Question

11.If (2,3), (4,5) and (a,2) are the vertices of a right triangle, find possible

values of a​

Answer 1

Answer:

Step-by-step explanation:

lety a(2,3),b(4,5),c(a,2) be the given points

Vertices of the triangle are A(2 , 3), B(4 , 5) , C(a , 2).

it is given that <B = 90°

so, ∆ABC is a right angle triangle

therefore, by Pythagoras theorem AC² = AB² + BC².

Distance between two points =  root of x2-xi the whole square +y2-y1 the whole square

AB = √[(4 - 2)² + (5 - 3)²] = √(4 + 4) = 2√2

BC = √[(a - 4)² + (2 - 5)²] = √(a² + 16 - 8a + 9) = √(a² + 25 - 10a )

AC = √[(a - 2)² + (2 - 3)²] = √(a² + 4 - 4a + 1) = √(a² + 5 - 4a)

Now, AC² = AB² + BC²

(a² + 5 - 4a) = a² + 25 - 8a + 8

5 - 4a = 25 - 8a +8

8a - 4a = 25 - 5 + 8

4a = 28

a = 28/4

a = 7

I HOPE ITS HELP YOU DEAR,

THANKS