Let A(1,2) and C(-3,-6) be two diagonally opposite vertices of a rhombus, whose sides AD a…

Question

Let $A(1,2)$ and $C(-3,-6)$ be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line $7x-y=14$. If $B(\alpha ,\beta )$ and $D(\gamma ,\delta )$ are the other two vertices, then $|\alpha +\beta +\gamma +\delta |$ is equal to :

Accepted Answer

Let $A=(1,2)$ and $C=(-3,-6)$ be diagonally opposite vertices of a rhombus.

In a rhombus, the diagonals bisect each other at a common point M..

Calculation of Midpoint $M$ :

$\begin{gathered} M=(\frac{1+(-3)}{2},\frac{2+(-6)}{2}) \ M=(\frac{-2}{2},\frac{-4}{2}) \ M=(-1,-2) \end{gathered}$

Since $M$ is also the midpoint of the other diagonal $BD$ where $B=(\alpha ,\beta )$ and $D=(\gamma ,\delta )$ :

$\begin{gathered} \frac{\alpha +\gamma }{2}=-1\implies \alpha +\gamma =-2 \ \frac{\beta +\delta }{2}=-2\implies \beta +\delta =-4 \end{gathered}$

$\begin{gathered} |\alpha +\beta +\gamma +\delta |=|(\alpha +\gamma )+(\beta +\delta )| \ |\alpha +\beta +\gamma +\delta |=|(-2)+(-4)| \ |\alpha +\beta +\gamma +\delta |=|-6| \ |\alpha +\beta +\gamma +\delta |=6 \end{gathered}$

Option (B) is correct.

Solutions · 1