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Q.
The points $A(2,1), B(3,-2)$ and $C(a, b)$ are vertices of the rectangle $A B C D$. If the point $P(3,4)$ lies on $C D$ produced, then $5 a+10 b=$
TS EAMCET 2020
Solution:
Given, $A B C D$ is a rectangle
Slope of $A B=\frac{-2-1}{3-2}=-3=m_{1} \ldots$.(i)
Slope of $B C=\frac{b+2}{a-3}=m_{2} \ldots$.(ii)
Now $A B \perp B C \Rightarrow m_{1} m_{2}=-1 \ldots$. (iii)
$\therefore $ Putting value of $m_{1}$ and $m_{2}$ in Eq. (iii)
$-3\left(\frac{b+2}{a-3}\right)=-1$
$-3 b-6=-a+3$
$\Rightarrow a-3 b-9=0 \ldots$ (iv)
Now, slope of line $C D=$ slope of $C P$
$m_{3}=\frac{b-4}{a-3}$
$\therefore $ line
$A B \| C D \Rightarrow m_{1}=m_{3}$
$\therefore -3=\frac{b-4}{a-3}$
$-3 a+9=b-4$
$\Rightarrow 3 a+b-13=0 \ldots( v )$
Now, solving Eqs. (iv) and (v),
$a-3 b-9=0 \Rightarrow 3 a+b-13=0$
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, $5 a+10 b=5 \times \frac{24}{5}+10 \times\left(-\frac{7}{5}\right)$
$=24-14=10$
