The straight line $3x + 4y + 5 = 0 $ and $4x - 3y - 10 = 0$ intersect at point $A$. Point $B$ on line $3x + 4y + 5 = 0 $ and point C on line $4x - 3y - 10 = 0$ are such that $d(A,B)=d(A,C)$.
Find the equation of line passing through line $\overline{BC}$ if it also passes through point $(1,2)$.
I have found out slope of both the lines through $A$: $\frac{-3}{4}$ and $\frac{4}{3}$.
I can't figure out how to solve it.
$\endgroup$ 33 Answers
$\begingroup$Note that the triangle $ABC$ is isosceles. Then, the bisector $b$ of the angle $A$ is perpendicular to the side $BC$.
To compute the equation of $b$ we must write:
$$\frac{|3x+4y+5|}{5}=\frac{|4x-3y-10|}{5}$$
That is, the eqution of $b$ is either $$x-7y-15=0$$ or $$7x+y-5=0$$
Thus, the equation of the line $BC$ is either $$7x+y=9$$ or $$x-7y=-13$$
$\endgroup$ 1 $\begingroup$Intersection point $A$ has coordinates $(1, -2)$.
The first line equation can be written as $y = -\frac34x - \frac54$, the second: $y = \frac43x-\frac{10}{3}$, so $k_1 = -\frac34, k_2 = \frac43$.
$AB$ would lie on the first line if $y_B - y_A = k_1(x_B - x_A)$.
$AC$ would lie on the second line if $y_C - y_A = k_2(x_C - x_A)$.
This gives us $y_B = -\frac34x_B - \frac54$, $y_C = \frac43x_C - \frac{10}{3}$. $AB = AC$ means that $\sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2}$, so $\sqrt{(x_B - 1)^2 + (y_B + 2)^2} = \sqrt{(x_C - 1)^2 + (y_C + 2)^2}$
Let $y = k_3x + b_3$ be the equation of line containing $BC$. If it passes through $D = (1,2)$, then it means that it contains segments $BD$ and $CD$, so $k_3 = \frac{y_B - y_D}{x_B - x_D} = \frac{y_D - y_C}{x_D - x_C}$ (as $D$ lies between $B$ and $C$), so his gives the last equation: $\frac{y_B - 2}{x_B - 1} = \frac{2 - y_C}{1 - x_C}$.
Finally, you have to solve the following system:
\begin{cases} y_B = -\frac34x_B - \frac54 \\ y_C = \frac43x_C - \frac{10}{3} \\ \sqrt{(x_B - 1)^2 + (y_B + 2)^2} = \sqrt{(x_C - 1)^2 + (y_C + 2)^2} \\ \frac{y_B - 2}{x_B - 1} = \frac{2 - y_C}{1 - x_C} \end{cases}
Its solution gives you coordinates of $B$ and $C$ from which you can obtain the line equation.
$\endgroup$ $\begingroup$inspection shows the lines are perpendicular and meet at $(-1,2)$. define variables $(w,z)$ by: $$ \begin{pmatrix} w \\ z \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix} \left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix}\right) $$ the equations of the lines are now $w=0$ and $z=0$, so the lines perpendicular to their bisector have equation $w+z=k$ for some $k$, i.e. $$ \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} w \\ z \end{pmatrix} = k $$ in the original coordinates the line is $$ \begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix} \left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix}\right) = k $$ or $$ 7x+y =k' $$ and since the line passes through $(1,2)$ we must have $k'=9$
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