Question

A$(-3,4)$, B$(5,4)$, C and D form a rectangle. If $x-4y+7=0$ is a diameter of the circumcircle of the rectangle ABCD then area of rectangle ABCD is

• A. $8$
• B. $16$
• C. $24$
• D. $64$

Answer 1

Answer: (C)

$24$

Given points

$A(-3,4),B(5,4)$

Equation of diagonal $x-4y+7=0$

Equation of side AB

$y-4=\frac{0}{5+3}(x+3)$

$y=4----\left(1\right)$

Equation of side AD perpendicular to AB

$x=\lambda$ this equation passes through $A(-3,4)$

$x=-3-----\left(2\right)$

On solving eq (2) and eq of diagonal $x-4y+7=0$ we get point D

$-3-4y+7=0$

$y=1$

$D(-3,1)$

Hence area of rectangle

$area=AB\times AD$

$area=8\times 3=24$