Find the coordinates of point $A'$

Question

Find the coordinates of the point $A'$ which is the symmetric point of $A(3,2)$ with respect to the line $(D):2x+y-12=0$

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For my trying

I find the length from the point $A(3,2)$ to the line $2x+y-12=0$. $d=\frac{|2\times 3+2-12|}{\sqrt{2^2+1^2}}=\frac{4}{\sqrt{5}}$.

Then I chose a line that through $A(3,2)$ and perpendicular to a line $(D)$.

At this point, I don't know how to do more.

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Please kindly help to give me a hint or some ideas about this problem.

Answer 1

I chose a line that through $A(3,2)$ and perpendicular to a line $(D)$.

So that's enough. Express the line in parametric equation $ \begin{cases} x=3+2t\\ y=2+t \end{cases}$ (we choose the directing vector $(2,1)$ to be parallel th the normal vector of $(D):\, (2,1)\cdot(x,y)-12=0$).

Then just plug $x,y$ from the parametric equation into your distance formula $\frac{|2x+y-12|}{\sqrt{5}}$ and equate to the desired distance: $$\frac{|2x+y-12|}{\sqrt{5}}= \frac{|2(3+2t)+(2+t)-12|}{\sqrt{5}}=\frac{4}{\sqrt{5}}$$ $$|5t-4|=4$$ $$5t-4=\pm4$$ $$t=\frac{4\pm4}{5}$$ $$\left[\begin{array}{ll}t=0&\hbox{ -- this is the point A}\\ t=\frac{8}{5}&\hbox{ -- this is not the point A, thus A'}\\\end{array}\right.$$ So substitute the $t=\frac{8}{5}$ back to get $A'=(x,y)= \left(\frac{31}{5},\frac{18}{5}\right)$.

Answer 2

The line perpendicular to $ax+by+c=0$ is of the form $bx-ay+k=0$ where $k$ is a constant to be determined .

You have the equation of the line $(D)$ .Then find the equation of the perpendicular line to $(D)$ ,let us say $(L)$ which passes through point $A$ (which actually helps in finding the constant $k$).

Now find the intersection point of $(L)$ and $(D)$.

Notice that the intersection point would be the mid-point of $A$ and the required symmetric point $A'$

Answer 3

What you've to find is image of $A$ in the line $D$. So, the image of the point $(x_1,y_1)$ in the line $ax+by+c=0$ is given by:$$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2(ax_1+by_1+c)}{a^2+b^2}$$ Check the proof here.