1,000,000 Question Thread :D

[kenhung123]

The point (h,k) lies on the line y=x+1 and is 5 units from the point (0,2). Write down the two equation connecting h and k and hence find the possible values of h and k.

I used the gradient and distance between 2 points equations and got

[TrueTears]

The point (h,k) lies on the line y=x+1 and is 5 units from the point (0,2). Write down the two equation connecting h and k and hence find the possible values of h and k.

I used the gradient and distance between 2 points equations and got

That's good, now you just have to solve it simultaneously.

[kenhung123]

I got the wrong answer, or is the textbook wrong?

[TrueTears]

Wait!! Sorry, I misread the first time, why do you have h = k-2 it should be k = h + 1 I think

[kenhung123]

How did you get that?

[TrueTears]

well (h,k) lies on the line y = x+1

so sub k for y and h for x

[kenhung123]

well (h,k) lies on the line y = x+1

so sub k for y and h for x

I did (2-k)/(0-h)=1
-h=2-k
h=k-2

But I guess it didn't say the point was on the same line hehe
Thanks TT

[TrueTears]

You can do that but it wouldn't be (0,2)

If you sub 0 for x you should get 1 for y

so it'd be (1-k)/(0-h) = 1

1-k = -h

1+h=k

[kenhung123]

P and Q are the points of intersection of the line y/2+x/3=1 with the x and y axes respectively. The gradient of QR is 1/2, where R is the point with x co ordinate 2a, a>0

Find:
The y co ordinate of R in terms of a.

I did
(0-y)/(3-2a)=1/2
-y=1/2(3-2a)
y=-1/2(3-2a)

[trinon]

Find P and Q from letting x and y equal 0:




We know that the gradient of QR is 1/2 and that there is a point (0,2) on it, so we simply create the equation and sub in the known co-ordinate:






Now sub in the x co-ord:

[kenhung123]

Thanks
ABCD is a parallelogram, lettered anticlockwise, such that A and C are points (-1,5) and (5,1) respectively.

Given that BC is perpendicular to AC, find:
The co ordinates of B

[kyzoo]

Gradient of AC = (5-1)/(-1-5) = -2/3

Therefore Gradient of BC = 3/2

Equation of straight line intersecting with BC is
y - 1 = 1.5(x-5)
y = 1.5x - 6.5

Let B = (x, y)

AC = sqrt((5+1)^2 + (5-1)^2) = sqrt(52)
BC = sqrt((5-x)^2 + (1-y)^2)
AB = sqrt((-1-x)^2 + (5 - y)^2) = sqrt((1+x)^2 + (5 - y)^2)

Now because ABC is a right-angled triangle

Therefore AC^2 + BC^2 = AB^2

52 + (5-x)^2 + (1-y)^2 = (1+x)^2 + (5 - y)^2
52 + 25 + 1 + x^2 + y^2 - 10x - 2y = 1 + 25 + x^2 + y^2 + 2x -10y
52 - 10x - 2y = 2x - 10y
52 + 8y = 12x
1.5x = (12x)/8 = 52/8 + y

Now we have simultaneous equations

y = 1.5x - 6.5
= 52/8 + y - 6.5

...Lol I can't figure out where I went wrong

[kenhung123]

I did this:
Since BC has gradient of 3/2 and AB has gradient of -2/3
Given A and C are points (-1,5) and (5,1)
(1-y)/(5-x)=3/2
(5-y)/(-1-x)=-2/3

I solve simultaneously and get false...

[Ilovemathsmeth]

I'd find coordinates of letter B by solving the these two inequations:

Equation of line BD and equation of line BC. Both of these were found in previous parts of the question. By solving these two simultaneously, you can find where they cross and thus the coordinates of B.

Hope that helps.

[kenhung123]

Thanks mathsmeth! Finally did it! But do you know why I could not have solved it the way I attempted?