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Question

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

A

$\left(3,1\right),\left(–7,11\right)$

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B

$\left(3,1\right),\left(7,11\right)$

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C

$\left(–3,1\right),\left(–7,11\right)$

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D

$\left(1,3\right),\left(–7,11\right)$

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Solution

## The correct option is A $\left(3,1\right),\left(–7,11\right)$The explanation for the correct answer.Solve for the points on the line $x+y=4$Let $P\left(h,k\right)$ lies on $x+y=4$$⇒h+k=4...\left(i\right)$We know the distance of the point $\left({x}_{1},{y}_{1}\right)$ from the line $ax+by+c=0$ is$d=\frac{\left|a{x}_{1}+b{y}_{1}+c\right|}{\sqrt{{a}^{2}+{b}^{2}}}$$⇒1=\frac{\left|4h+3k-10\right|}{\sqrt{{4}^{2}+{3}^{2}}}\phantom{\rule{0ex}{0ex}}⇒1=\frac{\left|4h+3k-10\right|}{5}\phantom{\rule{0ex}{0ex}}⇒4h+3k–10=±5\phantom{\rule{0ex}{0ex}}⇒4h+3k=15\dots \left(ii\right)\phantom{\rule{0ex}{0ex}}⇒4h+3k=5\dots \left(iii\right)$The distance of the point$P\left(h,k\right)$ from the line $4x+3y=10$is $1$unit.Solving $\left(i\right)$ and $\left(ii\right)$, we get $\left(h,k\right)=\left(3,1\right)$ $h+k=4...\left(i\right)$$4h+3k=15...\left(ii\right)$$⇒4h+3\left(4-h\right)=15\phantom{\rule{0ex}{0ex}}⇒h=15-12=3$Similarly,solving $\left(i\right)$ and $\left(iii\right)$, we get $\left(h,k\right)=\left(-7,11\right)$Hence, option(A) is the correct answer.

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