You are on the right tracks.
Since angle ABC is a right angle, that means lines AB and BC are perpendicular.
Therefore the gradient of BC = the negative reciprocal of the gradient of AB. We can use this to form an equation to find what K is.
You have already worked out the gradient of AB ( 1/2) (note it's easier to leave it as a fraction)
Now lets get the gradient of BC:
[tex]\frac{5-k}{6-4}= \frac{5-k}{2}[/tex]
Remember: The gradient of BC = the negative reciprocal of the gradient of AB. So:
[tex]\frac{5-k}{2} =negative..reciprocal..of..\frac{1}{2}[/tex]
So:
[tex]\frac{5-k}{2}=-2[/tex] (Now just solve for k)
[tex]5-k=-4[/tex]
[tex]-k=-9[/tex] (now just multiply both sides by -1)
[tex]k = 9[/tex]
That means the coordinates of C are: (4, 9)
We can now use this to work out the gradient of line AC, and thus the equation:
Gradient of AC:
[tex]\frac{1-9}{-2-4} =\frac{-8}{-6} = \frac{4}{3}[/tex]
Now to get the equation of the line, we use the equation:
y - y₁ = m( x - x₁)
Let's use the coordinates for A (-2, 1), and substitute them for y₁ and x₁ and lets substitute the gradient in for m:
y - y₁ = m( x - x₁)
[tex]y - 1=\frac{4}{3}(x +2)[/tex] (note: x - - 2 = x + 2)
Now lets multiply both sides by 3, to get rid of the fraction:
[tex]3y - 3 = 4(x+2)[/tex] (now expand the brackets)
[tex]3y - 3 = 4x+8)[/tex]
Finally, we just rearrange this to get the format: ay + bx = c
[tex]3y - 3 = 4x+8[/tex]
[tex]3y = 4x+11[/tex]
[tex]3y - 4x = 11[/tex]
And done!:
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Answer:
The equation of a line that passes through point A and C is:
[tex]3y - 4x = 11[/tex]
