Question

the ends of a diagonal of a square have coordinates (-2,p) and (p,2).Find p, if the area of the square is 40 square units.

plz answer this question quickly!!!! plz!​

Answer 1

Solution

Let the coordinates of diagonal be A(-2, p) and B(p, 2)

First find the length of the diagonal in terms p using Distance formula

$d = \sqrt{ {( x_2 - x_1)}^{2} + {( y_2 - y_1) }^{2} }$

A(-2, p) B(p, 2)

$\text{here } \\ \quad \rightarrow x_1 = - 2 \\ \quad \rightarrow x_2 = p \\ \quad \rightarrow y_1 = p \\ \quad \rightarrow y_2 = 2$

Substituting the values in formula

$\implies AB = \sqrt{ { \{( p - ( - 2) \}}^{2} + {(2 - p) }^{2} }$

$\implies AB = \sqrt{ { ( p + 2) }^{2} + {2}^{2} + {p}^{2} - 2(2)(p) }$

[ Because (a - b)² = a² + b² - 2ab ]

$\implies AB = \sqrt{ {p}^{2} + {2}^{2} + 2(p)(2) + {2}^{2} + {p}^{2} - 2(2)(p) }$

$\implies AB = \sqrt{2 {p}^{2} + 4 + 4p + 4 -4p }$

$\implies AB = \sqrt{2 {p}^{2} + 8}$

Now, Length of the diaganal of the square AB = √(2p² + 8)

Relation between area and diagonal of the square

$\text{Area of the square} = \dfrac{(Length \ of \ the \ diagonal)^{2}} {2}$

Given, Area of the square = 40 square units

$\implies \dfrac{ \big(\sqrt{2 {p}^{2} + 8} \big)^{2} }{2} = 40$

$\implies 2 {p}^{2} + 8 = 40 \times 2$

$\implies 2 {p}^{2} + 8 = 80$

$\implies 2 {p}^{2} = 80 - 8$

$\implies 2 {p}^{2} = 72$

$\implies {p}^{2} = \dfrac{72}{2}$

$\implies {p}^{2} = 36$

$\implies p = \pm \sqrt{36}$

$\implies p = \pm 6$

Hence, the value of p is 6 or - 6.