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secondary 3 | A Maths

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I can’t solve someone please help!!

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.

- 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)

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.

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.

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.

[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.

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