Question

Given that the four points A(1,2), B(2, -5), C(7,0) and D(p,q) are the vertices of the rhombus ABCD

Answer 1

Step-by-step explanation:

Three coordinates of a Rhombus is given to us. And the fourth coordinate is D(p,q) . Weneedtofind the value of p and q . We know that the diagonals of Rhombus bisect each other at right angles . So lets find the midpoint.

• Midpoint of AC :-

$\tt\to Midpoint_{(AC)}= \bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\tt\to Midpoint_{(AC)}= \bigg( \dfrac{1+7}{2},\dfrac{2+0}{2}\bigg)\\\\\tt\to Midpoint_{(AC)}= \bigg(\dfrac{8}{2},\dfrac{2}{2}\bigg)\\\\\tt\to \boxed{\orange{\tt Midpoint_{(AC)}= (4,1)}}$

_______________________________

• Midpoint of BD :-

$\tt\to Midpoint_{(BD)}= \bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\tt\to \boxed{\orange{\tt Midpoint_{(BD)}= \bigg( \dfrac{2+p}{2},\dfrac{q-5}{2}\bigg)}}$

_______________________________

Now since these two coordinates must be equal, therefore ,

$\tt\to \bigg( \dfrac{2+p}{2},\dfrac{q-5}{2}\bigg)= (4,1)\\\\\tt\to \dfrac{2+p}{2}=4 \qquad and \qquad \dfrac{q-5}{2}=1 \\\\\tt\to 2 + p = 8 \qquad and \qquad q-5 = 2 \\\\\tt\to p = 8 -2 \qquad and \qquad q = 2 +5 \\\\\tt\to\boxed{\orange{\tt (p,q) = (6,7) }}$