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 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

First, we need to know about the concept of the parallelogram and how to find its equations diagonal using the equations to pair opposite sides. The two equations are given above.
A parallelogram is the simple quadrilateral with the given two sides are the parallel sides, and opposite sides of the parallelogram are exactly equal in length and also the opposite angles of the parallelogram are equal measure.
Formula used:
The slope of the two vertices $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Equation of the diagonal points $y - {y_1} = m(x - {x_1})$ (where $x,y$ are fixed points)

Complete step by step answer:
Since we have two equations from the given, which are ${x^2} - 5x + 6 = 0$ and ${y^2} - 6y + 5 = 0$. We will find its zeroes of the polynomial using the quadratic method or simplification method.
Take the first equation ${x^2} - 5x + 6 = 0$. Now by the quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$where $a = 1,b = - 5,c = 6$. Thus, substituting the values in the above equation, we get $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{5 \pm \sqrt {25 - 4 \times 6} }}{2}$ and further solving we get $x = \dfrac{{5 \pm \sqrt {25 - 4 \times 6} }}{2} \Rightarrow \dfrac{{5 \pm 1}}{2} \Rightarrow 2,3$
Hence the values are $x = 2,3$
Similarly, we can also find the values of the equation ${y^2} - 6y + 5 = 0$
Now by the quadratic formula, we have $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$where $a = 1,b = - 6,c = 5$
Thus, we get $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{6 \pm \sqrt {36 - 4 \times 5} }}{2} \Rightarrow y = \dfrac{{6 \pm 4}}{2} \Rightarrow 1,5$
Hence, we get the values $y = 1,5$
Now we have the two equations points, let us make the vertices of the parallelogram with the four points, each point will make a pair of points with the corresponding values;
Thus, we have $x = 2,3$and $y = 1,5$. By acting the vertices with each other we get $A(2,1),B(3,1),C(3,5),D(2,5)$
Now we are going to find the slope; let the slope of $AC = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ where $A(2,1),C(3,5)$
Then we get $AC = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \Rightarrow m = \dfrac{{5 - 1}}{{3 - 2}} \Rightarrow m = 4$
Using the same method to find the slope for $BD = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \Rightarrow m = \dfrac{{5 - 1}}{{2 - 3}} \Rightarrow m = - 4$ where $B(3,1),D(2,5)$ [find slope points will be only applicable for the diagonal entries]
Hence to find the equation for AC, we have $y - {y_1} = m(x - {x_1}) \Rightarrow y - 1 = 4(x - 2)$ where $m = 4$
Further solving we get, $y - 1 = 4(x - 2) \Rightarrow 4x - y - 7 = 0 \Rightarrow y = 4x - 7$
Similar to finding the equation of BD, we have $y - {y_1} = m(x - {x_1}) \Rightarrow y - 1 = - 4(x - 3)$where $m = - 4$
Further solving we get $y - 1 = - 4(x - 3) \Rightarrow y - 1 + 4x - 12 = 0 \Rightarrow 4x + y = 13$
Therefore, the diagonal equations are $4x + y = 13,y = 4x - 7$

So, the correct answer is “Option C”.

Note: We solved the two-quadratic equation into the zeroes of the polynomial (finding its values) and then we will make the vertices with those points.
After getting the points we will make them into the vertices for the parallelogram.
Thus, we find its slope only for the diagonal which is AC and BD.
Hence, we solve using the equation of the diagonal formula with the slope and we get the required equations.