If no one answers this by 1 am, I will do the write up

thankyou!

BC's length

= √( (-2 - 1)² + (0 - 2)² )

= √13 units

Since 2AD = 5BC,
AD = 5/2 BC = 5/2 √13 units

Let the coordinates of D be (a,b)
AD also = √((-6 - a)² + (6 - b)²)

So 5/2 √13 = √((-6 - a)² + (6 - b)²)

(5/2 √13)² = (-6 - a)² + (6 - b)²

25/4 x 13 = 36 + 12a + a² + 36 -12b + b²
81¼ = 72 + 12a + a² - 12b + b²
9¼ = 12a + a² - 12b + b²

3y - 2x - 30 = 0
So 2x = 3y - 30
x = 3/2y - 15


Since D lies on the line, its coordinates satisfies the equation. So a = 3/2b - 15

So 9¼ = 12(3/2b - 15) + (3/2b - 15)² - 12b + b²

9¼ = 18b - 180 + 9/4 b² - 45b + 225 - 12b + b²

13/4 b² - 39b + 35¾ = 0

b² - 12b + 11 = 0
(b - 11)(b - 1) = 0

b = 11 or b = 1(rejected, as b should be greater than 6 since its higher up along the y axis than A)

so b = 11

a = (3/2 x 11) - 15
a = 1.5

So D(1.5,11) (shown)

Will start writing soon

I have sent in a word explanation for this question. This works perfectly well for two parallel lines. This is also the method for me to find the fourth coordinate of a square rather than finding the exact centre of the square (the midpoint of the diagonals).

thankyou j and eric! i get it now :)

Welcome. My method for ii) is much longer than the other, but will come in useful when we have a situation whereby the lines are not parallel. Exercise your discretion when you see 2 mark questions like this, use the faster method if possible.

Perpendicular lines can also be described by such word methods, but much harder.

Say ABCD is a square and A to B is 5 units rightwards and 1 unit upwards.

B to C will be 1 unit rightwards and 5 units downwards.

It’s much harder in this case.

For all other cases, there is no other choice but to use the long algebra way.

ohh okay! thankyou both!!