Question

Points on the line $x+y=4$ that lies at a unit distance from the line $4x+3y-10=0$, are

• A. $(3,1)$ and $(-7,11)$
• B. $(03,7)$ and $(2,2)$
• C. $(-3,7)$ and $(-7,11)$
• D. None of these

Answer 1

Answer: (A)

$(3,1)$ and $(-7,11)$

Any point on the line is $(t,4-t)$
According to the question,
$\frac{|4t+3(4-t)-10|}{\sqrt{4^2+3^2}}=1$
$\Rightarrow |t+2|=5$
$\Rightarrow t+2=\pm 5$
$\Rightarrow t=3-7$
Hence, the required points are $(3,1)$ and $(-7,11)$.