Question

Find points on the line x + y + 3 = 0 that are at a distance of 5 units
from the line x + 2y + 2 = 0

Answer 1

$\text{Let the required points on the line x+y=3=0 be (u,v) }$

$\text{Then, u+v+3=0}$......(1)

$\textbf{Given:}$

$\text{The distance of (u,v) from the line x+2y+2=0 = 5 units}$

$\implies\displaystyle|\frac{a\;x_1+b\;y_1+c}{\sqrt{a^2+b^2}}|=5$

$\implies\displaystyle|\frac{1(u)+2(v)+2}{\sqrt{1^2+2^2}}|=5$

$\implies\displaystyle|\frac{u+2v+2}{\sqrt{5}}|=5$

$\implies\;u+2v+2=\pm\;5\sqrt{5}$

$u+2v+2=5\sqrt{5}$......(2)

$u+2v+2=-5\sqrt{5}$......(3)

$\textbf{Solving (1) and (2):}$

$u+v+3=0$......(1)

$u+2v+2=5\sqrt{5}$......(2)

$\implies\;-v+1=5\sqrt{5}$

$\implies\;-v=5\sqrt{5}-1$

$\implies\;v=-5\sqrt{5}+1$

$(1)\implies\;u+(-5\sqrt{5}+1)+3=0$

$(1)\implies\;u=5\sqrt{5}-4$

$\textbf{Solving (1) and (3):}$

$u+v+3=0$......(1)

$u+2v+2=-5\sqrt{5}$......(3)

$\implies\;-v+1=-5\sqrt{5}$

$\implies\;-v=-5\sqrt{5}-1$

$\implies\;v=5\sqrt{5}+1$

$(1)\implies\;u+(5\sqrt{5}+1)+3=0$

$(1)\implies\;u=-5\sqrt{5}-4$

$\therefore\text{The required points are }\bf(5\sqrt{5}-4,-5\sqrt{5}+1)\text{ and }\bf(-5\sqrt{5}-4,5\sqrt{5}+1)$