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I'm given a point $P(1, 4)$ and its distance from a line passing through the intersection of the lines $x-2y+3=0$ and $x-y-5=0$ is 4 units. How do I find its equation?

Here is how I tried to solve it,

Solving the two equations gives their point of intersection as $(13, 8)$. Then if I use the two point formula between $(1, 4)$ and $(13, 8)$ the equation comes out as $4x-12+44=0$, but the answer in my textbook is not similar to mine. Can anyone tell me what am I doing wrong?

The question as in my textbook is

The distance of a point $(1, 4)$ from a line passing through the intersection of the lines $x-2y+3=0$ and $x-y-5=0$ is 4 units. Find its equation.

The answer of the above question as in my textbook is $3x-4y-7=0$, $y-8=0$

What am I doing wrong here?

What I think is that maybe I have to find the equation of line through the intersection of the given lines ($x-2y+3=0$ and $x-y-5=0$) and then use the perpendicular distance formula between that line, the point $(1, 4)$ (and the distance is also given) and then work out the equation. But how do I find the equation of line through the intersection of the given lines,

I know that it is given by $l_1 + kl_2 = 0$ where $l_1$ is equation of line one and $l_2$ is equation of line 2. But for this equation a condition is necessary to find $k$.

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Note that the problem specifies that the distance between point $P$ and the line passing through the point of intersection of the given lines is 4 units. Then it asks for the equation of the line passing through the point of intersection of the given lines which happens to be 4 units away from point P. What you seemed to aim for is finding the equation of the line passing through P and the point of intersection of the two lines (which isn't correct: one need only look at the y-coordindinates of the point of intersection, and point P, to see that that line has slope = 0 ... –  amWhy May 25 '11 at 2:21
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Yes...I just saw your edits, you're going to need make use of the fact that the shortest distance between a point P and line l (p not on l) is measured along the line intersecting P which is perpendicular to l. –  amWhy May 25 '11 at 2:35
    
@amWhy: But to find the equation of the line passing through the point of intersection of the given lines, I use the following formula i.e. $l_1 + kl_2 = 0$ but for this I need an additional condition to find the value of $k$, how do I do that? The point of intersection of the two given lines (as I found it by solving the equations) is $(13, 8)$ so should I put the values $x=13$ $y=8$ in $l_1 + kl_2 = 0$, get $k$ and then solve it? Is this correct? –  cardano May 25 '11 at 2:38
    
The slope of which line? I know that slope = $m = (y_2 - y_1)/(x_2 - x_1)$ and $m_1 * m_2 = -1$ for perpendicular lines. –  cardano May 25 '11 at 2:51
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Ross is correct: you'll want to use the perpendicular distance formula to solve for the slope of the line passing through the point of intersection of two given points, which happens to be 4 units away from P... –  amWhy May 25 '11 at 2:55

1 Answer 1

up vote 2 down vote accepted

The intersection of the two lines is not $(13,4)$. Plugging that point into the first line gives $8$ and into the second gives $4$ for the left hand side, not $0$, so you need to redo the solution for the point. The $x$ coordinate is correct, but not $y$. Then there are many lines through that point and you are to use the distance information to pick two of them. If the intersection point is $(a,b)$, the point slope formula would be $y-b=m(x-a)$ and you need to use your formula for the distance between $(1,4)$ and this line to pick $m$.

Added: I meant the line through $(13,8)$ of slope $m$, so $y-8=m(x-13)$ or $-mx+y+13m-8=0$. You need the distance from $(1,4)$ to this line to be $4$. Mathworld gives the distance from the point $(x_0,y_0)$ to the line $ax+by+c=0$ as $\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ so you have $4=\frac{|-m+4+13m-8|}{\sqrt{m^2+1}}$. Moving the $\sqrt{m^2+1}$ to the other side and squaring, you should find two values for $m$.

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Yes that was a mistake, I corrected it. Thanks :) –  cardano May 25 '11 at 2:38
    
you need to use your formula for the distance between (1,4) and this line to pick $m$. Which line are you referring to here? and which formula to get $m$? –  cardano May 25 '11 at 2:42
    
If you meant that $y-4=m(x-1)$, this becomes $mx-y+4-m=0$ and then use the perpendicular distance formula between this line and the point $(1,4)$, it becomes $4 = \frac{m-4+4-m}{\sqrt{m^2+1}}$, there are no solutions for this. $4 = \frac{0}{\sqrt{m^2+1}}$ and $4 = 0$ which is not true –  cardano May 25 '11 at 3:33
    
Thank you Ross, this helped and I got the correct answers :) –  cardano May 25 '11 at 3:54

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