Question

the vertices of the base of an isosceles triangle are (-1,-2) and (1,4). it the third vertex lies on the line 4x +3y = 12, find the area of the triangle

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given:

In an isosceles triangle, the vertices of the base of the triangle are (-1, -2) and (1, 4).

the third vertex lies on the line 4x + 3y = 12

To Find

area of the isosceles triangle

Solution:

Let the isosceles triangle be ABC, where the base of B and C are (-1, -2) and (1, 4) respectively, and A(x, y).

Then, 4x + 3y = 12 ..(1)

and AB = AC

      ⇒ (x + 1)² + (y + 1)² = (x - 1)² + (y - 1)²

      ⇒ (2x + 1) + (4y + 1) = (-2x + 1) - (8y + 16)

      ⇒ 4x + 12y = 12

      ⇒ x + 3y = 3 ..(2)

Solving (1) & (2),

       3x = 9

       ⇒ x = 3

and y = 0

Hence, A is the point (3, 0)

As, the altitude of ΔABC is drawn from A vertex forming an angle bisector and median to the base BC.

Now, we have

       (AC)² = (AD)² + (CD)²

     ⇒ (AD)² = (AC)² - $\frac{1}{4}$(BC)²

     ⇒ (AD)² = {(3+1)²+ (0+2)²} - $\frac{1}{4}${(1+1)² + (4, 2)²}

     ⇒ (AD)² = 16 + 4 - $\frac{1}{4}$(4 + 36)

     ⇒ $\sqrt{10}$

Area of ΔABC = $\frac{1}{2}$×base×height

                       = $\frac{1}{2}$×BC×AD

                       = $\frac{1}{2}$×√40× √10

                       = 10

Therefore, area of the triangle is 10.