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A Level Maths: Coordinate Geometry

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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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.
Reply 2
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Tia.Rose
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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.
(edited 5 years ago)
Reply 3
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razzor
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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?
Reply 4
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Tia.Rose
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Would it be root 34 = root (x^2 + y^2) ?
Reply 5
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LowIQ
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Original post by fortuneandero
Would it be root 34 = root (x^2 + y^2) ?


Yes
Reply 6
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Tia.Rose
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But I don’t understand where I would go from there to find two possible positions of point P
Reply 7
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razzor
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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.
Reply 8
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LowIQ
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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
Reply 9
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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.
Reply 10
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Tia.Rose
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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
170D78F6-33B3-45F4-BCD9-FA139E9E7B7C.jpg.jpeg17.2KB
Reply 11
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LowIQ
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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
Reply 12
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LowIQ
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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
Reply 13
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Tia.Rose
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So it would be root 34 = root (x^2 + (4-3x)^2) ?
Reply 14
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LowIQ
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Yes but forget the root as it’s the same on both sides
Reply 15
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Tia.Rose
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Like this?
Attachment not found
(edited 5 years ago)
Reply 16
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Tia.Rose
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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?
Reply 17
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LowIQ
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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
Reply 18
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Tia.Rose
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Original post by LowIQ
Yes and it will form a quadratic


So then I would just factorise the quadratic and solve for x?
Reply 19
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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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