Question

If $A(3,6)$ and $C(-1,2)$ are two vertices of a rhombus $ABCD$, then find the equation of straight line that lies along the diagonal $BD$.

• A. $x+y-5=0$
• B. $x+y-6=0$
• C. $x+y-7=0$
• D. None of these

Answer 1

The correct option is A $x+y-5=0$

Given by, $A(3,6)$ and $C(-1,2)$ are two vertices of a rhombus $ABCD$, then find the equation of straight line that lies along the diagonal $BD$

Since, in any rhombus two diagonals bisect each other and they are perpendicular to each other.

$Mid-pointofAC=Mid-pointofBD$

$mid-point=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$=(\frac{3-1}{2},\frac{6+2}{2})$

$=(\frac{2}{2},\frac{8}{2})$

$=(1,4)$

$SlopeofAC=\frac{y_2-y_1}{x_2-x_1}$

$=\frac{2-6}{-1-3}$

$=\frac{-4}{-4}=1$

So, $SlopeofBD=-\frac{1}{1}$

$=-1$

Equation of $BD$

$y-y_1=m(x-x_1)$

$y-4=-1(x-1)$

$y-4=-x+1$

$x+y-4-1=0$

$x+y-5=0$

Hence, this is the answer.