Find the coordinates of the foot of the perpendicular I'm quite confused about this question which states:
Find the coordinates of the foot of the perpendicular from the point $(5, 7)$ on the straight line which joins the points $(6, -1)$ and $(1, 6)$.

How would I go about doing this problem? I'm also curious about ways to improve understanding for the wording in mathematics problems, as I am struggling a bit with these type of problems(English is not my main language).
Many Thanks!
 A: The equation of the striaght line $BC$ with $B=(x_1,y_1)$ and $C=(x_2,y_2)$ is by the two point form
$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$
Putting the values
$\Rightarrow y+1=\frac 7{-5}(x-6)$
$7x+5y=37  \quad (1)$
Again equation of a straight line perpendicular to $ax+by+c=0$ is given by
$bx-ay+k=0$   where $k$ is a constant to be determined.
So, the equation of straight line perpendicular to $BC$ and passing through $A(5,7)$ is
$5x-7y+k=0$ and since this passes through $A$ , we have
$k=24$
So the line is $5x-7y+24=0 \quad (2)$
Can you find the intersection point of lines $(1)$ and $(2)$?
Note : Finding foot of the perpendicular from a given external  point to a given line is essentially finding the intersection point  between the given line and the perpendicular line passing through the given point.
A: There are several ways of obtaining the coordinates of the
from the point $A$ to the line $BC$. Here is a simple step-by-step approach.

Consider $\triangle ABC$. The sought point is $H$,
and $|AH|$ is the altitude of $\triangle ABC$.
Let's find the length $|AH|$ first.
To get it, we can use the formula for the area of triangle:
\begin{align}
S_{ABC}&=\tfrac12\,|BC|\cdot|AH|
\tag{1}\label{1}
,\\
\text{so }\quad
|AH|&=\frac{2\,S_{ABC}}{|BC|}
\tag{2}\label{2}
.
\end{align}
In \eqref{2} we know that
\begin{align}
|BC|&=\sqrt{(6-1)^2+(-1-6)^2}
=\sqrt{74}
\tag{3}\label{3}
.
\end{align}
And we can also use another formula for the area of triangle,
based on the coordinates of its vertices,
\begin{align}
S_{ABC}&=\tfrac12\,|(B_x-A_x)(C_y-A_y)-(C_x-A_x)(B_y-A_y)|
=\tfrac{33}2
\tag{4}\label{4}
.
\end{align}
Thus we get
\begin{align}
|AH|&=\frac{2\cdot \tfrac{33}2}{\sqrt{74}}
=\tfrac{33}{74}\,\sqrt{74}
\tag{5}\label{5}
.
\end{align}
To get the coordinates of the point $H$
we just need to add some vector $v$ of length $|AH|$
to the coordinates of the point $A$,
and we also know that this vector is perpendicular to $BC$,
that is, $v\perp BC$.
Given that vector $BC=(-5,7)$,
we have vector$(-7,-5)$
as $BC$ rotated $90^\circ$ counterclockwise.
To make it the vector $v$ we are looking for,
we need to normalize it (divide it by its length $=|BC|$)
and multply by $|AH|$, so
\begin{align}
v&=
(-7,-5)\cdot 
\frac{\tfrac{33}{74}\,\sqrt{74}}{\sqrt{74}}
=(-\tfrac{231}{74},\, -\tfrac{165}{74})
\tag{6}\label{6}
.
\end{align}
Finally,
\begin{align}
H&=
A+v=
(\tfrac{139}{74},\, \tfrac{353}{74})
\tag{7}\label{7}
.
\end{align}
A: There is another 2 way that is relevant to computing and drawing on computers. Of course, you have to know it mathematically before you can tell computers. Let a=BC, b=CA, c=AB as usual.
Way 1: The foot H on the segment BC of the altitude AH can be computed via
$$H=\dfrac{1}{a^2+c^2-b^2}B+\dfrac{1}{a^2+b^2-c^2}C$$
that is the barycentric coordinate of H with respect to BC is
$$\left(\dfrac{1}{a^2+c^2-b^2}:\dfrac{1}{a^2+b^2-c^2}\right).$$
Way 2: The foot K on the segment BC of the altitude AH (so $H\equiv K$) can be computed via dot product
$$K=C+\left<A-C,\frac{B-C}{BC}\right>\cdot \frac{B-C}{BC}$$
where $\frac{B-C}{BC}$ is the unit vector of $B-C$.
I illustrate below using Asymptote.

// http://asymptote.ualberta.ca/
import math;
pair barycentric(pair A=(0,0), pair B=(0,0), real a=1, real b=0){
return (a*A+b*B)/(a+b);}

unitsize(7mm);
pair A=(5,7), B=(6,-1), C=(1,6);
real a=abs(B-C), b=abs(C-A), c=abs(A-B);
pair H=barycentric(B,C,1/(a^2+c^2-b^2),1/(a^2+b^2-c^2));

pair K=C+dot(A-C,unit(B-C))*unit(B-C); 


draw(A--K,blue+1pt);
draw(A--H,red);

draw(A--B--C--cycle);
label("$A$",A,plain.E);
label("$B$",B,plain.E);
label("$C$",C,plain.W);
label("$H$",H,plain.SW,red);
label("$K$",K,plain.N,blue);
shipout(bbox(5mm,invisible));
// write out approximately Cartesian coordinates 
// of H and K
write("Two ways of computing the foot of the altitude");
write("H = ("+ string(H.x,10)+","+string(H.y,10)+") approximately");
write("K = ("+ string(K.x,10)+","+string(K.y,10)+") approximately");

