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Equation of three lines: $$X=0$$ $$Y=-2x+12$$ $$Y=0.5x+4.5$$ Therefore the vertices of the triangle are: $(3,6), (0,12), (0,4.5)$ But i dont know what to do next to work out the area because i dont have a length? Thanks in advance

BLAZE
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1 Answers1

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Lets say $A = (3,6), B = (0,12), C = (0,4.5)$.

Now find the vectors $BA = A-B=(3-0,6-12)^T=(3,-6)^T$ and $BC = C-B=(0-0,4.5-12)^T=(0-7.5)^T$.

Now use the Formula for the Area: $$A = \frac{1}{2}|\det{[BA,BC]}|$$

Can you make it from here?

You can also use the formula $A=\frac{1}{2}|BA||BC|\sin(\alpha)$. You can get the $\alpha$ from the dot product $<BA,BC>=|BA||BC|\cos(\alpha)$.

You can also calculatr the magnitude of the cross product (add zero als third component for both vectors) and then multiply by 1/2 to get the area.

MrYouMath
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