$P$ and $Q$ are two points on the line...

Question

$P$ and $Q$ are the two points on the line $x-y+1=0$ such that each of them is $5$ units from the origin. Find the co ordinates of two points.

My Attempt;

Let $P(a,b)$ and $Q(a,b)$ be the co ordinates of two points.

Then $$x-y+1=0$$ We can write:

$$a-b+1=0$$ $$a-b=-1$$

And,

$$c-d+1=0$$ $$c-d=-1$$

Thus,

$$a-b=c-d$$.

I got stuck at here. Please help me to continue.

Thanks in advance.

Answer 1

I think you are not understanding the problem:

You have to find points on the line, which have distance $5$ from the origin.

This boils down to solving the system \begin{align}& x-y+1=0\text{ : being on the line}\\ &x^2+y^2=5^2 \text{ : having distance $5$ from the origin}\end{align} Which can be easily solved by plugging the first equation into the second.

Graphically the two points are the instersections of the above mentioned line with the circle of radius $5$:

Answer 2

Consider that the line is passing through $(-1,0)$ and $(0,1)$, so the midpoint is $Q=(-1/2,1/2)$ which has a distance from the origin $h=\sqrt{2} /2$. Note that the segment from the origin to $Q$ ($OQ$) is orthogonal to the line.
Take a point $P$ on the line, indicate with $d$ its distance from $Q$, while the distance from the origin shall be five. Then $d= \sqrt{(25-h^2)}= 7\sqrt{2} /2$.

Hence $P=(-1/2\pm d\sqrt{2} /2, 1/2\pm d\sqrt{2} /2)=(-1/2\pm 7/2,1/2\pm 7/2)$.