The coordinates of the vertices of △ABC are A(1,4), B(-2,-1), and C(-3,-2). The height h to side BC is √2, and b is the length of BC. What is the area of △ABC?

Question

I don't really know how to approach or start this problem, so if you can help, please do! Thanks!

Answer 1

From what you have here, it looks like h is given as sqrt(2) and that base length b must be determined from the points given. The area of a triangle being simply (base length * height from base)/2 or:

area = 1/2 * b * h

_|_|_|_|_|_|_|A|_|_|_|_| 4

_|_|_|_|_|_|_|_|_|_|_|_| 3

_|_|_|_|_|_|_|_|_|_|_|_| 2

_|_|_|_|_|_|_|_|_|_|_|_| 1

_|_|_|_|_|_|_|_|_|_|_|_| 0

_|_|_|_|B|_|_|_|_|_|_|_| -1

_|_|_|C|_|_|_|_|_|_|_|_| -2

-5 -4 -3 -2 -1 0 1 2 3 4 5

After plotting the points, and knowing that the triangle BCD where D is (-2,-2) gives each base as 1 unit and the hypotenuse as sqrt(2), The height from a base should be perpendicular to that base, so b looks like sqrt(2) and h = sqrt(2), so solve for your area with the 1/2 b*h.