Find the co ordinate of the foot of perpendicular from $(a,0)$ on the line $y=mx+\dfrac {a}{m}$
My Attempt:
$$y=mx+\dfrac {a}{m}$$
$$mx-y+\dfrac {a}{m}=0$$
The equation of the line perpendicular to $mx-y+\dfrac {a}{m}=0$ is $x+my+k=0$ where $k$ is an arbitrary constant. Since the line passes through $(a,0)$ , we have:
$$a+0+k=0$$
$$k=-a$$
Then, $x+my-a=0$ is the required equation. Now, how do I proceed further?
Best Answer
To continue from where you stopped, all you need to do is solve the system of equations $$y=mx+\frac am\\x+my-a=0$$ which, if you don’t know how to do already, you can find any number of explanations on the Internet with a simple search. You might start by substituting for $y$ in the second equation and solving it for $x$.
Another approach is to use the formula for the (signed) distance from a point to a line. Using $(m,-1)$ as the normal vector of the line (which you can read directly from its equation), we get $$d=-{ma-0+a/m\over\sqrt{m^2+(-1)^2}}=-\frac am\sqrt{m^2+1}.$$ The required point is this distance in the direction of the normal from the point $(a,0)$: $$(a,0)-\frac am\sqrt{m^2+1}{(m,-1)\over\|(m,-1)\|}=(a,0)-\frac am(m,-1)=\left(0,\frac am\right).$$