Question

In an isosceles triangle, the ends of the base are (2a, 0) and (0, a) and one side is parallel to y-axis. The third vertex is​

Answer 1

Given:

In isosceles triangle the end base are (2a, 0), ( 0, a).

One side is parallel to Y-axis.

To find:

Here, we need to find the third vertex.

Solution:

Midpoint = $(\frac{x1 + x2}{2}, \frac{y1+y2}{2})$

Points (2a, 0) (0,a)

Now, x1 = 2a ; x2 =0 ; y1 = 0 ; y2 = a

Midpoint = $(\frac{x1 + x2}{2}, \frac{y1+y2}{2})$

               = $(\frac{2a+0}{2}, \frac{0+a}{2})$

               = ( 2, $\frac{a}{2}$)

Here, the perpendicular bisector passes through the point ( 2, $\frac{a}{2}$).

The slope is 2.

Now we can put in the equation of line.

Equation of line ⇒  y - y1 = 2(x - x1)

Substitute the values

y - $\frac{a}{2}$ = 2 (x - a)

y - $\frac{a}{2}$ = 2x -a

2y- a = 4x -4a

2y- 4x = - 3a

Now, the line intersect the line X = 2a

Put X = 2a in equation

2y - 4(2a) = -3a

2y - 8a = -3a  

2y = -3a + 8a

2y = 5a

y = $\frac{5a}{2}$

Here, the third vertex is (2a,$\frac{5a}{2}$ )