Question

The points A(2,9) B(k,5)and C(5,5) are the vertices of triangle ABC right angled at B.find the values of K and hence the area of triangle.

Answer 1

Given,
Vertices  = A(2,9) , B(k,5) and C(5,5)
Angle ABC= 90 degree
By Distance formula=√ [ (X2-X1)² + (Y2-Y1)²)]
AB = √[{5-9}²+ {k-2}²]
=√[ 16 + K² - 4k + 4]
=√(k² - 4k + 20)units 
AC = √[{5-9}²+{5-2}²]
=√[ 16 + 9]
= 5 units
BC = √[{5-5]²+{5-k}²]
=√[ 0+25-10k+k²]
=√( k²-10k+25) units
By pythagoras theorem(AC)² = (AB)² + (BC)²
25=k²-4k+20 + k²-10k+25
2k² -14k + 20=- 7k + 10 = k^2 - (5k+2k) +10 = 0 
(By Middle term factorization)
k² - 5k - 2k + 10 =k(k-5)-2(k-5)
=(k-5)(k-2)
 k-5=0                                                     k - 2 = 0 
k=5                                                       
k = 2 
if we put the value of k in BC, we get 0.
Hence k hould be taken as 2 
BC=3 units
AB=4 units
Area  = 1/2 * AB * BC(from 1/2 * base * height)
= 1/2 * 4 * 3=6 (unit) ²

Answer 2

Answer:

Step-by-step explanation: