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

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(B) Let the point on the line $x+y+3=0$ be $(k, -3-k)$.Given that the perpendicular distance from this point to the line $x+2y+2=0$ is $\sqrt{5}$.Usin

Vedclass · Accepted Answer

$(1, -4)$

Vedclass · Answer

(B) Let the point on the line $x+y+3=0$ be $(k, -3-k)$.Given that the perpendicular distance from this point to the line $x+2y+2=0$ is $\sqrt{5}$.Using the distance formula $d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$,we have:$\frac{|k + 2(-3-k) + 2|}{\sqrt{1^2+2^2}} = \sqrt{5}$$\frac{|k - 6 - 2k + 2|}{\sqrt{5}} = \sqrt{5}$$|-k - 4| = 5$$|k + 4| = 5$This gives two cases:$k + 4 = 5 \Rightarrow k = 1$$k + 4 = -5 \Rightarrow k = -9$For $k = 1$,the point is $(1, -3-1) = (1, -4)$.For $k = -9$,the point is $(-9, -3-(-9)) = (-9, 6)$.

The coordinates of the point on the line $x+y+3=0$,whose distance from the line $x+2y+2=0$ is $\sqrt{5}$ units,are

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