Question

The points on the line $x+y=4$ which lie at a unit distance from the line $4x+3y=10$, are?

• A. $(3,1),(-7,11)$
• B. $(7,11),(2,2)$
• C. $(7,-11),(-3,7)$
• D. $(1,3),(-5,9)$

Answer 1

The correct option is A $(3,1),(-7,11)$

Let Point $P(x,y)$ is ar the line

$x+y=\mathrm{4...}\left(1\right)$

& have unique distance from the

line $4x+3y=10$

So, $|P|=\frac{|A{x}_1+B{y}_1+c|}{\sqrt{A^2+B^2}}$

here P is distance (which is parallel)

$A=4\&B=3,x_1=x\&y_1=y$

$c=-10$

$1=\frac{|4x+3y-10|}{\sqrt{4^2+3^2}}$

$|4x+3y-10|=5$

$4x+3y-10=\pm 5$

$4x+3y-10=5$ $4x+3y-10=-5$

$4x+3y-15=0\to 2\left(i\right)$ $4x+3y-5=0\to 2\left(ii\right)$

From (1) & 2(i)$\to x=3\&y=1$

From (1) & 2(ii)$\to x=-7\&y=11$

So, the points $(3,1)\&(-7,11)$

are on the line $x+y=4$ &

having unit distance from the

line $4x+3y=10$