reflection of a line on another line

Question

Show that the reflection of the line $ax+by+c=0$ on the line $x+y+1=0$ is the line $bx+ay+(a+b-c)=0$

Attempt:

Consider a point $(h,k)$ on the line $ax+by+c=0$

Thus, $ah+bk+c=0 $ \\ $\mathbb{1}$

Now note that the image $(x_2,y_2)$ of a point$(x_1,y_1)$ with respect to a line $ax +by + c= 0$ is given by:

$$\dfrac{x_2-x_1}{a}= \dfrac{y_2-y_1}{b}= \dfrac{-2(ax_1+by_1+c)}{a^2+b^2}$$

Here, we can use this formula to get:

$$x_2 - h = y_2 - k = - h-k-c$$

Now, for $x_2$:

$bx_2 = -bk - c \\ay_2 = - ah-c$

Add the above two equations:

$ay_2 +bx_2 = -bk - ah - bc - ac$

From equation 1, $-bk-ah= c$

Thus, $ay_2 + bx_2 + ac + bc - c = 0 \implies bx+ay + c(a+b-1)= 0$ (Substitute $x$ for $x_2$ and $y$ for $y_2$ to get the locus).

This equation is close to the answer but incorrect. Where have I gone wrong?

Answer 1

You’ve substituted for $a$ and $b$ in your formula for the image of a point, but not for $c$. Then, for some reason you didn’t multiply this extraneous $c$ (which should be $1$) by $b$ or $a$ in the expressions that you derived for $bx_2$ and $ay_2$.

You should have $x_2+k+1=y_2+h+1=0$, from which $bx_2+ay_2+ah+bk+a+b=0$ and the desired result follows.

To avoid errors like the first one in the future, I recommend not using the same parameter names in general formulas as you have in the problem that you’re trying to solve.

Answer 2

Hint:

As $ah+bk+c=0, k=-\dfrac{ah+c}b$

So, $P\left(h,-\dfrac{ah+c}b\right)$ will represent any point on $ax+by+c=0$

Now if $Q(p,q)$ is the reflection of $P,$

the midpoint $\left(\dfrac{p+h}2,\dfrac{bq-ah-c}{2b}\right)$ will lie on $$x+y+1=0\ \ \ \ (1)$$

Again $PQ$ will be perpendicular to $x+y+1=0,$ hence its gradient will be $\dfrac{-1}{-1}\ \ \ \ (2)$

Use $(1),(2)$ to find $p,q$ in terms of $h$

Equate the two values of $h$

Answer 3

HINTS: $$ ax +by+c=0$$ $$ px + q y+ r =0$$ have angle bisector

$$ \dfrac{ ax +by+c}{\sqrt{a^2+b^2}}-\dfrac{ px +qy+r}{\sqrt{p^2+q^2}}=0$$

which is same as

$$x+y+1 =0 $$

comparing coefficients of the straight line

$$ \dfrac{a}{\sqrt{a^2+b^2}}-\dfrac{ p}{\sqrt{p^2+q^2}}=1$$ $$ \dfrac{b}{\sqrt{a^2+b^2}}-\dfrac{ q}{\sqrt{p^2+q^2}}=1$$ $$ \dfrac{c}{\sqrt{a^2+b^2}}-\dfrac{ r}{\sqrt{p^2+q^2}}=1$$

simplify and evaluate three $p,q,r$ from three above equations.

EDIT1:

Note also the correct sign of $\pm$ sign to choose quadrant and normal position properly.