Problem solving coordinate geometry question

**Ahmadtaz** · January 8th 2012, 10:32 AM

Hello All,

I wonder if anyone can help me with the question below.

A line is drawn through the point A(1,2) to cut the line 2y = 3x - 5 in P and the line x + y = 12 in Q.

If AQ = 2AP, find the coordinates P and Q.

Try as I might, and I even have the answer to the value of P and Q, but I cant understand how anyone would be able to find out four unknowns (ie the cordinates of P and Q) with just those three equations.

Thanks in advance!

**Ahmadtaz** · January 8th 2012, 10:36 AM

Spoiler Alert:

The answer to P and Q for anyone who might be able to help if given the oppurtunuity to work backwards is...
$(\frac{2}{5}, \frac{-19}{10})$
and
$(2\frac{1}{5}, 9\frac{4}{5})$

**Plato** · January 8th 2012, 10:54 AM

Originally Posted by Ahmadtaz

A line is drawn through the point A(1,2) to cut the line 2y = 3x - 5 in P and the line x + y = 12 in Q.
If AQ = 2AP, find the coordinates P and Q.

I would not like to solve this system. BUT.

If $P=(p_x,p_y)~\&~Q=(q_x,q_y)$ then

$2p_y=3p_x-5~\&~q_x+q_y=12$ from incidence.

$4[(p_x-1)^2+(p_y-2)^2]=(q_x-1)^2+(q_y-2)^2$ from distance.

$\frac{p_y-2}{p_x-1}= \frac{q_y-2}{q_x-1}$ from slope.

**Ahmadtaz** · January 9th 2012, 11:20 AM

Originally Posted by Plato

I would not like to solve this system. BUT.

If $P=(p_x,p_y)~\&~Q=(q_x,q_y)$ then

$2p_y=3p_x-5~\&~q_x+q_y=12$ from incidence.

$4[(p_x-1)^2+(p_y-2)^2]=(q_x-1)^2+(q_y-2)^2$ from distance.

$\frac{p_y-2}{p_x-1}= \frac{q_y-2}{q_x-1}$ from slope.

Sorry Plato, I see that you've put P and Y into the two equations for the lines, but could you explain how you get to this please?
$4[(p_x-1)^2+(p_y-2)^2]=(q_x-1)^2+(q_y-2)^2$

**Ahmadtaz** · January 9th 2012, 11:29 AM

Apologies, I understand $4[(p_x-1)^2+(p_y-2)^2]=(q_x-1)^2+(q_y-2)^2$ however, I don't understand how knowing the slope (your final equation) will help me solve the problem. Can you elaborate please?

**Plato** · January 9th 2012, 12:05 PM

Originally Posted by Ahmadtaz

Apologies, I understand $4[(p_x-1)^2+(p_y-2)^2]=(q_x-1)^2+(q_y-2)^2$ however, I don't understand how knowing the slope (your final equation) will help me solve the problem. Can you elaborate please?

That gives you four equations in four unknowns.

**mithgar** · January 9th 2012, 06:16 PM

Originally Posted by Ahmadtaz

Spoiler Alert:

The answer to P and Q for anyone who might be able to help if given the oppurtunuity to work backwards is...
$(\frac{2}{5}, \frac{-19}{10})$
and
$(2\frac{1}{5}, 9\frac{4}{5})$

The answer should be P(4, 3.5) and Q(7,5). And it can be done in a easier method.

Since AQ = 2AP, P is the mid-point of AQ.

Let point Q=(x, 12-x)

P, which is the mid-point, = $(\frac{x+1}{2} , \frac{12-x+2}{2})$

Substitute P into the line 2y=3x-5, and you have

$2(\frac{14-x}{2}) = 3(\frac{x+1}{2}) - 5$

solve and you will get the points.

**mithgar** · January 10th 2012, 05:14 PM

Originally Posted by earboth

I've done this question in a completely different way:

1. Draw a sketch of the lines $a:y = \frac32 x -5$ and $b:y=-x+12$ and the point A.

2. Determine the point of intersection of a and b: $a\cap b = \{S\}$

3. Determine the equation of a line parallel to a through A $p:y=\frac32 x +\frac12$

4. Determine the point of intersection of p and b: $p\cap b = \{R\}$

5. You reach S if you go from R 2.2 units right and 2.2 units down. Now go from S 4.4 units right and 4.4 units down and you are at Q: $Q\left(\frac{56}5\ ,\ \frac45 \right)$ .

6. Determine the equation of the line AQ = l: $y = -\frac2{17} x + \frac{36}{17}$ . Determine the point of intersection of $l \cap a=\{P\}$

7. I've got $P\left(\frac{22}5\ ,\ \frac85 \right)$

Check your first equation. Should be. $y = \frac{3}{2} x - \frac{5}{2}$

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