Question

If the lines $y=4-3x,$ $ay=x+10,$ $2y+bx+9=0$ represent three consecutive sides of a rectangle and the fourth side passes through $(1,-2),$ then its area (in sq. units) is equal to

• A. $\frac{289}{20}$
• B. $\frac{225}{36}$
• C. $\frac{289}{24}$
• D. $\frac{225}{16}$

Answer 1

The correct option is A $\frac{289}{20}$

$y=4-3x\cdots \left(1\right)$
$ay=x+10\cdots \left(2\right)$
$2y+bx+9=0\cdots \left(3\right)$


$\left(1\right)\perp \left(2\right)\Rightarrow (-3)\left(\frac{1}{a}\right)=-1\Rightarrow a=3$
$\left(2\right)\perp \left(3\right)\Rightarrow \left(\frac{1}{a}\right)\left(\frac{-b}{2}\right)=-1\Rightarrow b=2a=6$
$\left(4\right)||\left(2\right)\Rightarrow$ Slope of fourth side is $\frac{1}{3}$ and it passes through $(1,-2)$
So, equation of the fourth side is
$y+2=\frac{1}{3}(x-1)$
$\Rightarrow x-3y-7=0$

Length of rectangle $=$ Distance between $\left(2\right)$ and $\left(4\right)$
$\left|\frac{10+7}{\sqrt{1^2+3^2}}\right|=\frac{17}{\sqrt{10}}$
Breadth of rectangle $=$ distance between $\left(1\right)$ and $\left(3\right)$
$\left|\frac{4+\frac{9}{2}}{\sqrt{3^2+1^2}}\right|=\frac{17}{2\sqrt{10}}$
$\therefore$ Area $=\frac{17}{\sqrt{10}}\times \frac{17}{2\sqrt{10}}=\frac{289}{20}$ sq. units