Question

The straight lines $3x + 4y = 5$ and $4x - 3y = 15$ intersect at the point A. On these lines, the points B and C are chosen so that $AB = AC$ . Find the possible equation of the line BC passing through the point $\left( {1,2} \right)$ .

Answer 1

To find the possible equations, we find out the type of triangle the line BC makes, to find its internal angles. Using that we write the general equation of line BC. Then we use the formula of slope of bisectors to find the slope, substituting the slope in the general equation gives us the equation of line BC.

Complete step-by-step answer:
Given data, $AB = AC$
In order to find the type of triangle let us find the slope of the two lines.
For the first line.
$3x + 4y = 5 \\\ \Rightarrow 4y = - 3x + 5 \\\ \Rightarrow y = - \dfrac{3}{4}x + \dfrac{5}{4} \\\$
Slope = $- \dfrac{3}{4}$
For the second line.
$4x - 3y = 15 \\\ \Rightarrow 3y = 4x - 15 \\\ \Rightarrow y = \dfrac{4}{3}x - \dfrac{{15}}{3} \\\$
Slope = $\dfrac{4}{3}$
As the product of the slopes is -1. So these lines are perpendicular to each other.
The given lines are perpendicular and as $AB = AC$ , Therefore $\Delta ABC$ is a right angled isosceles triangle as shown in the figure.

Hence the line BC through point $\left( {1,2} \right)$ will make angles of ${45^0}$ with the given lines.
As we know that the general equation of any line is of the form $y - b = m\left( {x - a} \right)$, where $\left( {a,b} \right)$ is the point it passes through.
Hence our line becomes, $y - 2 = m\left( {x - 1} \right)$ as it passes through $\left( {1,2} \right)$ .
There can be two possible equations of the line BC. As the line BC may lie on the left or the right side of the line AC and the figure we have taken is intuitional.
So let us find the slope of the line BC with the help of the line AC.
As we know that the slope of line AC is ${m_1} = - \dfrac{3}{4}$
Also we know the angle between the line AC and BC is ${45^0}$
Where $\tan \left( {{{45}^0}} \right) = 1$
As we know the formula for finding the slope of two lines with the angle between them is given as:
${m_2} = \dfrac{{{m_1} \pm \tan \theta }}{{1 \mp {m_1}\tan \theta }}$
So, let us now use the formula to find the slope of the line BC.

$$\because {m_2} = \dfrac{{{m_1} \pm \tan \theta }}{{1 \mp {m_1}\tan \theta }} \\\ \Rightarrow {m_2} = \dfrac{{\left( {\dfrac{{ - 3}}{4}} \right) \pm \left( {\tan {{45}^0}} \right)}}{{1 \mp \left( {\dfrac{{ - 3}}{4}} \right)\left( {\tan {{45}^0}} \right)}} \\\$$

Let us now substitute the values and solve for the slope of BC.

$$\Rightarrow {m_2} = \dfrac{{\dfrac{{ - 3}}{4} \pm 1}}{{1 \mp \left( {\dfrac{{ - 3}}{4}} \right) \cdot 1}} \\\ \Rightarrow {m_2} = \dfrac{{ - 3 \pm 4}}{{4 \mp \left( { - 3} \right)}} \\\ \Rightarrow {m_2} = \dfrac{1}{7}, - 7 \\\$$

So, the slope of the line BC are $\dfrac{1}{7}{\text{ or }} - 7$
Therefore, the equation of the line BC is:
$\Rightarrow y - 2 = \dfrac{1}{7}\left( {x - 1} \right){\text{ or }}y - 2 = - 7\left( {x - 1} \right) \\\ \Rightarrow 7y - 14 = x - 1{\text{ or }}y - 2 = - 7x + 7 \\\ \Rightarrow x - 7y + 13 = 0{\text{ or }}7x + y - 9 = 0 \\\$
Hence, the equations of the line BC is:
$x - 7y + 13 = 0{\text{ or }}7x + y - 9 = 0$

Note: In order to solve this type of problems the key is to identify that the line BC makes a right angled isosceles triangle. It is important to notice that the line equation of BC could be any one of the perpendicular lines with slopes $\dfrac{1}{7}{\text{ or }} - 7$ respectively. Adequate knowledge in the concepts of straight lines is required.