JEE Main 2023 — Mathematics Coordinate Geometry

Given,

The equations of two adjacent sides of a parallelogram $ABCD$ABCD be $2x-3y=-23$2x−3y=−23 and $5x+4y=23$5x+4y=23,

So, $AB\equiv 2x-3y=-23$AB≡2x−3y=−23 and $BC\equiv 5x+4y=23$BC≡5x+4y=23

Also given, $AC\equiv 3x+7y=23$AC≡3x+7y=23

Solving the above lines we get, $A(–4,5),B(–1,7),C(3,2)$A(–4,5),B(–1,7),C(3,2)

We know that,

Diagonal of parallelogram have same midpoint,

So $AC$AC and $BD$BD have same mid-point and let point $D$D be $(x,y)$(x,y),

So midpoint formula we get,

$\frac{x-1}{2}=\frac{-4+3}{2}\Rightarrow x=0$2x−1​=2−4+3​⇒x=0 and $\frac{y+7}{2}=\frac{2+5}{2}\Rightarrow y=0$2y+7​=22+5​⇒y=0

Hence, point $D$D is $(0,0)$(0,0)

Now Equation of $BD$BD will be $7x+y=0$7x+y=0

Now finding the distance of $A(-4,5)$A(−4,5) from $7x+y=0$7x+y=0 we get,

$$d=|\frac{7(-4)+5}{\sqrt{{7}^{2}+{1}^{2}}}|=\frac{23}{\sqrt{50}}$$d=∣72+12​7(−4)+5​∣=50​23​

Hence, $50{d}^{2}={23}^{2}=529$50d2=232=529