Question

M∠B = 90 in the triangle whose vertices are A(2,3), B(4,5) and C(a,2). Find a.

Answer 1

HELLO DEAR,

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 = $\sf{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$

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

Answer 2

Dear Student,

Solution:

As it is a right angle triangle.

So we can apply Pythagoras Theorem : (Hypotenuse) ^2 = (Base)^2 + (Perpendicular)^2

(AC)^2 = (BC)^2 + (AB)^2

Now calculate the distance between given points and put into the above equation.

AC = √ (2-a)^2 +(3-2)^2

$AC^{2} = (2-a)^{2} +1$

AB =$\sqrt{(2-4)^{2}+(3-5)^{2}} \\ =\sqrt{(8)}$

$AB^{2} = 8$

BC =$\sqrt{(4-a)^{2}+(5-2)^{2}} \\ =\sqrt{(4-a)^{2}+9}$

$BC^{2} = (4-a)^{2} +9$

$(2-a)^{2} +1$ = $(4-a)^{2} +9$ + 8

$(4+a^{2} -4a +1) = ( 16 +a^{2} -8a +9 +8)\\ \\ -4a+8a = 33-5\\ \\ 4a = 28\\ \\ a = \frac{28}{4}$
a = 7
Hope it helps you.