Answer :
Given:
The equation of line is
[tex]4x+3y=12[/tex]
To find:
The points on y-axis which are at a distance of 4 units from the given line.
Solution:
The point lie on the y-axis. So, their x-coordinate must be zero.
Let the points are in the form of (0,k).
The distance between a point [tex](x_0,y_0)[/tex] and a line [tex]ax+by+c=0[/tex] is
[tex]d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}[/tex]
The given equation can be written as
[tex]4x+3y-12=0[/tex]
The distance between (0,k) and [tex]4x+3y-12=0[/tex] is 4 units.
[tex]4=\dfrac{|4(0)+3(k)-12|}{\sqrt{4^2+3^2}}[/tex]
[tex]4=\dfrac{|0+3k-12|}{\sqrt{16+9}}[/tex]
[tex]4=\dfrac{|3k-12|}{\sqrt{25}}[/tex]
[tex]4=\dfrac{|3k-12|}{5}[/tex]
Multiply both sides by 5.
[tex]20=|3k-12|[/tex]
[tex]\pm 20=3k-12[/tex]
[tex]12\pm 20=3k[/tex]
[tex]\dfrac{12\pm 20}{3}=k[/tex]
Now,
[tex]k=\dfrac{12-20}{3}\text{ and }k=\dfrac{12+20}{3}[/tex]
[tex]k=\dfrac{-8}{3}\text{ and }k=\dfrac{32}{3}[/tex]
Therefore, the two points are [tex]\left(0,\dfrac{-8}{3}\right)\text{ and }\left(0,\dfrac{32}{3}\right)[/tex].