This is the question:
Find the line through $(-1,4)$ for which the distance to $(6,3)$ is $5$
The answer is:
$y-4=-4/3(x+1)$ and $y-4=3/4(x+1)$
I do not know how to get this answer.
Started with finding a line through $(6,3)$ and perpendicular to the line through $(-1,4)$
Tried to find the intersection between the two lines
Used the equation for the distance between a point and a line.
But I cannot solve it as there are too many unknowns.
On
The equation of any line passing through $(-1,4)$ is $$\frac{y-4}{x+1}=m\iff mx-y+m+4=0$$
The distance of $(6,3)$ from this straight line is $$\frac{|6m-3+m+4|}{\sqrt{m^2+(-1)^2}}=\frac{|7m+1|}{\sqrt{m^2+1}}$$ which is $=5$(here)
Squaring we get $$(7m+1)^2=25(m^2+1)\iff 24m^2+14m-24=0$$
Solve for $m$
On
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\begin{align} y &=4 + m(x + 1)\ \imp\ \min\braces{\root{\pars{x - 6}^{2} + \braces{\bracks{m\pars{x + 1} + 4}- 3}^{2}}} = 5 \end{align}
\begin{align} \min\braces{\pars{x - 6}^{2} + \pars{mx + m + 1}^{2}} = 25 \end{align} It determines the value of $x$ as a function of $m$ such that: $$ x = -\,{m^{2} + m - 6 \over m^{2} + 1} $$
Then $$ {\pars{7m + 1}^{2} \over m^{2} + 1} = 25\quad\imp\quad \left\lbrace\begin{array}{rcrcr} m_{-} & = & -\,{4 \over 3} & \approx & -1.3333 \\[3mm] m_{+} & = & {3 \over 4} & = & 0.75 \end{array}\right. $$
Obviously, there are two lines.
Another way to think about this is to consider the circle with $(6,3)$ as the center and radius $5$. $$(x-6)^2+(y-3)^2=25$$ Now the lines you are seeking will be tangent lines to this circle from the point $(-1,4)$. So it boils down to finding the equation of the two tangent lines to this circle.