Question

The diagonals of a parallelogram \[PQRS\] are along the lines \[x+3y=4\]and \[6x-2y=7\]. Then \[PQRS\] must be a
A. Rectangle
B. Square
C. Cyclic quadrilateral
D. Rhombus

Answer 1

Hint: To get the answer of this question you should have the basic knowledge of parallelograms and their properties. Firstly find out the slope of both the lines. After finding the slope, figure out the relation of the two slopes. The relation will give us the idea of the desired parallelogram.

Complete step by step answer:
Parallelogram is the closed figure which contains four sides and their opposite sides are parallel and equal. There are three major types of parallelogram and they are square, rectangle and rhombus.
Let us discuss all the parallelograms and their properties one by one.
Square: Square is the parallelogram in which all the sides are equal and opposite sides are parallel. All the interior angles are equal to \[{{90}^{0}}\]. And the diagonals of the square bisect each other at \[{{90}^{0}}\].
Rectangle: A parallelogram in which opposite sides are equal and parallel to each other is known as rectangle. Their all interior angles are equal to \[{{90}^{0}}\]. Diagonals do not bisect each other at \[{{90}^{0}}\]
Rhombus: Rhombus is a parallelogram in which all the four sides are equal and opposite sides are parallel. Their interior angles are not equal to \[{{90}^{0}}\]. The diagonals bisect each other and divide each diagonal into two equal parts.
The given equations of the diagonals of parallelogram are
\[x+3y=4\]
\[6x-2y=7\]
If we are supposed to find the slope of the given lines, the formula used is
\[y=mx+c\]\[........(1)\]
Where,
\[m\] represents the slope of the line
Now to find the slope of the given lines compare from equation \[(1)\]. For comparing first we have to convert into the slope intercept form
\[x+3y=4\]
\[\begin{align}
  & \Rightarrow 3y=-x+4 \\
 & \Rightarrow y=-\dfrac{1}{3}x+\dfrac{4}{3} \\
\end{align}\]
Comparing from equation \[(1)\], we can say that
\[{{m}_{1}}=-\dfrac{1}{3}\]
Where \[{{m}_{1}}\]is the slope of line \[x+3y=4\]
Similarly, compare the equation of line \[6x-2y=7\]from equation \[(1)\]
\[\begin{align}
  & \Rightarrow 2y=6x-7 \\
 & \Rightarrow y=3x-\dfrac{7}{2} \\
\end{align}\]
We can say that, \[{{m}_{2}}=3\]
Where, \[{{m}_{2}}\]is the slope of the line \[6x-2y=7\]
Now find \[{{m}_{1}}\times {{m}_{2}}\], we get
\[\begin{align}
  & \Rightarrow -\dfrac{1}{3}\times 3 \\
 & \Rightarrow -1 \\
\end{align}\]
If \[{{m}_{1}}{{m}_{2}}=-1\]then we can say that both the lines are perpendicular.
This means that both the diagonals of the parallelogram bisect each other \[{{90}^{0}}\].
Hence by the properties of the parallelogram we can conclude that option \[(B)\]and \[(D)\]both are correct. \[PQRS\]must be a square or rhombus.

So, the correct answer is “Option B and D”.

Note: All parallelograms are quadrilateral but reverse is not true i.e. all quadrilaterals are not parallelograms. In a parallelogram opposite angles are equal. “The sum of the squares of the sides is equal to the sum of the squares of the diagonals”. This law is known as parallelogram law.