how can i find a missing coordinate of one vertex of a triangle when given the other coordinates

Question

the points A,B and C have co-ordinates (2,1), (b,3) and (5,5) respectively, where b>3 and angle ABC = 90 degrees, find the value of the x co-ordinate labelled b.

Answer 1

Given coordinates are A(2,1),B(b,3),C(5,5)

Now find the distances of each side.

AC = √[(5-2)^2+(5-1)^2]

= √(9+16) = √25 = 5

AB = √[(b-2)^2+(3-1)^2 = √[(b-2)^2+4]

BC = √[(5-b)^2+(5-3)^2 = √[(5-b)^2+4]

ABC is right angle traingle.

5^2 = (√[(b-2)^2+4])^2+(√[(5-b)^2+4])^2

25 = (b-2)^2+4+(5-b)^2+4

25 = b^2+4-4b+4+25+b^2-10b+4

Separate variables and constants.

25-25-4-4-4 = 2b^2-14b

-12 = 2b^2-14b

Add 12 to each side.

2b^2-14b+12 = 0

Divide to each side by 2.

b^2-7b+6 = 0

b^2-6b-b+6 = 0

b(b-6)-1(b-6) = 0

(b-6)(b-1) = 0

(b-6) = 0 and b-1 = 0

b = 6,1

Given b > 3

Missing coordinate  is (6,3).