Hi,
I am a high school student and I've been trying to solve the problem shown in the image for an hour or so now. The question requires me to basically find the area of triangle using the coordinates of straight lines. If helpful, the unit has been focusing on equations of straight lines. I would really appreciate some guidance on how I might tackle this question.
The slope of the line $L$ thru $A=(-3,12)$ and $B=(6,0)$ is $S=\frac {12-0} {(-3)-6}=-4/3.$ So the slope of the line $M, $ perpendicular to $L$ and passing thru $A,$ has slope $\frac {-1}{S}=3/4.$
$M$ meets the $x$-axis at the un-labelled point that I will call $C.$
As you travel from $C$ to $A$ along $M,$ your $y$-coordinate's change, which is $12,$ is $3/4$ of your $x$-co-ordinate change because $M$'s slope is $3/4.$ So the $x$-co-ordinate change from $C$ to $A$ is $\frac{12}{3/4}=16.$ So $C=((-3)-16,0)=(-19,0).$
So in $\triangle ABC$ the length of the base $AC$ is $6-(-19)=25.$ And the height of the point $B$ above $AC$ is $12.$ So the area of $ABC$ is $(25)(12)/2=150.$
Now $\triangle A'B'C',$ with $A'=(0,0), B'=B=(6,0),$ and with $C'$ being the intersection of $AB$ with the $y$-axis, is similar to $\triangle ABC,$ as they have equal sets of interior angles. We have length $A'B'=6.$ We have length $AB=\sqrt {(6-(-3))^2+(0-12)^2}=\sqrt {3^2} \sqrt {3^2+4^2}=15.$
So $A'B'/AB=6/15.$
So the area of $A'B'C'$ is $(6/15)^2$ times the area of $ABC.$ That is, the area of $A'B'C'$ is $(6/15)^2\cdot 150 =(4/25)\cdot 150=24.$
So the shaded area is $150-24=126.$