# A Level Maths: Coordinate Geometry

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#1
Could someone help me with this?

A point P lies on the line with equation y = 4 - 3x. The point P is a distance of square root 34 from the origin. Find the two possible positions of point P.

I’ve been trying to work it out for so long and I just can’t get my head around it.
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2 years ago
#2
(Original post by fortuneandero)
Could someone help me with this?

A point P lies on the line with equation y = 4 - 3x. The point P is a distance of square root 34 from the origin. Find the two possible positions of point P.

I’ve been trying to work it out for so long and I just can’t get my head around it.
Draw the line on a graph, there will be two points root 34 away from the origin on that line.
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#3
(Original post by Howie_2114)
Draw the line on a graph, there will be two points root 34 away from the origin on that line.
We’re not meant to use graphs, we were given the distance formula though.
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2 years ago
#4
Let p have coordinates (x,y). We want to find the possible values of x and y.

First set up an equation using the distance formula.
Can you work from there?
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#5
Would it be root 34 = root (x^2 + y^2) ?
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2 years ago
#6
(Original post by fortuneandero)
Would it be root 34 = root (x^2 + y^2) ?
Yes
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#7
But I don’t understand where I would go from there to find two possible positions of point P
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2 years ago
#8
(Original post by fortuneandero)
Would it be root 34 = root (x^2 + y^2) ?
Good so far, Now square both sides to get rid of the square roots and rearrange to make y the subject.

As P(x.y) lies on the line y = 4 - 3x, the coordinates satisfy that equation.
So now, because they both have y as the subject, you can equate your distance equation with the equation of the straight line.
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2 years ago
#9
(Original post by razzor)
Good so far, Now square both sides to get rid of the square roots and rearrange to make y the subject.

As P(x.y) lies on the line y = 4 - 3x, the coordinates satisfy that equation.
So now, because they both have y as the subject, you can equate your distance equation with the equation of the straight line.
Yes but that’s gonna take longer than just substituting values and getting 34
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2 years ago
#10
(Original post by LowIQ)
Yes but that’s gonna take longer than just substituting values and getting 34
Trial and error will take much longer as there is more than one possible set of coordinates for P.
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#11
(Original post by razzor)
Good so far, Now square both sides to get rid of the square roots and rearrange to make y the subject.

As P(x.y) lies on the line y = 4 - 3x, the coordinates satisfy that equation.
So now, because they both have y as the subject, you can equate your distance equation with the equation of the straight line.
I’ve done this so far
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2 years ago
#12
(Original post by razzor)
Trial and error will take much longer as there is more than one possible set of coordinates for P.
The coordinates for this problem are actually easier to find by substitution
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2 years ago
#13
(Original post by fortuneandero)
I’ve done this so far
Like someone said, you could substitute y=4-3x into y^2 in the distance formula
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#14
So it would be root 34 = root (x^2 + (4-3x)^2) ?
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2 years ago
#15
Yes but forget the root as it’s the same on both sides
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#16
Like this?Attachment 771632
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#17
(Original post by LowIQ)
Yes but forget the root as it’s the same on both sides
Oh I see, so I simply expand the y substitution and collect the like terms?
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2 years ago
#18
(Original post by fortuneandero)
Oh I see, so I simply expand the y substitution and collect the like terms?
Yes and it will form a quadratic
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#19
(Original post by LowIQ)
Yes and it will form a quadratic
So then I would just factorise the quadratic and solve for x?
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2 years ago
#20
(Original post by fortuneandero)
So then I would just factorise the quadratic and solve for x?
Don’t forget to subtract 34 from both sides then factorise
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