The coordinates of a point on the line x+y+1= | vedclass

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(B, D) Let $(h, k)$ be a point on the line $x+y+1=0.$ Since the point lies on the line,we have $h+k+1=0,$ which implies $h = -k-1.$ The perpendicular

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$(-1, 0)$

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(B, D) Let $(h, k)$ be a point on the line $x+y+1=0.$ Since the point lies on the line,we have $h+k+1=0,$ which implies $h = -k-1.$ The perpendicular distance from $(h, k)$ to the line $3x+4y+2=0$ is given by $\frac{|3h+4k+2|}{\sqrt{3^2+4^2}} = \frac{1}{5}.$ Substituting $h = -k-1$ into the distance formula: $\frac{|3(-k-1)+4k+2|}{5} = \frac{1}{5}$ $|-3k-3+4k+2| = 1$ $|k-1| = 1$ This gives two cases: $k-1 = 1 \implies k=2,$ then $h = -2-1 = -3.$ Point is $(-3, 2).$ $k-1 = -1 \implies k=0,$ then $h = 0-1 = -1.$ Point is $(-1, 0).$ Thus,the required points are $(-3, 2)$ and $(-1, 0).$ Both options $B$ and $D$ are correct.

The coordinates of a point on the line $x+y+1=0,$ which is at a distance of $\frac{1}{5}$ unit from the line $3x+4y+2=0,$ are

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