Eric Nicholas K's answer to Grace's Secondary 3 E Maths Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Part (c), but lengthy working
Date Posted:
4 years ago
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.
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!
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!