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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.

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)

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

Johnny Mccrum
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2 Answers2

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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 $$y-1=\frac{1}{2}(x-(-1))\iff y=\frac 12x+\frac 32.$$

mathlove
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  • Could you take me through the process slower? I get the graident part, but the second part with the line equation I simply don't understand. Where is the half coming from? A(1, -1)? – Johnny Mccrum Oct 14 '14 at 10:33
  • @JohnnyMccrum: Sure. If you understand the slope of the line we want is $1/2$, then note that our line passes through the point $A(-1,1)$. The line whose slope is $a$ and which passes through a point $(b,c)$ can be represented as $y-c=a(x-b)$. So, in our case, we have $y-1=(1/2)(x-(-1))$. – mathlove Oct 14 '14 at 10:38
  • "the slope of the line we want is 1/2" Why is this the case? – Johnny Mccrum Oct 14 '14 at 10:39
  • @JohnnyMccrum: I used the following fact : Two lines $y=m_1x+b_1,y=m_2x+b_2$ cross perpendicularly if and only if $m_1m_2=-1$. Since we know the slope of the line $BC$ is $-2$, the slope of the line we want is $(-1)/(-2)=1/2$. – mathlove Oct 14 '14 at 10:43
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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 $y-y_1 = m(x-x_1)$.

taninamdar
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