Finding the set of points $(x,y)$ that have a distance of $7$ from line

Question

I have a line $y=x$ , and I have to find the set of points $(x,y)$
that have a distance of $7$ or less from this line.

So
$\sqrt{(x-a)^2 +(y-a)^2 }\leq 7$
$x^2 $$-2xa$ + $a^2 + $$y^2 $$-2ya + $$a^2\leq 7$
what should I do next?

Answer 1

Pick any point $A = (a,b)$ in the plane. Then draw lines parallel to $x-$axis and $y-$axis. They intersect the line at point $B = (a,a), C = (b,b)$. Note that the triangle $ABC$ is right-angled and isoscelec, as $AB = AC = |b-a|$. Now draw a perpendicular line to the line $x=y$ from $A$ and let it intersect the line at $D$. Now the distance from $A$ to the line $x=y$ is given by $AD$. A little bit of algebra gives us that: $AD = \frac{|b-a|}{\sqrt{2}}$. Therefore $A$ is in the wanted subset if $|b-a| \le 7\sqrt{2}$. Therefore the required set is:

$$\{(x,y) \big| |x-y| \le 7\sqrt{2}\}$$

Geometrically it corresponds to region bounded by the lines $y = x + 7\sqrt{2}$ and $y = x - 7\sqrt{2}$