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Need help to solve As levels Maths P1 Question

Its a 10 mark question (attached) of GCE As levels Pure Maths 1 Chapter Coordinate Geometry.
Would appreciate any help :smile:
Reply 1
What have you tried so far?
Original post by ahsanshah199
Its a 10 mark question (attached) of GCE As levels Pure Maths 1 Chapter Coordinate Geometry.
Would appreciate any help :smile:


I did some rough calculations and this is what i did
Too bad handwriting I know
Reply 3
Original post by ahsanshah199
I did some rough calculations and this is what i did
Too bad handwriting I know


Find the equation of the line that passes through points A and B.
Original post by ahsanshah199
I did some rough calculations and this is what i did
Too bad handwriting I know


There are 4 steps:

1. Finding co-ordinates of A and B
2. Constructing the line which is a perpendicular bisector of AB
3. Finding the intersections, C and D, between this perpendicular bisector and the curve
4. Using Pythagoras to calculate the distance between the C and D

Unfortunately, you have failed to find the co-ordinates of A and B correctly.

x−y+2=0⇒y=x+2x-y+2=0 \Rightarrow y=x+2

⇒2x2−(x+2)2+2x+1=0\Rightarrow 2x^2-(x+2)^2+2x+1=0

⇒2x2−(x2+4x+4)+2x+1=0\Rightarrow 2x^2-(x^2+4x+4)+2x+1=0

⇒2x2−x2+2x−4x+1−4=x2−2x−3=0\Rightarrow 2x^2-x^2+2x-4x+1-4=x^2-2x-3=0

Can you see where you went wrong? Can you carry on from there?
(edited 7 years ago)
Original post by B_9710
Find the equation of the line that passes through points A and B.


Thats where im stuck at
I m making some mistake here
Can you solve the rest for me
Original post by RDKGames
There are 4 steps:

1. Finding co-ordinates of A and B
2. Constructing the line which is a perpendicular bisector of AB
3. Finding the intersections, C and D, between this perpendicular bisector and the curve
4. Using Pythagoras to calculate the distance between the C and D

Unfortunately, you have failed to find the co-ordinates of A and B correctly.

x−y+2=0⇒y=x+2x-y+2=0 \Rightarrow y=x+2

⇒2x2−(x+2)2+y2x+1=0\Rightarrow 2x^2-(x+2)^2+y2x+1=0

⇒2x2−(x2+4x+4)+2x+1=0\Rightarrow 2x^2-(x^2+4x+4)+2x+1=0

⇒2x2−x2+2x−4x+1−4=x2−2x−3=0\Rightarrow 2x^2-x^2+2x-4x+1-4=x^2-2x-3=0

Can you see where you went wrong? Can you carry on from there?

Brother you might be doing wrong..look putting value of y, from line one that is the straight line that intersect, into equation of curve line is wrong way
You get it
Reply 7
Original post by ahsanshah199
Thats where im stuck at
I m making some mistake here
Can you solve the rest for me

Actually you don't need to do that. You need to find the midpoint of AB and find a line that passes through the midpoint and is perpendicular to the line AB. I assume you can find gradients of lines and know the conditions for lines to be perpendicular.
Original post by ahsanshah199
Brother you might be doing wrong..look putting value of y, from line one that is the straight line that intersect, into equation of curve line is wrong way
You get it


What??? Are you saying substituting the equation of the line into the curve is a 'wrong' way???

I'd hate to break it to you but you're doing the exact same thing with your working out, you're just finding what y2y^2 is first before substituting it into the curve's equation.

Also, having looked over your working out, you said y2=x2+4y^2=x^2+4 which is incorrect because
y2=(x+2)2≠x2+4y^2=(x+2)^2 \not= x^2+4
(edited 7 years ago)
Original post by RDKGames
What??? Are you saying substituting the equation of the line into the curve is a 'wrong' way???

I'd hate to break it to you but you're doing the exact same thing with your working out, you're just finding what y2y^2 is first.

Also, having looked over your working out, you said y2=x2+4y^2=x^2+4 which is incorrect because
y2=(x+2)2≠x2+4y^2=(x+2)^2 \not= x^2+4


I got it
Gonna try it myself again
Original post by ahsanshah199
I got it
Gonna try it myself again


I'm just...i don't know
I find cords for A and B
Solved their gradient and their midpoint
Got gradient of Line CD
Got equation of line CD
I substitute equation of curve to equation of line CD
But still im doing it the wrong way
Original post by ahsanshah199
I'm just...i don't know
I find cords for A and B
Solved their gradient and their midpoint
Got gradient of Line CD
Got equation of line CD
I substitute equation of curve to equation of line CD
But still im doing it the wrong way


Show your working. You have the right approach.
Can someone help me with a solution
Original post by ahsanshah199
Can someone help me with a solution


Show your working and I'll tell you exactly where you're going wrong.

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