Q.Find the points on the x-axis, whose distances from the line $\dfrac{x}{3} + \dfrac{y}{4} = 1$ are $4$ units.

Write a general point on the line, force the vector from the given point to it to be perpendicular to the line's direction, solve for the parameter, then read off the foot, distance, and perpendicular's equation.

The line is $\dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+10}{8}=t$, passing through $(1,-1,-10)$ with direction ratios $(2,-3,8)$. A general point on it is

$Q=(1+2t,\,-1-3t,\,-10+8t).$

The vector from the given point $P(1,0,0)$ to $Q$ is

$\vec{PQ}=(2t,\;-1-3t,\;-10+8t).$

For $Q$ to be the foot of the perpendicular, $\vec{PQ}$ must be perpendicular to the line's direction $(2,-3,8)$, i.e. their dot product is zero:

$2(2t)+(-3)(-1-3t)+8(-10+8t)=0$

$4t+3+9t-80+64t=0$

$77t-77=0\ \Rightarrow\ t=1.$

Foot of perpendicular: substituting $t=1$: $Q=(1+2,\,-1-3,\,-10+8)=(3,-4,-2)$.

Distance: $\vec{PQ}=(3-1,\,-4-0,\,-2-0)=(2,-4,-2)$, so

$|PQ|=\sqrt{2^2+(-4)^2+(-2)^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt6.$

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Find the foot of the perpendicular from the point to the line (by making the vector from a general point on the line to the given point perpendicular to the line's direction), then the image is the point such that this foot is the midpoint of the given point and its image.

Line: $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}=t$, so a general point on the line is $Q(t)=(t,\,1+2t,\,2+3t)$, with direction ratios $(1,2,3)$.

Let $P=(1,6,3)$.

Find the foot of the perpendicular $F$: The vector $\overrightarrow{QP} = (1-t,\ 6-(1+2t),\ 3-(2+3t)) = (1-t,\ 5-2t,\ 1-3t)$ must be perpendicular to the direction $(1,2,3)$:

$(1-t)(1)+(5-2t)(2)+(1-3t)(3)=0$

$1-t+10-4t+3-9t=0$

$14-14t=0 \;\Rightarrow\; t=1$

So the foot of the perpendicular is $F=Q(1)=(1,\,1+2,\,2+3)=(1,3,5)$.

Find the image $P'$: Since $F$ is the midpoint of $P$ and its image $P'$:

$F=\dfrac{P+P'}{2} \;\Rightarrow\; P'=2F-P$

$P' = (2(1)-1,\ 2(3)-6,\ 2(5)-3) = (1,\,0,\,7)$

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