the points a (2, 9) B(k, 5) and C (5, 5) are the vertices of the triangle ABC right angled at B find the value of k and hence area of the triangle
Answers
Answer:
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
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thank you so much for your answer
