Question

The equations to a pair of opposite sides of a parallelogram are $x^2-5x+6=0$ and $y^2-6y+5=0$. The equations to its diagonals are

• A. $x+4y=13$, $y=4x-7$
• B. $4x+y=13$, $4y=x-7$
• C. $4x+y=13$, $y=4x-7$
• D. $y-4x=13$, $y+4x=7$

Answer 1

The correct option is C

$4x+y=13$, $y=4x-7$

Explanation for the correct option

Step 1: Solve for the vertices of the given parallelogram

The lines describing the pair of opposite sides of the parallelogram are,
$x^2-5x+6=0$ and $y^2-6y+5=0$

Solving the above two equations,
$\begin{array}{cccc}& x^2-5x+6=0& & y^2-6y+5=0\\ \Rightarrow & x^2-2x-3x+6=0& \Rightarrow & y^2-1y-5y+5=0\\ \Rightarrow & x\left(x-2\right)-3\left(x-2\right)=0& \Rightarrow & y\left(y-1\right)-5\left(y-1\right)=0\\ \Rightarrow & \left(x-2\right)\left(x-3\right)=0& \Rightarrow & \left(y-1\right)\left(y-5\right)=0\\ \Rightarrow & x=2,3& \Rightarrow & y=1,5\end{array}$

Therefore, the vertices of the parallelogram are $A\left(2,1\right)$, $B\left(3,1\right)$, $C\left(3,5\right)$, and $D\left(2,5\right)$

Step 2: Solve for the equations of diagonals

Diagonals are lines that connect opposite vertices of a parallelogram.
The slope of the diagonal can be found by the two-point form and the equation of the line can be found by the point-slope form.

For the points $\left(x_1,y_1\right),\left(x_2,y_2\right)$the two-point form is given as,
$\frac{y_2-y_1}{x_2-x_1}=m$; where $m$ is the slope.

The point-slope form is given as $y-y_1=m\left(x-x_1\right)$

The equation of the diagonal connecting $A$ and $C$ is,
$\begin{array}{cc}& y-1=\left(\frac{5-1}{3-2}\right)\left(x-2\right)\\ \Rightarrow & y-1=4x-8\\ \Rightarrow & y=4x-7\end{array}$

The equation of the diagonal connecting $B$ and $D$ is,
$\begin{array}{cc}& y-1=\left(\frac{5-1}{2-3}\right)\left(x-3\right)\\ \Rightarrow & y-1=-4x+12\\ \Rightarrow & 4x+y=13\end{array}$

Therefore, the lines describing the diagonals of the given parallelogram are $4x+y=13$ and $y=4x-7$.

Hence, option(C) i.e. $4x+y=13$, $y=4x-7$ is correct.