If (2 1) (-2 5) are two opposite vertices of a square then the area of the square is

Question 14 Coordinate Geometry Exercise 19.4

Answer:

Let A(3,4), B(x,y), C(1,-1), and D be the vertices of the square.

For a square, diagonal = √2×side

\begin{aligned} &\therefore \mathrm{AC}=\sqrt{2 \times \mathrm{AB}}\\ &\left.\left.\therefore \sqrt{[}(1-3)^{2}+(-1-4)^{2}\right]=\sqrt{2} \times \sqrt{[}(x-3)^{2}+(y-4)^{2}\right]\\ &\text { Squaring both sides, }\\ &\therefore\left[(1-3)^{2}+(-1-4)^{2}\right]=2 \times\left[(\mathrm{x}-3)^{2}+(\mathrm{y}-4)^{2}\right]\\ &-2^{2}+(-5)^{2}=2\left[(x-3)^{2}+(y-4)^{2}\right]\\ &4+25=2\left[(x-3)^{2}+(y-4)^{2}\right]\\ &29=2\left[(x-3)^{2}+(y-4)^{2}\right]\\ &\left[(x-3)^{2}+(y-4)^{2}\right]=29 / 2 \quad . .(i) \end{aligned}

AB = BC [sides of a square]

When x = 9/2 , y = (23-4×9/2)/10 = (23-18)/10 = 5/10 = ½

When x = -1/2 , y = (23-4×-1/2)/10 = (23+2)/10 = 25/10 =5/2

So the coordinates of the remaining vertices of the square are (9/2, 1,2) and (-1/2, 5/2)

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It 2, 2 and 5,2 are opposite vertices of a square, then the length of the side of the square is

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