Question

Find the equations of the lines through the point of intersection of the lines x − y + 1 = 0 and 2x − 3y + 5 = 0, whose distance from the point(3, 2) is 7/5. [NCERT EXEMPLAR]

Answer 1

The equations of the lines through the point of intersection of the lines x − y + 1 = 0 and 2x − 3y + 5 = 0 is given by
x − y + 1 + a(2x − 3y + 5) = 0
⇒ (1 + 2a)x + y(−3a − 1) + 5a + 1 = 0 .....(1)
The distance of the above line from the point is given by
$\frac{3\left(2a+1\right)+2\left(-3a-1\right)+5a+1}{\sqrt{{\left(2a+1\right)}^2+{\left(-3a-1\right)}^2}}$

$$\begin{gathered} \therefore \frac{\left|3\left(2a+1\right)+2\left(-3a-1\right)+5a+1\right|}{\sqrt{{\left(2a+1\right)}^2+{\left(-3a-1\right)}^2}}=\frac{7}{5} \\ \Rightarrow \frac{\left|5a+2\right|}{\sqrt{13a^2+10a+2}}=\frac{7}{5} \\ \Rightarrow 25{\left(5a+2\right)}^2=49\left(13a^2+10a+2\right) \\ \Rightarrow 6a^2-5a-1=0 \\ \Rightarrow a=1,-\frac{1}{6} \end{gathered}$$
Substituting the value of a in (1), we get
3x − 4y + 6 = 0 and 4x − 3y + 1 = 0