Question

Points on $x+y=4$ which are at unit distance from $4x+3y=0$ are

• A. $(3,1),(-7,11)$
• B. $(3,1),(-9,13)$
• C. $(2,2),(-8,12)$
• D. $(-7,11),(-17,21)$

Answer 1

The correct option is D $(-7,11),(-17,21)$

Let point as line $(x+y=4)$ be $A(t,4-t)$

distance of A from $4x+3y=0$ is 1

$\frac{\left|4\right(t)+3(4-t\left)\right|}{\sqrt{4^2+3^2}}=1$

$|4t+12-3t|=5$

$|t+12|=5$

$t+12=5$

$t=-7$

So $A=(-7,11)$

$t+12=-5$

$t=-17$

$A=(-17,21)$

So points are (-7, 11) and (-17, 21)