Question

$ABCD$ is a rhombus. Its diagonals $AC$ and BD intersect at the point $M$ and satisfy $BD=2AC$. If the coordinates of $D$ and $M$ are $(1,1)$ and $(2,-1)$ respectively, then the coordinates of $A$ are

• A. $(1,3)$ or $\left(-\frac{3}{2},-\frac{1}{2}\right)$
• B. $\left(1,-\frac{3}{2}\right)$ or $\left(3,-\frac{1}{2}\right)$
• C. $(1,2)$ or $(1,-1)$
• D. $\left(1,-\frac{1}{2}\right)$ or $\left(3,-\frac{3}{2}\right)$

Answer 1

Answer: (B)

Let the vertex $A$ be $(x,y)$
Since the diagonals of a rhombus are perpendicular,
we have, S
Slope of $AM\times$ slope of $MD=-1$

$\frac{y-1}{x-2}\times \frac{1+1}{1-2}=-1$
$\Rightarrow x-2y=4\dots$(i)
Given that $BD=2AC$
$\Rightarrow MD=2AM$
[Since the diagonals of a parallelogram bisect]
$\therefore M{D}^2=4A{M}^2$
$\Rightarrow (2-1{)}^2+(-1-1{)}^2$
$=4\left[(x-2{)}^2+(y+1{)}^2\right]$
$\Rightarrow 4x^2+4y^2-16x+8y+15=0$
$\Rightarrow 4(2y+4{)}^2+4y^2-16(2y+4)+8y+15=0$
$[$ from(i) $]$
$\Rightarrow 4y^2+8y+3=0$
$\Rightarrow y=-\frac{3}{2}$ or $-\frac{1}{2}$
If $y=-\frac{3}{2},x=2\left(-\frac{3}{2}\right)+4=1$
If $y=-\frac{1}{2},x=2\left(-\frac{1}{2}\right)+4=3$
$\therefore$ A is $\left(1,-\frac{3}{2}\right)$ or $\left(3,-\frac{1}{2}\right)$