Finding the length of the altitude of a triangle

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Might be a dublicate but I didn't quite find what I was looking for in the other questions.

We have $A(1, 1)$, $B(3,2)$, $C(4, 4)$

Find the length of the height which comes from A

Solution:

We have that $BC: y=2x-4$

Let $H(h1, h2)$; AH be the altitude from A

The slope of BC is $1/2 =>$ the slope of AH is $-2$

the equation of $AH:((1-h2)x+h2-h1)/(1-h1)$

We get that $-2=(1-h1)/(1-h2)$

But ... I can proceed from here. Can someone solve this problem and give me the cordinates of H, or the length of AH(the answer is AH = 3/sqrt(5))

PS. Sorry for the bad LaTex

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2 Answers

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The slope of BC is $1/2$

No, the slope of $BC$ is $2$.

So, you should have $$\color{red}{-\frac 12}=\frac{1-h_2}{1-h_1}\tag1$$

Note here that $H$ is on the line $BC$, so $$h_2=2h_1-4\tag2$$

Now solve $(1)(2)$ to get the coordinates of $H$.

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From $BC : -2x + y = – 4$, we get N, the normal to it is $–x – 2y = k$; for some $k$.

If N passes through A, then $k = … = -3$ and $N: -x – 2y = -3$.

Solving BC and N, we get H = (2.2, 0.4)

Length of the required altitude follows by applying the distance formula to A and H.

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