Question

The two adjacent sides of parallelogram are given by the following equations
\[y=0\text{ , }y=\sqrt{3}\left( x-1 \right)\]
One of the diagonals is given by the below equation. Find the other diagonal.
\[\sqrt{3}y=(x+1)\]
 (a) \[\sqrt{3}y=\left( x-1 \right)\text{ }\]
(b) \[y=\sqrt{3}\left( x+1 \right)\]
(c) \[y=-\sqrt{3}\left( x-1 \right)\]
(d) \[\sqrt{3}y=-\left( x+1 \right)\]

Answer 1

Hint: Try to find points of intersections of sides and diagonal lines to find the 3 vertices of parallelogram. Then find the intersection of diagonals.
In a parallelogram the diagonals intersect at mid-point of opposite vertices.
Use this condition to solve the remaining question.

Complete step-by-step answer:
First we need to find intersection points of the sides B.
Let us name the intersection point of two sides as B.
Sides are given by equations:
 \[\begin{align}
  & y=0\text{ }.....\left( 1 \right) \\
 & y=\sqrt{3}\left( x-1 \right)\text{ }.....\left( 2 \right) \\
\end{align}\]
By substituting equation (2) in equation (1), we get:
\[\sqrt{3}\left( x-1 \right)=0\]
By simplifying, we get:
x – 1 = 0
x = 1.
By equation (1) we can say y = 0.
Thus the intersection point B is given by:
B = (1, 0)
Next we need to find the intersection point of one of the sides and diagonal.
Let us name the intersection point of one of the sides and diagonal as A.
Their are given by equations:
 \[\begin{align}
  & y=0\text{ }.....\left( 3 \right) \\
 & \sqrt{3}y=(x+1)\text{ }.....\left( 4 \right) \\
\end{align}\]
By substituting equation (4) in equation (3), we get:
\[\left( x+1 \right)=0\times \sqrt{3}=0\]
By simplifying, we get:
x + 1 = 0
x = -1.
By equation (1) we can say y = 0.
Thus the intersection point A is given by:
A = (-1, 0)
Next we need to find the intersection point of the other side and diagonal.
Let us name the intersection point of the other side and diagonal as C.
Their are given by equations:
 \[\begin{align}
  & y=\sqrt{3}\left( x-1 \right).....\left( 5 \right) \\
 & \sqrt{3}y=(x+1)\text{ }.....\left( 6 \right) \\
\end{align}\]
By substituting equation (4) in equation (3), we get:
\[\left( x+1 \right)=\sqrt{3}\left( x-1 \right)\times \sqrt{3}\]
By simplifying, we get:
x + 1 = 3x-3
By simplifying, we get:
2x=4
x = 2.
By substituting the value of x in (5), we get:
\[y=\sqrt{3}\]
Thus the intersection point C is given by:
\[C=\left( 2,\sqrt{3} \right)\]

By finding mid-point of AC we get the intersection point of diagonals.
Co-ordinates of mid-point of AC = average of the co-ordinates of A, C.
Let the intersection of diagonals be O.
\[O=\left( \dfrac{-1+2}{2},\dfrac{0+\sqrt{3}}{2} \right)\]
\[O=\left( \dfrac{1}{2},\dfrac{\sqrt{3}}{2} \right)\]
As the intersection point of diagonals also lies on other diagonal and it passes through B.Now the equation OB is the equation of other diagonal.If a line passes through 2 points (a, b) and (c, d) then its equation is given by
\[\left( y-b \right)=\dfrac{d-b}{c-a}\left( x-a \right)\]
Here we need line equation passing through B and O
\[B\text{ }\left( 1,0 \right)\text{ and O }\left( \dfrac{1}{2},\dfrac{\sqrt{3}}{2} \right)\]
By substituting these points in the condition, we get:
\[\left( y-0 \right)=\dfrac{\dfrac{\sqrt{3}}{2}-0}{\dfrac{1}{2}-1}\left( x-1 \right)\]
By simplifying, we get:
\[y=-\sqrt{3}\left( x-1 \right)\]
This is the equation of another diagonal.
So option (c) is correct.
Note: Alternate methods –
 You can use elimination method to find intersection points of two given lines.
For calculating mid-point you can assume (x, y) and then apply the distance formula and equate the distance from both points to get x, y values. By this also you will get the same result.