Question

The points $A(2,1),B(3,-2)$ and $C(a,b)$ are vertices of the rectangle $ABCD$. If the point $P(3,4)$ lies on $CD$ produced, then $5a+10b=$

• A. 41
• B. 10
• C. 45
• D. -15

Answer 1

Answer: (B)

Given, $ABCD$ is a rectangle

Slope of $AB=\frac{-2-1}{3-2}=-3=m_1\dots$.(i)
Slope of $BC=\frac{b+2}{a-3}=m_2\dots$.(ii)
Now $AB\perp BC\Rightarrow m_1{m}_2=-1\dots$. (iii)
$\therefore$ Putting value of $m_1$ and $m_2$ in Eq. (iii)
$-3\left(\frac{b+2}{a-3}\right)=-1$
$-3b-6=-a+3$
$\Rightarrow a-3b-9=0\dots$ (iv)
Now, slope of line $CD=$ slope of $CP$
$m_3=\frac{b-4}{a-3}$
$\therefore$ line
$AB\Vert CD\Rightarrow m_1=m_3$
$\therefore -3=\frac{b-4}{a-3}$
$-3a+9=b-4$
$\Rightarrow 3a+b-13=0\dots (v)$
Now, solving Eqs. (iv) and (v),
$a-3b-9=0\Rightarrow 3a+b-13=0$
Eq. (iv) $\times (3)$ and subtracting Eq. (iv) and Eq. (v),

These value is putting in Eq. (iv)
$a-3\left(-\frac{7}{5}\right)-9=0$
$\Rightarrow a+\frac{21}{5}-9=0$ $a+\left(\frac{24}{5}\right)=0$
$\Rightarrow a=\frac{24}{5}$
Now, $5a+10b=5\times \frac{24}{5}+10\times \left(-\frac{7}{5}\right)$
$=24-14=10$