Find the height of the triangle

Question

We are given the equation of the following lines: $$AB: x - 2y + 3 = 0$$ $$AC: 2x - y - 3 = 0$$ $$BC: 3x + 2y + 1 = 0$$

Find the equation of the height corresponding to $BC$ of the $ABC$ triangle.

Here is what I did so far:

Let $h: ax + by +c = 0$ be the equation of the height corresponding to $BC$. Because $h$ and $BC$ are perpendicular, the product of their slopes is $-1$. I got $3a = -2b$.

Now, $h$ contains the point $A(3, 3)$, so $3a + 3b +c = 0$. Plugging in $3a = -2b$, I got $c = -b$.

At this point I have no idea what should be done next.

So, if you know, please help me!

Thank you!

Answer 1

Assume height $h_{BC}$ corresponding to $BC$ of the $ABC$ triangle has equation $y=kx+n$.

Since it is perpendicular on line $BC$ we know that $$(k,-1)(3,2)=0,$$ from where we get $$k=\frac{2}{3}.$$ So height corresponding to $BC$ has equation $y=\frac{2}{3}x+n$.

From condition $A\in h_{BC}$, we get $$3=\frac{2}{3}3+n=2+n$$ and $$n=1.$$

Height $h_{BC}$ has equation $$-2x+3y-3=0.$$

Answer 2

You have the equation of your "height line" (2 equations in 3 unknowns $a,b,c$ but that doesn't matter, such equations are only defined up to a constant factor anyway).

Find the intersection between your "height line" and the base $BC$, call it $X$.

Find the intersection between $AB$ and $AC$, i.e. the location of point $A$.

The height is the distance $|A-X|$

Answer 3

From the equations you obtained, $3a=-2b$ and $c=-b$, you already have the equation $$h:\frac{-2}{3}bx+by-b=0.$$ You can put any nonzero value of $b$ you want and the equation is essentially the same, i.e. the equation represent the same line no matter which nonzero $b$ you put in. This is known as linear dependence.

Answer 4

Any line through $A$ can be written as \begin{align*} x-2y+3 + k(2x-y-3) = 0 \end{align*} This must be perpendicular to $BC$ for the altitude. Thus \begin{align*} -\frac{3}{2} \times \frac{2k+1}{2+k} = -1 \end{align*} and $k=\frac{1}{4}$. Thus the altitude is $2x-3y+3=0$