The coordinates of the two points lying on x+ | vedclass

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(B) Let the point on the line $x+y=4$ be $P(h, 4-h)$.The perpendicular distance from $P(h, 4-h)$ to the line $4x+3y-10=0$ is given by $d = \frac{|4h +

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

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(B) Let the point on the line $x+y=4$ be $P(h, 4-h)$.The perpendicular distance from $P(h, 4-h)$ to the line $4x+3y-10=0$ is given by $d = \frac{|4h + 3(4-h) - 10|}{\sqrt{4^2 + 3^2}}$.Given $d=1$, we have $\frac{|4h + 12 - 3h - 10|}{5} = 1$.$|h + 2| = 5$.This gives two cases:Case $1$: $h+2 = 5 \implies h = 3$. The point is $(3, 4-3) = (3, 1)$.Case $2$: $h+2 = -5 \implies h = -7$. The point is $(-7, 4-(-7)) = (-7, 11)$.Thus, the coordinates are $(3, 1)$ and $(-7, 11)$.

The coordinates of the two points lying on $x+y=4$ and at a unit distance from the straight line $4x+3y=10$ are

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