Question

if p(2,4),q(0,3),r(3,6),and s(a,b) are vertices of a parallelogram then the value of a+b is? ​

Answer 1

The value of a + b is 12.

• Given vertices of parallelogram are

        P(2, 4), Q(0, 3), R(3, 6), S(a, b)

• We know that, diagonals of a parallelogram bisect each other.
• Midpoint of PR = Midpoint of QS

        $(\frac{x_{1}+x_{3}}{2},\frac{y_{1}+y_{3}}{2})=(\frac{x_{2}+x_{4}}{2},\frac{y_{2}+y_{4}}{2})$

        $(\frac{2+3}{2},\frac{4+6}{2})=(\frac{0+a}{2},\frac{3+b}{2})$

         $(\frac{5}{2},\frac{10}{2})=(\frac{a}{2},\frac{3+b}{2})$

• Equating x and y coordinates, we get

           $\frac{5}{2}=\frac{a}{2}, \frac{10}{2}=\frac{3+b}{2}$ ⇒ b = 10 - 3

          ⇒ a = 5, b = 7

• Therefore, the value of a + b = 5 + 7 = 12.