Question

Sides of a rhombus are parallel to the lines $x+y-1=0$ and $7x-y-5=0$. It is given that diagonals of the rhombus intersect at $(1,3)$ and one vertex, '$A$' of the rhombus lie on the line $y=2x$. Then the coordinates of the vertex are

• A. $(\frac{8}{5},\frac{16}{5})$
• B. $(\frac{7}{15},\frac{14}{15})$
• C. $(\frac{6}{5},\frac{12}{5})$
• D. $(\frac{4}{15},\frac{8}{15})$

Answer 1

Answer: (A), (C)

$(\frac{8}{5},\frac{16}{5})$
$(\frac{6}{5},\frac{12}{5})$

From the above questions, it can be observed that the diagonals of the rhombus will be parallel to the bisectors of the line $x+y-1=0$ and $7x-y-5=0$
Therefore
$\frac{x+y-1}{\sqrt{2}}=\pm \frac{7x-y-5}{5\sqrt{2}}$
$5x+5y-5=7x-y-5$ ...(i) and $5x+5y-5=-7x+y+5$...(ii)
$6y=2x$ and $12x+4y=10$
Therefore the equations of the diagonals will be $x-3y+c=0$ and $6x+2y+d=0$
Since they pass through $(1,3)$, therefore
$1-9+c=0$ and $6+6+d=0$
Hence $c=8$ and $d=-12$.
Therefore the equation for the diagonals will be
$x-3y+8=0$ and $3x+y-6=0$
Thus the required vertex will be the point where these lines meet the line $y=2x$
Substituting $y=2x$ in the above equations we get
$x-6x=-8$ and $5x=6$
$-5x=-8$ and $x=\frac{6}{5}$
$x=\frac{8}{5}$ and $x=\frac{6}{5}$
Thus the coordinates are $(\frac{8}{5},\frac{16}{5})$ and $(\frac{6}{5},\frac{12}{5}).$