Question

find points on the line y=x which are at a distance of 5 units from the line 4x +3y-1=0
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Answer 1

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$x+y=4\rightarrow y=4-x$

$4x+3y=10\rightarrow 4x+3y-10=0$

The distance between the general line $ax+by+c=0$ and an arbitrary point $M=(x_M,y_M)$ is:

${|ax_M+by_M+c|\over \sqrt{a^2+b^2}}$

And we know that M is located on $y=4-x \rightarrow y_M=4-x_M$ or $M=(x_M,4-x_M)$.

And have in mind that we want the distance to be 1 unit. Putting all these together:

${|ax_M+by_M+c| \over \sqrt{a^2+b^2}}=1 \rightarrow {|4x_M+3y_M-10| \over \sqrt{4^2+3^2}}=1$

$\rightarrow |4x_M+3y_M-10|=5 \rightarrow |4x_M+3(4-x_M)-10|=5$

$\rightarrow |x_M+2|=5$

Case A:

$x_M+2=5 \rightarrow x_M=3 \rightarrow y_M=1 \rightarrow A=(3, 1)$

Case B:

$x_M+2=-5 \rightarrow x_M=-7 \rightarrow y_M=11 \rightarrow B=(-7, 11)$

So the results are $A=(3, 1)$ and $B=(-7, 11)$.