Question

the area of the given triangle. 2. The two opposite vertices of a square are (-1,2) and (3.2). Find the
coordinates of the other two vertices.
3. Find the centre of a circle passing through the points (6.-6) (3.-7) and (3,3).​

Answer 1

Let ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinates of vertex B be (x, y).

AB = BC (As ABCD is a square)

AB2 = BC2

[x – (–1)] 2 + (y – 2)2 = (x – 3)2 + (y – 2)2

(x + 1)2 = (x – 3)2

x 2 + 2x + 1 = x 2 – 6x + 9

2x + 6x = 9 – 1

8x = 8

x = 1

In ΔABC, we have:

AB2 + BC2 = AC2 (Pythagoras theorem)

2AB2 = AC2 (Since, AB = BC)

2[(x – (–1))2 + (y – 2)2] = (3 – (–1))2 + (2 – 2)2

2[(x + 1)2 + (y – 2)2] = (4)2 + (0)2

2[(1 + 1)2 + (y – 2)2] = 16 ( x = 1)

2[ 4 + (y – 2)2] = 16

8 + 2 (y – 2)2 = 16

2 (y – 2)2 = 16 – 8 = 8

(y – 2)2 = 4

y – 2 = ± 2

y – 2 = 2 or y – 2 = –2

y = 4 or y = 0

Thus, the other two vertices of the square ABCD are (1, 4) and (1, 0).

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3)Let A (6, –6), B (3, –7) and C (3, 3) be the points on the circle and the centre of the circle be P (h, k). ∴ PA = PB = PC (Radius of the circle) Thus, the centre of the circle is (3, –2).Read more on Sarthaks.com - https://www.sarthaks.com/2447/find-the-centre-of-a-circle-passing-through-the-points-6-6-3-7-and-3-3

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