Question

abc is an isosceles triangle.if the coordinates of the base are b(1,3)and c(-2,7) then find the coordinates of the vertex

Answer 1

Given that ABC is an isosceles triangle.
The two vertices are B (1,3) = (x₂,y₂) and C (-2,7) = (x₂,y₂)
We have to find the co-ordinates of the vertex A
Since the triangle is an isosceles triangle So,
AB = AC   ............. (1)
And also vertex A lies on y-axis.
Let A be (0,y) = (x₁,y₁)
Equation (1) ⇒ AB = AC
√ [(x₂ - x₁)² + (y₂ - y₁)²] = √ [(x₂ - x₁)² + (y₂ - y₁)²]
(1 - 0)² + (3 - y)² = (- 2 - 0)² + (7 - y)²
1 + (3 - y)² = 2 + (7 - y)²
1 + 9 + y² - 6y = 2 + 49 + y² -14y
10 - 6y = 51 - 14y
14y - 6y = 51 - 10
9y = 41
y = 41/9
y = 4.6
So, co-ordinates of the vertex A are A (0,4.6)
This is the required answer.
Thanks.

Answer 2

Answer:Given that ABC is an isosceles triangle.

The two vertices are B (1,3) = (x₂,y₂) and C (-2,7) = (x₂,y₂)

We have to find the co-ordinates of the vertex A

Since the triangle is an isosceles triangle So,

AB = AC   ............. (1)

And also vertex A lies on y-axis.

Let A be (0,y) = (x₁,y₁)

Equation (1) ⇒ AB = AC

√ [(x₂ - x₁)² + (y₂ - y₁)²] = √ [(x₂ - x₁)² + (y₂ - y₁)²]

(1 - 0)² + (3 - y)² = (- 2 - 0)² + (7 - y)²

1 + (3 - y)² = 2 + (7 - y)²

1 + 9 + y² - 6y = 2 + 49 + y² -14y

10 - 6y = 51 - 14y

14y - 6y = 51 - 10

9y = 41

y = 41/9

y = 4.6

So, co-ordinates of the vertex A are A (0,4.6)

This is the required answer.

Thanks.

Step-by-step explanation: