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# Straight Lines Algebra Problem watch

1. The question in full is:

15) A line is drawn through the point to cut the line in and the line in .
If , find the coordinates of and .

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I have done a sketch of the lines and and plotting can see there are two solutions. One instance is in the centre of and and in the other it is not (i.e. for one solution and for the other).

I let and and therefore and . Then doing distance between two points equations produced:

and as then .

Using fact that gradients must be the same can produce another equation in a and c, . The problem I have is that if I then substitute these together I end up with a quartic in c which doesn't seem like the correct path to continue on. I feel I'm likely missing something fundamental. Appreciate any pointers on the routes I should be following.
2. (Original post by Franzen)
I feel I'm likely missing something fundamental. Appreciate any pointers on the routes I should be following.
I would consider APQ to be the only valid combination.

As before you have a general point on the first line,

NOTE: IF AQ=2AP, then the difference in x-coordinates of A and Q will be twice the difference in x-coordinates of A and P. Similarly the y-coordinates.

So, you can work out the coordinates of Q in terms of "a".

This must lie on the 2nd line, so sub into the eqn of 2nd line and solve for a.

Edit: I get

Spoiler:
Show

a=4

3. (Original post by ghostwalker)
I would consider APQ to be the only valid combination.

As before you have a general point on the first line,

NOTE: IF AQ=2AP, then the difference in x-coordinates of A and Q will be twice the difference in x-coordinates of A and P. Similarly the y-coordinates.

So, you can work out the coordinates of Q in terms of "a".

This must lie on the 2nd line, so sub into the eqn of 2nd line and solve for a.

Edit: I get

Spoiler:
Show

a=4

Tyvm ghostwalker Got through to same solution but back of book gives other case too. Will try to apply similar rationale for those pair of P and Q.
4. (Original post by Franzen)
Tyvm ghostwalker Got through to same solution but back of book gives other case too. Will try to apply similar rationale for those pair of P and Q.
Fair enough. I wouldn't have consider the other case valid, unless they said the length of AP, or used |AP|. But same methodology will work.

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