Area of triangle given 2 coordinates and a line equation

Question

Problem: The vertices of the base of an isosceles triangle are $(-1,-2)$ and $(1,4)$. If the third vertex lies on the line $4x + 3y = 12$, find the area of the triangle.

Attempt 1 : Convert $4x + 3y = 12$ to point slope form which is $(y-0) = \frac{-4}{3}(x-3)$ then use (0,3) and the two given coordinates to solve the area by using the area by its coordinates formula but got the wrong answer.

Attempt 2: get the distances between $(-1,-2)$ and $(1,4)$, between $4x + 3y = 12$ and $(-1,-2)$ and between $4x + 3y = 12$ and $(1,4)$ to get sides a,b, and c to get the area by heron's formula but again resulted to the wrong answer.

Question: How to answer this? I think it has just something to do with converting $4x - 3y =12$ to standard form to get its coordinates or something. Any help would be appreciated.

Answer 1

hint

$$A=(-1,-2) $$ $$B=(1,4) $$ $$C=(a,b) $$

$$AC^2=BC^2\implies$$

$$(a+1)^2+(b+2)^2=(a-1)^2+(b-4)^2$$

$$\implies 4a+12b=12$$

on the other hand

$$4a+3b=12$$

thus $$a=3,b=0$$

the middle of $[A,B] $ is $$J=(0,1) $$

the area is $$S=\frac {AB.CJ}{2} $$

Answer 2

How would I approach such problem?

We know that the third vertex, call it $C$ lies on the perpendicular bisector of $AB$. The midpoint of $AB$ is $((-1+1)/2, (-2+4)/2)=(0, 1)$. The slope of $AB$ is $3$, thus the perpendicular bisector is determined by line $y = -x/3 + 1$. Compute the intersection with $4x+3y=12$ to get $(3,0)$. The area is thus $|A - B|\cdot|C - (0, 1)|/2 = \sqrt{2^2+6^2}\cdot\sqrt{1^2+3^2}/2 = 10$.

Indeed, the result is that pretty, since the given triangle is right, which can be seen after drawing the points and lines. Always start with drawings, they help alot.