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secondary 3 | A Maths
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Henry Kor
Henry Kor

secondary 3 chevron_right A Maths chevron_right Singapore

Could someone please help with part (iii)? Thanks!

Date Posted: 4 years ago
Views: 334

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Kayden Kiew
Kayden Kiew's answer
2 answers (A Helpful Person)
1st
Tldr: sub equation AB with BC to find B
Henry Kor
Henry Kor
4 years ago
Hey but they asked you to find the coordinates of C and D as well. How would you find it?
Kayden Kiew
Kayden Kiew
4 years ago
A=(-1,4)
B=(3.1,9.125)

In general, to solve for C and D Coordinates, need to look at the X and Y coordinates independently

To solve for coordinate C, need to look at the increase in x from A to B first.

Increase in x from A to B= (3.1-(-1))=4.1

Since it is a rectangle(or look at diagram) , means from B to C, the x increases by 2 times from B to C as well.

X coordinate of C= 3.1+4.1*2=11.3

Now lets look at the y increase from A to B.

From A to B, the y increase is from 4 to 9.125,so the nett increase is 5.125.
However, from B to C, it is a downward slopiny so Y will decrease by 2times.

Therefore, y coordinate of C = 9.125 - (5.125*2)= - 1.125

C(11.3,-1.125)

For coordinate D,
The length and gradient mirrors from B to A. Hence they are similar lines.

So from B to A,
X decrease by 4.1
So x coordinate of D=x-coodinate of C - 4.1=11.3-4.1=7.2

Similarly,
From B to A, y decrease by 5.125.
So y coodinate or D =y-coordinate of C-5.125=-1.125-5.125=-6.25