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 $k$.
Best Answer
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 $m$.