Question

The points a(2,9), b(k,5),and c(5,5) are the vertices of a triangle abc right angled at b,find the value of k and the area of triangle

Answer 1

As the triangle is right angles at B... we have,

AB^2 + BC^2 = CA^2

= [ (2-k) ^2 + (9-5)^2] + [ (k-5)^2 + (5-5)^2 = (2-5)^2 + (9-5)^2

=4+k^2 - 4k+16+k^2+25-10k = 9+16

= 2k^2 - 14k+20 = 0

= 2(k^2-7k+10)=0

=k^2-7k+10 = 0

= k^2 - 5k-2k+10 = 0

= k(k-5) - 2(k-5) = 0

= (k-5)*(k-2)

k=5 k=2

but k cannot be 5 as the points will coincide to each other

;K = 2

area = 1/2 [2(5-5)+2(5-9)+5(9-5)]

= 1/2(0-8+20)

= 1/2*12

= 6 square units

hope it helps:-):-)