Thank u!

Hi shouldn’t it be -8+13=5? So A should be (0,5)? Since the starting point of the base is -8?

Will take a look later by 3 am, could be a miscomputation of mine

I have taken a look. In the case of my triangle, I have cut the two triangles into two equal parts. Where the line BC cuts the y-axis, we call this point P.

In other words, I cut the triangle ABC into two equal parts ABP and ACP. At P, y = -6. We take this into consideration when calculating the coordinates of the unknown point.

AP is therefore the base, with a starting y-value of P of -6.

You can also draw a line rightwards from B. However, this method is much tougher. Hence, I considered the y-intercept as the areas triangles ABP and ACP would be easy to compute.

Recall that the area of a triangle is 1/2 x base x perpendicular distance, from primary school knowledge. For this question, I took the base to be the “vertical distance” between A and P, whereas the height would be the “horizontal distance” from P to B and C respectively.

This is easier than taking lengths PB and PC as bases or heights since horizontal lines are always perpendicular to vertical lines.

If necessary, I can later approach the question using A Maths techniques of finding areas to verify my answers. Will post this by 3 am.

Let me know if you have further questions or doubts and I will explain them again!

Thank you!