Q.Find the image of the point $(1, 2)$ in the straight line $3x + 4y - 1 = 0$.

The foot of the perpendicular from a point to a line is found using the standard formula, moving along the line's normal direction by the right proportional amount.

For a line $ax+by+c=0$ and point $(x_1,y_1)$, the foot of perpendicular $(x_0,y_0)$ is given by:

$\dfrac{x_0-x_1}{a} = \dfrac{y_0-y_1}{b} = \dfrac{-(ax_1+by_1+c)}{a^2+b^2}$

Here, line: $3x-4y+12=0$ ($a=3,b=-4,c=12$), point $(x_1,y_1)=(4,1)$.

$ax_1+by_1+c = 3(4)-4(1)+12 = 12-4+12=20$

$a^2+b^2 = 9+16=25$

$k = \dfrac{-(20)}{25} = -\dfrac{4}{5}$

…

Since $PQ\parallel$ normal $(a,b)$, set $h-x_1=\lambda a$, $k-y_1=\lambda b$; forcing $Q$ onto the line fixes $\lambda=-\frac{ax_1+by_1+c}{a^2+b^2}$.

The segment $PQ$ is perpendicular to the line $ax+by+c=0$, hence parallel to the normal direction $(a,b)$. So there is a scalar $\lambda$ with:

$h-x_1=\lambda a,\qquad k-y_1=\lambda b$,

giving $\dfrac{h-x_1}{a}=\dfrac{k-y_1}{b}=\lambda$.

Since $Q(h,k)$ lies on the line, $ah+bk+c=0$. Substitute $h=x_1+\lambda a$, $k=y_1+\lambda b$:

$a(x_1+\lambda a)+b(y_1+\lambda b)+c=0$ …

Use the reflection (image) formula $\dfrac{x'-x_1}{a}=\dfrac{y'-y_1}{b}=\dfrac{-2(ax_1+by_1+c)}{a^2+b^2}$ for the line $ax+by+c=0$.

For the line $3x+4y-1=0$: $a=3,\ b=4,\ c=-1$, and the point is $(x_1,y_1)=(1,2)$.

$ax_1+by_1+c = 3(1)+4(2)-1 = 3+8-1 = 10$

$a^2+b^2 = 9+16=25$

$k = \dfrac{-2(10)}{25} = \dfrac{-20}{25} = -\dfrac{4}{5}$

…