The vertices of triangle ABC are A(-4,0), B(2,4), and C(4,0). What is its area?

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Answer 1 · 2017-10-18T04:23:16

#16un.^2#

Don"t be intimidated by the points. Graph them and find your base and height

This is easy because your base is just the distance from A to A on the horizontal plane, 8.
The height is defined by the vertical distance of B, 4.

Now use the area of a triangle formula (A = #1/2*b*h#)
A = #1/2(4)(8)#
A = #1/2 * 32#
A = 16

Your area is 16 sq. units.

Answer 2 · 2017-10-18T13:24:23

# 16" sq. units."#

Let us denote, by #[ABC],# the Area of a #DeltaABC.#

We know from the Co-ordinate Geometry, that, if the vertices of

#DeltaABC# are #A(x_1,y_1), B(x_2,y_2), and, C(x_3,y_3),# then,

#[ABC]=1/2*|D|," where, "D=|(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)|.#

Here, #D=|(-4,0,1),(2,4,1),(4,0,1)|,#

#=-4(4xx1-0xx1)-0+1(2xx0-4xx4),#

#=-4(4)+1(-4),#

# rArr D=-32.#

#"Therefore, "[ABC]=1/2*|-32|=16" sq.units."#