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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.

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  • $\begingroup$ I'm guessing that $AB=AC$ means the distance from $A$ to $B$ equals the distance from $A$ to $C$. The notation is a little confusing since you use $BC$ to refer not to a distance but to a line. Labelling the slopes $M_1$ and $M_2$ is not a way to clearly connect either of them to the lines passing through $A$, though the Reader can probably work that out. Note that this makes the two lines through $A$ perpendicular. $\endgroup$ – hardmath Oct 9 '14 at 9:58
  • $\begingroup$ @hardmath I have written line BC as i didn't know how to put line above BC(Can you edit that?) and Slopes of line is to show my efforts regarding to problem(I will remove M1 and M2) $\endgroup$ – Freddy Oct 9 '14 at 10:02
  • $\begingroup$ Hint: The family of lines like $\overline{BC}$ that intercept the lines through $A$ at equal distances from $A$ all share the same slope. Find one of those lines, take its slope, and then use the fact that point $(1,2)$ falls on the line $\overline{BC}$ to get the equation. $\endgroup$ – hardmath Oct 9 '14 at 10:09
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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 $$x-7y=-13$$

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  • $\begingroup$ I could not get last step, can you please help me! $\endgroup$ – Freddy Oct 11 '14 at 5:27
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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.

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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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