Find the point on the line 6x+5y+5=0 which is closest to the point (1,-5) I tried putting y alone and got y=(-6x-5)/5. Which I then put into the distance formula sqrt((x-1)^2+(y+5) and substitute the number above in for y but my answer never comes out correct.. Wondering if I could get some help.
 A: Method$\#1:$
As the perpendicular distance is the shortest  find the equation of the perpendicular  of $$6x+5y+5=0$$ passing through $(1,-5)$
Find the intersection of the two lines.
Method$\#2:$
$$6x+5y+5=0\iff\dfrac x5=\dfrac{y+1}{-6}$$ $=k$(say)
$\implies x=\cdots, y=\cdots$
We need to minimize $$\sqrt{(x-1)^2+(y+5)^2}$$
$\iff $  to minimize $$(x-1)^2+(y+5)^2$$ which will be quadratic in $k$
Do you know how to find the possible extreme values of a quadratic in real?
A: Another method. Take a circle centered in $(1,-5)$ that is
$$(x-1)^2+(y+5)^2=d^2$$
The slope of tangent line is
$$y'=-\dfrac{f_x}{f_y}=-\dfrac{x-1}{y+5}$$
will be $-\dfrac65$, the slope of given line. Then it's sufficient to solve the system
\begin{cases}
6x+5y+5=0,\\
5x-6y=35.
\end{cases}
A: By C-S
$$\sqrt{(6^2+5^2)\left((x-1)^2+(y+5)^2\right)}\geq6(x-1)+5(y+5)=14.$$
Thus, $$\sqrt{(x-1)^2+(y+5)^2}\geq\frac{14}{\sqrt{61}}.$$
The equality occurs for $(x-1,y+5)||(6,5),$ which gives the needed point.
I got $\left(\frac{145}{61},-\frac{235}{61}\right).$
A: Hint:-
$(1,-5)$ projected on the point $(x, \frac{-5-6x}{5})$. Line passes through $(0,-1)$. Vector joining points $(0,-1)$and $(x, \frac{-5-6x}{5})$ is perpendicular to the Vector joining  $(x, \frac{-5-6x}{5})$ to $(1,-5)$. Find $x$ from the condition. 
