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What are the points on the x-axis whose perpendicular distance from the

Question:

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

 

Solution:

Given: perpendicular distance is 4 units and line

$\frac{x}{3}+\frac{y}{4}=1$

To find : points on the x-axis

Formula used:

We know that the length of the perpendicular from (m,n) to the line ax + by + c = 0 is given by,

$D=\frac{|a m+b n+c|}{\sqrt{a^{2}+b^{2}}}$

The equation of the line is $4 x+3 y-12=0$

Any point on the $x$-axis is given by $(x, 0)$

Here $m=x$ and $n=0, a=4, b=3, c=-12$ and $D=4$ units

$D=\frac{|4(x)+3(0)-12|}{\sqrt{4^{2}+3^{2}}}=4$

$D=\frac{|4 x-12|}{\sqrt{16+9}}=\frac{|4 x-12|}{\sqrt{25}}=\frac{|4 x-12|}{5}=4$

$|4 x-12|=4 \times 5=20$

$4 x-12=20$ or $4 x-12=-20$

$4 x=20+12$ or $4 x=-20+12$

$4 x=32$ or $4 x=-8$

$x=32 / 4=8$ or $x=(-8) / 4=-2$

(8,0) and (2,0)are the points on the x-axis whose perpendicular distance from the line is 4 units

 

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