 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 equation of diagonal can be:A. $x+4y=13$B. $4x+y=13$C. $y=4x-7$D. $y=4x+7$

Hint: Both equations are quadratic equations. Solve them and get the values of x and y. Parallelogram has 4 vertices. Using x and y from the 4 vertices, find the equation of the diagonal formed in the parallelogram.

We have been given two equations which are a pair of opposite sides of a parallelogram.

\begin{align} & {{x}^{2}}-5x+6=0......(1) \\ & {{y}^{2}}-6y+5=0......(2) \\ \end{align}

Now let us solve both the equations and get the values of x and y.

Let us first solve equation (1).

${{x}^{2}}-5x+6=0$ is similar to the general equation $a{{x}^{2}}+bx+c=0$.

Now let us compare both the equations and we get the following values.

a = 1, b = -5, c = 6.

Now let us substitute these values in the quadratic formula.

\begin{align} & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-(-5)\pm \sqrt{{{(-5)}^{2}}-4\times 1\times 6}}{2\times 1} \\ & =\dfrac{5\pm \sqrt{25-24}}{2}=\dfrac{5\pm 1}{2} \\ \end{align}

Thus we got $x=\dfrac{5+1}{2}$ and $x=\dfrac{5-1}{2}$.

$x={}^{6}/{}_{2}=3,x={}^{4}/{}_{2}=2$.

Thus we got the values of x as 2, 3 by solving ${{x}^{2}}-5x+6=0$.

$\therefore x=2,x=3.......(3)$

Now let us solve the equation (2).

${{y}^{2}}-6y+5=0$ is similar to the general equation $a{{y}^{2}}+by+c=0$.

Now let us compare both the equations and get the values as,

a = 1, b = -6, c = 5.

Now let’s substitute these values in the quadratic formula.

\begin{align} & y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-(-6)\pm \sqrt{{{(-6)}^{2}}-4\times 1\times 5}}{2\times 1} \\ & =\dfrac{6\pm \sqrt{36-20}}{2}=\dfrac{6\pm \sqrt{16}}{2}=\dfrac{6\pm 4}{2} \\ \end{align}

Thus we got $y=\dfrac{6+4}{2}$ and $y=\dfrac{6-4}{2}$.

$y={}^{10}/{}_{2}=5,y={}^{2}/{}_{2}=1.$

Thus we got the values of y as 1, 5 by solving ${{y}^{2}}-6y+5=0$.

$\therefore y=1,y=5.......(4)$

Now let us compare equation (3) and (4).

x = 2, x = 3 and y = 1, y = 5.

We know that a parallelogram has 4 vertices. Thus let us consider the 4 vertices of a parallelogram ABCD. Frame the vertices with the help of values x and y.

Thus the vertices are A (2, 1), B (3, 1), C (3, 5), D (2, 5).

Thus from the figure, you can see the parallelogram ABCD with vertices A (2, 1), B (3, 1), C (3, 5), D (2, 5).

We were asked to find the equation of a diagonal. So let us join AC.

AC is a diagonal of parallelogram ABCD. The equation of the line formed by 2 points A and C can be written as,

$\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}$

here, $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 2,1 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,5 \right)$.

$\therefore$Equation of line AC, i.e.,

$\dfrac{y-1}{5-1}=\dfrac{x-2}{3-2}\Rightarrow \dfrac{y-1}{4}=\dfrac{x-2}{1}$

Let us cross multiply and simplify the above expression.

\begin{align} & y-1=4(x-2) \\ & y-1=4x-8 \\ & \Rightarrow 4x-y-8+1=0 \\ & 4x-y-7=0 \\ & \therefore y=4x-7 \\ \end{align}

Thus we got the equation of the parallelogram as $y=4x-7$.

Option C is the correct answer.

Note: The most important part of working with lines is understanding the relation between the form. Thus we can use the equation for finding the equation of line when 2 points are given. Pick either point and we can get the equation of the line.