What are the points on the y-axis whose dista | vedclass

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Question

(A) Let $(0, b)$ be the point on the $y$-axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is $4$ units.The given line can be written as $

Vedclass · Accepted Answer

$\left(0, \frac{32}{3}\right)$ and $\left(0, -\frac{8}{3}\right)$

Vedclass · Answer

(A) Let $(0, b)$ be the point on the $y$-axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is $4$ units.The given line can be written as $4x + 3y - 12 = 0$ $(1)$.On comparing equation $(1)$ to the general equation of a line $Ax + By + C = 0$,we obtain $A = 4$,$B = 3$,and $C = -12$.The perpendicular distance $(d)$ of a line $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Substituting the values,we get:$4 = \frac{|4(0) + 3(b) - 12|}{\sqrt{4^2 + 3^2}}$$4 = \frac{|3b - 12|}{5}$$20 = |3b - 12|$This implies $3b - 12 = 20$ or $3b - 12 = -20$.Case $1$: $3b = 32 \Rightarrow b = \frac{32}{3}$.Case $2$: $3b = -8 \Rightarrow b = -\frac{8}{3}$.Thus,the required points are $\left(0, \frac{32}{3}\right)$ and $\left(0, -\frac{8}{3}\right)$.

What are the points on the $y$-axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is $4$ units?

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