Co-ordinate Geometry Help (1 Viewer)

[rolror88]

I'm stuck on a coordinate Geometry Question. Your help would be appreciated.

Given 2 points A(-3,1), B(2,3). The line segment cuts AB cuts y-axis at P, find the ratio APB. Hence find the coordinates of P.

If you could show me the working out it would help me a lot.

Thank in advance

 

[Aznmichael92]

I'm stuck on a coordinate Geometry Question. Your help would be appreciated.

Given 2 points A(-3,1), B(2,3). The line segment cuts AB cuts y-axis at P, find the ratio APB. Hence find the coordinates of P.

If you could show me the working out it would help me a lot.

Thank in advance

I will attempt it. Not sure if it will help. It is a lot of working out though.

We know that the equation fo y axis is equal to x = 0
We will then use the x value of each point to find the ratio of AP:BP
Distance from X(a) to 0 = 3
Distance from X(b) to 0 = 2

Therefore ratio of AP:BP is 3:2

Right now i cant remember the ratio formula but i might bring it back to mind soon. What I did instead is find the gradient of line AB so I can find the equation of the line.

m of line AB = rise/run
= 1 – 3/ - 3 – 2
= -2/-5
= 2/5


Equation of Line AB: y – y1 = m(x – x1) I will use Point A but you could use Point B if you want
y – 1 = 2/5(x + 3)
5y – 5 = 2x + 6
5y = 2x + 11
y = (2x + 11)/5


Point P: If x = 0, y = 11/5


Therefore coordinates of P is (0,11/5).

If you dont trust me about the ratio, you can test it again if you want using the distance formula since you have coordinates of all 3 points.
Distance of AP to Distance of BP
Hope that helps

 

[Aznmichael92]

Oh here is the formula i just remembered where you can just sub the ratio and points in.

Let distance from AP be a and let distance from BP be b
Coordinates of A are (x1,y1) B are (x2,y2)

The equation to find a point given the ratio and coordinates are

(ax2 + bx1 , ay2 + by1)
aaaa + baaaa a + b

so it will be
(3x2 + 2x-3 , 3x3 + 2x1)
aaa3 + 2aaaa 3 + 2
= (0 , 11)
aa 5aa5
= (0 , 11)
aaaaa 5

Therefore Coordinates of P is (0,11/5)

 

[rolror88]

Thanks for your help Michael.

I was stumped on the first bit of the Question (the ratio bit). I get how to do it now.

 

[Aznmichael92]

Thanks for your help Michael.

I was stumped on the first bit of the Question (the ratio bit). I get how to do it now.

nps glad to help it is also good pratice for me. Incase i get it in future