Question

PLEASE someone help me with maths ​

Answer 1

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:

`(5-k)/(6-4)= (5-k)/(2)`

Remember: The gradient of BC = the negative reciprocal of the gradient of AB. So:

`(5-k)/(2) =negative..reciprocal..of..(1)/(2)`

So:

`(5-k)/(2)=-2` (Now just solve for k)

`5-k=-4`

`-k=-9` (now just multiply both sides by -1)

`k = 9`

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:

`(1-9)/(-2-4) =(-8)/(-6) = (4)/(3)`

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)

`y - 1=(4)/(3)(x +2)` (note: x - - 2 = x + 2)

Now lets multiply both sides by 3, to get rid of the fraction:

`3y - 3 = 4(x+2)` (now expand the brackets)

`3y - 3 = 4x+8)`

Finally, we just rearrange this to get the format: ay + bx = c

`3y - 3 = 4x+8`

`3y = 4x+11`

`3y - 4x = 11`

And done!:

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Answer:

The equation of a line that passes through point A and C is:

`3y - 4x = 11`