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.
<u>•</u><u> </u><u>Midpoint</u><u> </u><u>of</u><u> </u><u>AC</u><u> </u><u>:</u><u>-</u><u> </u>
![\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)}}](https://tex.z-dn.net/?f=%5Ctt%5Cto%20Midpoint_%7B%28AC%29%7D%3D%20%5Cbigg%28%5Cdfrac%7Bx_1%2Bx_2%7D%7B2%7D%2C%5Cdfrac%7By_1%2By_2%7D%7B2%7D%5Cbigg%29%5C%5C%5C%5C%5Ctt%5Cto%20Midpoint_%7B%28AC%29%7D%3D%20%5Cbigg%28%20%5Cdfrac%7B1%2B7%7D%7B2%7D%2C%5Cdfrac%7B2%2B0%7D%7B2%7D%5Cbigg%29%5C%5C%5C%5C%5Ctt%5Cto%20Midpoint_%7B%28AC%29%7D%3D%20%5Cbigg%28%5Cdfrac%7B8%7D%7B2%7D%2C%5Cdfrac%7B2%7D%7B2%7D%5Cbigg%29%5C%5C%5C%5C%5Ctt%5Cto%20%5Cboxed%7B%5Corange%7B%5Ctt%20Midpoint_%7B%28AC%29%7D%3D%20%284%2C1%29%7D%7D)
<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>
<u>• Midpoint of </u><u>BD</u><u> :- </u>
![\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)}}](https://tex.z-dn.net/?f=%5Ctt%5Cto%20Midpoint_%7B%28BD%29%7D%3D%20%5Cbigg%28%5Cdfrac%7Bx_1%2Bx_2%7D%7B2%7D%2C%5Cdfrac%7By_1%2By_2%7D%7B2%7D%5Cbigg%29%5C%5C%5C%5C%5Ctt%5Cto%20%5Cboxed%7B%5Corange%7B%5Ctt%20Midpoint_%7B%28BD%29%7D%3D%20%5Cbigg%28%20%5Cdfrac%7B2%2Bp%7D%7B2%7D%2C%5Cdfrac%7Bq-5%7D%7B2%7D%5Cbigg%29%7D%7D)
<u>_______________________________</u>
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) }}](https://tex.z-dn.net/?f=%5Ctt%5Cto%20%5Cbigg%28%20%20%5Cdfrac%7B2%2Bp%7D%7B2%7D%2C%5Cdfrac%7Bq-5%7D%7B2%7D%5Cbigg%29%3D%20%284%2C1%29%5C%5C%5C%5C%5Ctt%5Cto%20%5Cdfrac%7B2%2Bp%7D%7B2%7D%3D4%20%5Cqquad%20and%20%5Cqquad%20%5Cdfrac%7Bq-5%7D%7B2%7D%3D1%20%5C%5C%5C%5C%5Ctt%5Cto%202%20%2B%20p%20%3D%208%20%5Cqquad%20and%20%5Cqquad%20q-5%20%3D%202%20%5C%5C%5C%5C%5Ctt%5Cto%20p%20%3D%208%20-2%20%5Cqquad%20and%20%5Cqquad%20q%20%3D%202%20%2B5%20%5C%5C%5C%5C%5Ctt%5Cto%5Cboxed%7B%5Corange%7B%5Ctt%20%28p%2Cq%29%20%3D%20%286%2C7%29%20%7D%7D)