Let the equations of two adjacent sides of a parallelogram ABCD be $2 \mathrm{x}-3 \mathrm{y}=-23$ and $5 \mathrm{x}+4 \mathrm{y} =23$. If the equation of its one diagonal AC is $3 \mathrm{x}+ 7 \mathrm{y}=23$ and the distance of A from the other diagonal is d , then $50 \mathrm{~d}^{2}$ is equal to

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Straight Lines

Let the equations of two adjacent sides of a parallelogram ABCD be

2 x - 3 y = - 23

and

5 x + 4 y = 23

. If the equation of its one diagonal AC is

3 x + 7 y = 23

and the distance of A from the other diagonal is d , then

50 d^{2}

is equal to

Answer: 529

Solution

The two adjacent sides

2 x - 3 y = - 23

and

5 x + 4 y = 23

meet at

B (- 1, 7)

. Vertex

A

is where

2 x - 3 y = - 23

meets the diagonal

3 x + 7 y = 23

, giving

A (- 4, 5)

, and

C

is where

5 x + 4 y = 23

meets the diagonal, giving

C (3, 2)

. The centre is the mid-point of

A C

, namely

(- \frac{1}{2}, \frac{7}{2})

, which is also the mid-point of

B D

, so

D (0, 0)

. The other diagonal

B D

through

(0, 0)

and

(- 1, 7)

7 x + y = 0

, and the distance of

A (- 4, 5)

from it is

d = \frac{∣7 ( - 4 ) + 5∣}{50} = \frac{23}{50}

. Therefore

50 d^{2} = 529

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Integer answer

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