Question

The sum of squares of the abscissas of all the points on the line $x+y=4$ that lie at a unit distance from the line $4x+3y-10=0$ is

Answer 1

Any point on the line $x+y=4$ can be taken as $\left(x_{1}, 4-x_{1}\right)$.
As it is at a unit distance from the line $4 x+3 y-10=0$,
we get, $\frac{\left|4 x_{1}+3\left(4-x_{1}\right)-10\right|}{\sqrt{\left(4^{2}+3^{2}\right)}}=1$
$\Rightarrow\left|x_{1}+2\right|=5$
$\Rightarrow x_{1}+2=\pm 5 $
$\Rightarrow x_{1}=3 $ or $-7$
$\therefore$ Required sum of the squares $=9+49=58$