Question

find the values of a for which the points are a(2,9),(a,5)andc(5,5)from vertices of a right angle triangle right angled at b

Answer 1

1 ) finding distance between A & B:
using sectional formula,
AB = √[(2-a)²+(9+5)²] = √(4+a²-4a+16)
=√a²-4a+20

2) distance BC using sectional formula:
BC = √[(5-a)²+(5-5)²] = √(25+a²-10a+0²)
=√a²-10a+25

3) AC by sectional formula:
AC = √[(2-5)²+(9-5)²] = √(9+16) = √25 = 5

by Pythagoras theorem:
AC² = AB² + BC²
5² = (a²-10a+25) + (a²-4a+20)
25 = 2a² - 14a + 45
2a² - 14a + 20 = 0
a² - 7a + 10 = 0
a² - 5a - 2a = 0
a(a-5)-2(a-5) = 0
(a-2)(a-5) = 0
a = 2, 5

therefore a can be 2 or 5