BAC is a right angle. So BA is perpendicular to AC.

Since BA is part of the x-axis, it is a horizontal line.

Which means that AC must be a vertical line.

In Sec 2 you would have learnt that the equation of a vertical line is x = c, where c is a constant.

Why ? Because on this line, the x-value is always c, no matter how the y-value changes when you move along the line.

So equation of AC is x = -10, since we can clearly see that the intersection with the x-axis is at (-10,0)

i.e the x-intercept is -10

AB is 12 units long.

Since it is a horizontal line, the length is just the absolute value or difference between the x-values of A and B.

A has x-coordinate of -10 (it is in the negative region of the x-axis). Simply add 12 to get B's x-coordinate.

-10 + 12 = 2

Since B is the x-intercept of the line BC,
(where y = 0 on the x-axis)

B(2,0)

Thank you .

Now C lies on AC and BC. It is the intersection point of both lines. So its coordinates must satisfy both their equations.

We already know x = -10 on AC, so C must have x-coordinate of -10.

Next, recall the gradient formula :

m = (y2 - y1) / (x2 - x1)

We can rewrite this as y2 - y1 = m(x2 - x1)

Use it to solve for C's y-coordinate by substituting m = -½ (the gradient of BC) and the coordinates of B(2,0).

Letting C be the second point, sub B(2,0) as (x1,y1) and -10 as x2,

y2 - 0 = -½(-10 - 2)

y2 = 6


Therefore, C(-10,6)

Alternatively,

Since AB is 12 units long, and realising that the run from C to B (horizontal) is actually the same length,

And knowing that gradient = rise/run,

We just need to solve for C's y-coordinate by substituting into the rise portion.

-½ = (0 - y)/12
-½ × 12 = -y
-6 = -y
y = 6