Find the co ordinates of the foot of perpendicular from $(a,0)$ on the line $y=mx +\dfrac {a}{m}$.

Question

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?

Answer 1

You are looking for $x$ and $y$ so that $$ y = mx + \frac{a}{m} $$ (point lies on original line) and $$ x+my-a=0 $$ (point lies on the perpendicular through $a$). To find $x,y$ that satisfy both of these equations, solve them simultaneously.

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Let's start by eliminating $y$. The first equation gives a value for $y$ in terms of $x$, and we can put this into the second equation and solve it for $x$: $$ x+m\left(mx + \frac{a}{m} \right) - a = 0 \\ x(1+m^2) = 0, $$ so $x=0$ (since $m^2 \geq 0$ the other factor cannot be zero). But then substituting this value of $x$ into the first equation, $y=a/m$, so the point is $(0,a/m)$ (this is also consistent with the second equation; it's always worth back-substituting to check you have the right solution).

Answer 2

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 $(a,0)$: $$(a,0)-\frac am\sqrt{m^2+1}{(m,-1)\over\|(m,-1)\|}=(a,0)-\frac am(m,-1)=\left(0,\frac am\right).$$

Answer 3

Ok you've found out the equation which is $\perp$ to $y=mx+\dfrac{a}{m}$ and the equation is $x-my+a=0\text{ or, } y=\dfrac{x}{m}+\dfrac{a}{m}$.

From these two equation we can write: $$mx+\dfrac{a}{m}=\dfrac{x}{m}+\dfrac{a}{m}\Rightarrow x\left(m-\dfrac{1}{m}\right)=0\Rightarrow x=0$$ When $x=0$ then form $y=mx+\dfrac{a}{m}$ or, $y=\dfrac{x}{m}+\dfrac{a}{m}$ we get, $y=\dfrac{a}{m}$.

So the point is: $(x,y)=\left(0,\dfrac{a}{m}\right)$.

I've solved these two straight line equations and found out $(x,y)=\left(0,\dfrac{a}{m}\right)$; now the question is which point is that? This is the point where the two straight lines intersect with each other and that is what you've asked (The coordinates of the foot of the perpendicular). Try to make the diagram and I'm sure that you can find it out.