The vertices of a triangle are A(-1, 1) B(4,0) and C(1,6) Find the equation of the altitude of the triangle ABC drawn from A.

Question

I need some help understanding the process of how you go about answering this question:

The vertices of a triangle are A(-1, 1) B(4,0) and C(1,6) Find the equation of the altitude of the triangle ABC drawn from A.

Not going to lie, in a higher maths course and still struggling with some basic stuff like this, this isn't my homework persay it's pretty much me starting my revision for my assessments next year.

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I know I need to use y2 - y1 / x2 - x1 to find the graident.

my problem is, which points do I use?

Do I use B and C, C and A? do I use all three?

I don't uderstand the process past this point.

My straight line equation is Y-b=m(x - a)

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Like I've said, if you could help me out I'd appreciate it.

Answer 1

Since the slope of the line $BC$ is $$\frac{0-6}{4-1}=-2,$$ the line you want, whose slope is $\frac{-1}{-2}=\frac 12$, is $$y-1=\frac{1}{2}(x-(-1))\iff y=\frac 12x+\frac 32.$$

Answer 2

Gradient of $BC = \dfrac{6-0}{1-4} = -2$.

Since altitude on $BC$ will be perpendicular to $BC$, it will have gradient $\dfrac{-1}{-2} = \dfrac12$.

Now, altitude passes through $A$ and has gradient $\dfrac12$. Use $y-y_1 = m(x-x_1)$.