If two opposite vertices of a square are (5,4) and (1,-6), find the co ordinates of its remaining two vertices - Maths - Coordinate Geometry

Question

If two opposite vertices of a square are (5,4) and (1,-6), find
the co ordinates of its remaining two vertices

Harleen Birdi · Accepted Answer

Let A (5,4) and C (1, â€“6) Let B (x, y) $We have AB = BC \Rightarrow {AB}^{2} = {BC}^{2} \Rightarrow (x - 5)^{2} + {(y - 4)}^{2} = {(x - 1)}^{2} + {(y + 6)}^{2} \Rightarrow x^{2} + 25 - 10 x + y^{2} + 16 - 8 y = x^{2} + 1 - 2 x + y^{2} + 36 + 12 y \Rightarrow - 10 x - 8 y + 25 + 16 = - 2 x + 12 y + 1 + 36 \Rightarrow - 10 x + 2 x - 8 y - 12 y = 37 - 41 \Rightarrow - 8 x - 20 y = - 4 \Rightarrow 2 x + 5 y = 1 \Rightarrow y = \frac{1 - 2 x}{5} (1)$ In right âˆ†ABC, by Pythagoras theorem, ${AC}^{2} = {{AB}^{2} + BC}^{2} \Rightarrow {AC}^{2} = 2 {AB}^{2} \Rightarrow 116 = 2 [{(x - 5)}^{2} + {(y - 4)}^{2}] \Rightarrow 58 = x^{2} + y^{2} - 10 x - 8 y + 25 + 16 \Rightarrow 17 = x^{2} + y^{2} - 10 x - 8 y x^{2} + {(\frac{1 - 2 x}{5})}^{2} - 10 x - 8 \times (\frac{1 - 2 x}{5}) = 17 [U \sin g (1)] \Rightarrow x^{2} + \frac{{(1 - 2 x)}^{2}}{25} - 10 x - \frac{8}{5} (1 - 2 x) = 17 \Rightarrow 25 x^{2} + 1 + 4 x^{2} - 4 x - 250 x - 40 + 80 x = 17 \Rightarrow 29 x^{2} - 174 x - 56 = 0 D = (- 174)^{2} - 4 (29) (- 56) = 30276 + 6496 = 36772 ∴ x = \frac{174 \pm \sqrt{36772}}{2 \times 29} = 6 3, - 0 3 ∴ y = \frac{1 - 2 x}{5} = - 2 32, 0 32$ Thus, points are (6 3, â€“2 32) & (â€“0 3, 0 32)

If two opposite vertices of a square are (5,4) and (1,-6), find the co ordinates of its remaining two vertices.