core 1 coordinate geometry help!!!

cn anyone plz help how to work out the following qs:
1. A(-1,-2) B(7,2) and C(k,4), where k is constant, are the vertices of triangle ABC.
Angle ABC is a right angle. the gradient of AB is 1/2x
a) calculate the value of k

I hv no clue

cn anyone plz help how to work out the following qs:
1. A(-1,-2) B(7,2) and C(k,4), where k is constant, are the vertices of triangle ABC.
Angle ABC is a right angle. the gradient of AB is 1/2x
a) calculate the value of k

I hv no clue

As we have a right angle triangle, AB and AC are perpendicular to one another. You have the gradient of AB as 1/2. You can work out the gradient of the perpendicular as 1/2 X m=-1 so m=-2 Now you know that the perpendicular to AB which is AC is a line with gradient -2. The equation of AC needs to be worked out. This is because you need the x value (k) at which y=4. You know that AC has gradient -2 and passes through point A, whose coordinates you already have as (-1, -2)

Angle C cuts the y axis meaning x=0 y+0=8 y=8

A= (0,-2)
C= (0,8)
D=(4,4)

AREA= BASE TIME HEIGHT/ 2

Base= 10
Height=4

40/2 = 20

thnxx a lot

Thats fine- are you teaching yourself AS maths?

yep, preparing for A level maths, gona start it from this september

C1 and C2 are pretty good- Its S1 that most people seem to struggle with

ohh, my school does C1, C2 and D1. Wt about D1??

Oh we do D1 at A2, it sounds easy from what ive heard- the only tricky area of AS core was probably trig as it does take a while to get your head around sometimes

You can use $y-y_1=m(x-x_1)$ or rearrange $y=mx+c$ to get $c=y-mx$.

Using the first option you will have $y-3\sqrt{3}=\sqrt{3}(x-1)$ which you will need to rearrange.

With the second option $c=3\sqrt{3} -\sqrt{3}\times 1$.

the answer is y=√3x+2√3, I still don't get it

the answer is y=√3x+2√3, I still don't get it