A perpendicular bisector is a line which cuts a line segment at right angles and divides it into two equal smaller segments.

In particular, the perpendicular bisector of AB is a line which meets AB at right angles and passes through the mid-point of AB.

As you have probably learnt by now, the product of the gradients of two perpendicular lines is equal to -1.

To find the equation of this perpendicular bisector (which is equivalent to finding the equation of a straight line), we need to
- find the gradient of this bisector and
- find the midpoint of AB.

The gradient of AB
= change in y / change in x
= [1 - 7] / [5 - (-1)]
= -6 / 6
= -1

Since gradient AB x gradient bisector = -1, the gradient of the bisector must be 1.

The midpoint of AB
= (average of x, average of y)
= [(-1 + 5)/2, (1 + 7)/2]
= (2, 4)

Now that we have the gradient of the line (1) and a point on the line (2, 4), we are ready to form the equation of the line (i.e. the bisector).

y = mx + c
y = x + c (since m = 1)

Substituting (2, 4) into the equation,
4 = 2 + c
2 = c

So, the equation of the perpendicular bisector must be y = x + 2.

To check that the line passes through E (-4, -2), we simply check that the x- and y- coordinates of E satisfy the equation.

Clearly -4 + 2 = -2, so this is satisfied and E must lie along the line. Therefore, the line passes through E.

E to A is a direct movement of
- 3 units to the right (from -4 to -1)
- 9 units upwards (from -2 to 7)

Since AF : FE = 2, it is clear that EF : EA = 2 : 3, so the movements of E to F will be 2/3 of the movements of E to A.

E to F is a direct movement of
- 2 units to the right
- 6 units upwards

So, the coordinates of F
= (x-co. of E + 2, y-co. of E + 6)
= (-4 + 2, -2 + 6)
= (-2, 4)

For part (c), one way to do is to consider similar triangles, but this might not have been covered in your school yet (as E Maths).

Another way to do this is to find the coordinates of D and then perform the Gaussian Area Formula (i.e. the "shoelace" method), but this is much trickier.

I will not do this part on the assumption that you have yet to learn similar triangles.

thank you for your help! I think I definitely do understand it better. I’m going to re-try the question, and if I have any more questions I’ll ask u. thank u! I really appreciate it<3

as for part c) I have done similar triangles but I’m a bit confused as to how I can apply them? I’ve tried to find the coordinate of D but I can’t seem to find it :<

I have seen your other posts as well and found that you are in the midst of learning similar triangles.

If you recall in Sec 2 Maths, for any two similar triangles, their corresponding length ratios are the same.

[The concepts in Sec 2 are the basic ones which assume that the triangles or other shapes are already congruent or similar, while the concepts in Sec 3 aim to prove the congruency/simililarity rather than assume them]

We see that AF : FE = 1 : 2, so needless to say, AE : FE = 3 : 2.

Triangles ABE and FDE are obviously similar (this is normally what happens when lines such as AB and FD are parallel).

[I assume you have recently learned to write the proof for similarity, correct?]

In such situations, the other corresponding ratios such as AB : FD or BE : DE would also be 3 : 2 as a result.

We do not exactly need to find the coordinate of D in solving this part. However, if you wish to find the coordinates of D, we use the fact that BE : DE is equal to 3 : 2.

BE : DE = 3 : 2
EB : ED = 3 : 2

Note that E and B have coordinates (-4, -2) and (5, 1) respectively.

So, E to B is a movement
==> 9 units to the right
==> 3 units upwards

E to D is a movement which is 2/3 of this (that's what the ratio 3 : 2 is all about!)

So, E to D is a movement
==> 6 units to the right
==> 2 units upwards

So, we can see that the coordinates of D
= (-4 + 6, -2 + 2)
= (2, 0)

Yes, I know that D does not appear to be on the x-axis, but I know that some diagrams are misleading. In the actual O Levels, the diagrams are usually accurate to the scale.

alright I understand now! the diagram was just a little off lol. But I was wondering the question gave the area of ABE as 36 square units is there any way we can use that in this question? was just curious because I feel like we can use the 36 for something but I’m not sure how

If you have actually noticed by now, triangles ABE and FDE are similar.

When two triangles are similar, their area ratio would be the square of their length ratio.

[This is because areas are two-dimensional figures; the bases are in a ratio and the heights are in the same ratio, so collectively the area ratio takes these two identical ratios into account]

Recall that length ratio EF : EA = 2 : 3.

Correspondingly, area ratio FDE : ABE = 2² : 3² = 4 : 9.

We can use this ratio 4 : 9 for area comparisons. Since the area of ABE is 36 units², the area of FDE would be 16 units².

Now, the challenge would be to find the area of triangle BCD.

I will outline this next.

In actual fact, triangles FDE and CDB are also similar! Can you prove it?

[Hint: Lines EF and BC are parallel since they are sides of a parallelogram]

Their length ratio would be the ratio of their corresponding lengths, say FE : CB.

But obviously CB = FA because their sides of a parallelogram!

We know very well that EF : EA = 2 : 3 from our previous workings, so FE : FA = 2 : 1 and therefore FE : CB = 2 : 1.

With this, we see that the area ratio FDE : CDB = 2² : 1² = 4 : 1.

Since the area of FDE is 16 units², the area of CDB must be 4 units².

Therefore, the area of triangle BCD is 4 units².

alright I see! thank you for explaining <3