The points $(1,3),(5,1)$ are opposite vertices of a diagonal of a rectangle. If the other two vertices lie…
- (1,-2)
- (0,-4)
- (2,0)
- (3,2)
Solution
Diagonals of rectangle bisect each other.
$\therefore \quad$ Midpoint of $(1,3)$ and $(5,1)$ is $(3,2)$.
Also, $y=2 x+$ c passes through $(3,2)$.
$\begin{array}{rlrl}
\therefore & 2 & =2(3)+c \\
& \therefore & c & =-4
\end{array}$
$\therefore \quad$ Other two vertices lie on $y=2 x-4$.
Let co-ordinates of $\mathrm{B}$ be $(x, y)$ i.e., $(x, 2 x-4)$
slope of $\mathrm{AB} \times$ slope of $\mathrm{BC}=-1$
$\begin{aligned}
& \Rightarrow\left(\frac{2 x-4-3}{x-1}\right)\left(\frac{2 x-4-1}{x-5}\right)=-1 \\
& \Rightarrow\left(\frac{2 x-7}{x-1}\right)\left(\frac{2 x-5}{x-5}\right)=-1 \\
& \Rightarrow x^2-6 x+8=0 \\
& \Rightarrow x=4,2
\end{aligned}$
When $x=4, y=4$
When $x=2, y=0$
$\therefore \quad$ Vertex of the other diagonal is $(2,0)$.Asked in: MHT CET 2023 (10 May Shift 1)