Find the point on the line 6x+5y+5=0 which is closest to the point (1,-5)

Question

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.

Answer 1

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?

Answer 2

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}

Answer 3

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 $\left(\frac{145}{61},-\frac{235}{61}\right).$

Answer 4

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.