Question

The height of the isosceles triangle is twice the length of its base what are the coordinates of the undefined vertex ?

Answer 1

Answer-

The coordinate of the vertex is (7, 12)

Solution-

The coordinates of two end points of the base are, (4, 0), (10, 0)

As the triangle is an isosceles triangle, so the other vertex will be on the perpendicular bisector of the base.

The mid-point of the base is,

$=(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2})$

$=(\dfrac{4+10}{2},\ \dfrac{0+0}{2})$

$=(\dfrac{14}{2},\ \dfrac{0}{2})$

$=(7,\ 0)$

As given that the length of the height is twice that of the base,

Length of the base is,

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

$=\sqrt{(10-4)^2+(0-0)^2}$

$=\sqrt{(6)^2}$

$=6$

Length of the height will be $=6\times 2=12$

So the vertex is 12 units vertically up from the mid-point (7, 0). Then the coordinate of the point will be (7, 12)