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Ratios Question (1 Viewer)
- Thread starter Ambility
- Start date
What was your working? Maybe we can see what you did wrong (if anything).I believe the answer to this question is 1:2, but the answers say it is 2:1. Could someone tell me what they get as their answer?
$ and Q is the point $(-1,-2)$. Find the ratio of PC:CQ.$)
I'm pretty sure I'm right, but I'll give you my working just so you can check.
http://i.imgur.com/jlpR81R.jpg
Your answer is right, as the others have also said.
Another approach without finding the intersection point C.
Signed distances from P and Q to the given straight line in the ratio:
 + 3(2) -4}{\sqrt {etc etc}} : \frac {2(-1)+3(-2)-4}{\sqrt{etc etc}} = 6 : -12 = 1 : -2 )
The opposite signs show that P and Q are on opposite sides of the given line, as we would require.
Now the ratio above, ignoring signs, is for the perpendicular distances of P and Q from the line. But this is the same ratio as PC:CQ by virtue of similar triangles formed by the 2 dropped perpendiculars and PC and QC.
Signed distances from P and Q to the given straight line in the ratio:
The opposite signs show that P and Q are on opposite sides of the given line, as we would require.
Now the ratio above, ignoring signs, is for the perpendicular distances of P and Q from the line. But this is the same ratio as PC:CQ by virtue of similar triangles formed by the 2 dropped perpendiculars and PC and QC.
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