ∣ ∣ \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} ∣ ∣ ∣ \Rightarrow tan α = ∣ ∣ ∣ \frac{m_{2} - (- 2)}{1 + (- 2) m_{2}} ∣ ∣ ∣ \Rightarrow 1 = ∣ ∣ ∣ \frac{m_{2} + 2}{1 - 2 m_{2}} ∣ ∣ ∣$
$∴ 1 = \frac{m_{2} + 2}{1 - 2 m_{2}} \Rightarrow 1 - 2 m_{2} = m_{2} + 2 \Rightarrow m_{2} = \frac{- 1}{3}$
$∴ 1 = \frac{- (m_{2} + 2)}{1 - 2 m_{2}} \Rightarrow - 1 + 2 m_{2} = m_{2} + 2 \Rightarrow m_{2} = 3$
$∴$ for $m_{2} = \frac{- 1}{3}$ and point P (2, 1)
$\Rightarrow y - 1 = \frac{- 1}{3} (x - 2) \Rightarrow 3 y - 3 = - x + 2 \Rightarrow x + 3 y - 5 = 0$
$∴$ for $m_{2} = 3$ and point P (2, 1)
$\Rightarrow y - 1 = (3) (x - 2) \Rightarrow y - 1 = 3 x - 6 \Rightarrow 3 x - y - 5 = 0$
$∴$ The equation representing the pair of line PQ and PR
$\Rightarrow (x + 3 y - 5) (3 x - y - 5) = 0 \Rightarrow 3 x^{2} - x y - 5 x + 9 x y - 3 y^{2} - 15 y - 15 x + 5 y + 25 = 0 \Rightarrow 3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0$
The correct answer is B.

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Similar questions

Let $P Q R$ be a right angled isosceles triangle right angled at $P (2, 1)$ . If the equation of the line $Q R$ is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is

Q. Let

A B C

be a right angled triangle with angle at

A (2, 1)

being

90^{o}

. If the equation of pair of the line

B C

2 x + y = 3

then the joining equation of lines

A B

and

A C

3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0

Q. The equations of the diagonals of the square formed by the pairs of straight lines

3 x^{2} + 8 x y - 3 y^{2} = 0

and

3 x^{2} + 8 x y - 3 y^{2} + 2 x - 4 y - 1 = 0

are?

Q. Prove that the lines represented by

3 x^{2} - 8 x y - 3 y^{2} = 0

and

x + 2 y = 3

form the sides of an isosceles right angled triangle.

Q. The equation

3 x^{2} - 8 x y - 3 y^{2} = 0

and

x + 2 y = 3

represent the sides of

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Let PQR be a right angled isoceles triangle, right angled at P(2,1). If the equation of the line QR is 2x+y=3, then the equation representing the pair of lines PQ and PR is -

Let $P Q R$ be a right angled isoceles triangle, right angled at $P (2, 1)$ . If the equation of the line QR is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is -