Question

the two vertices of a triangle are (0,6) and (0,12). if the are of the triangle is 12 square units, where can the third vertex be? What if the triangle is also isosceles?

Answer 1

Answer:

1)  (4,6),(-4,6), (4,12),(-4,12)

2) (4,9) or (-4,9)

Step-by-step explanation:

Observe that the two vertices given, (0,6) and (0,12) are on the y-axis.

The distance between this points is |12-6|=6

We know area of a triangle is $A=\frac{1}{2}bh$

We were given area as 12.

$\frac{1}{2}*b*6=12$

$3b=12$

b=4 units

We can find a point 4 units horizontally away from any of the vertex.

So (4,6),(-4,6), (4,12),(-4,12) are feasible vertices.

For the triangle to be isosceles, then the third vertex must be on the horizontal line through the midpoint of (0,6) and (0,12) and must be 4 units away from the line x=0.

Therefore if thr triangle is isosceles, then third vertex must be at (4,9) or (-4,9)