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, $D=|(-4,0,1),(2,4,1),(4,0,1)|,$

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

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

$rArr D=-32.$

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