Question

If points \[P\left( {a,3} \right),Q\left( {6,1} \right),R\left( {8,2} \right)\] and \[S\left( {9,4} \right)\] are vertices of parallelogram \[PQRS\], find the value of \[a\].

Answer 1

Hint: In the given question, we have been given a parallelogram with the coordinates of its vertices. We have to find the value of an unknown of the vertices of the given parallelogram. First, we are going to draw the figure of the parallelogram. Then we are going to express any two opposite vectors. Then we are going to equate them using the distance formula and find the value of the unknown.

Complete step by step solution:

The given parallelogram \[PQRS\] has vertices \[P\left( {a,3} \right),Q\left( {6,1} \right),R\left( {8,2} \right)\] and \[S\left( {9,4} \right)\].
\[\overrightarrow {PQ} = \left( {6 - a, - 2} \right)\] and \[\overrightarrow {RS} = \left( {1,2} \right)\]
Now, since these two sides are opposite, they are equal in length, so
\[\left| {\overrightarrow {PQ} } \right| = \left| {\overrightarrow {RS} } \right|\]
Hence, \[\sqrt {{{\left( {6 - a} \right)}^2} + {{\left( { - 2} \right)}^2}} = \sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} \]
Squaring both sides and solving the expressions inside, we have,
\[36 + {a^2} - 12a + 4 = 1 + 4\]
Bringing all the terms to one side,
\[{a^2} - 12a + 35 = 0\]
Solving by splitting the middle term,
\[{a^2} - 7a - 5a + 35 = 0\]
Taking out commons,
\[a\left( {a - 7} \right) - 5\left( {a - 7} \right) = 0 \Rightarrow \left( {a - 5} \right)\left( {a - 7} \right) = 0\]
Hence, \[a = 5,7\]

Note: In this question, we had been given a parallelogram with the coordinates of its vertices. We had to find the value of an unknown used in the representation of one of the vertices of the given parallelogram. To do that, we first drew the figure of the parallelogram. Then we expressed two opposite vectors. Then we equated them using the distance formula and found the value of the unknown.