Question

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) is,
$\left( A \right)$ 3x – 4y + 6 = 0 and 4x – 3y + 1 = 0
$\left( B \right)$ 3x + 4y + 6 = 0 and 4x + 3y + 1 = 0
$\left( C \right)$ 3x – 4y - 6 = 0 and 4x + 3y + 1 = 0
$\left( D \right)$ None of these

Answer 1

Hint:In this question first use the substitution method to solve the two given equations that is $x - y + 1 = 0$ and $2x - 3y + 5 = 0$ to get the value of x and y, then simply write the equation of line passing through these points and having some slope m, use the concept $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$. Then use the concept that the perpendicular distance from any general point $\left( {{x_1},{y_1}} \right)$ on the any line ax + by + c = 0 is given as $d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$. This will help approaching the problem.

Complete step-by-step answer:
Given equation of lines are
$x - y + 1 = 0$.............................. (1)
And $2x - 3y + 5 = 0$...................... (2)
Now we solve these equations by substitution method so we have from equation (1)
$x = y - 1$...................... (3)
Now substitute this value in equation (2) we have,
$ \Rightarrow 2\left( {y - 1} \right) - 3y + 5 = 0$
Now simplify this we have,
$ \Rightarrow 2y - 2 - 3y + 5 = 0$
$ \Rightarrow - y + 3 = 0$
$ \Rightarrow y = 3$
Now from equation (3) we have,
$ \Rightarrow x = 3 - 1 = 2$
Therefore the point of intersection of the given equation is (x, y) = (2, 3)
Now we have to find the equation of lines passing through this point of intersection.
Now as we know that the equation of line passing through point (2, 3) having slope m is given as,
$ \Rightarrow \left( {y - 3} \right) = m\left( {x - 2} \right)$
Now simplify this we have,
$ \Rightarrow mx - y - 2m + 3 = 0$........................ (4)
Now it is given that the distance from the point (3, 2) on the above line is (7/5).
Now as we know that the perpendicular distance from any general point $\left( {{x_1},{y_1}} \right)$ on the any line $\left( {ax + by + c} \right) = 0$ is given as,
$d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
Therefore distance from point (3, 2) on the line $\left( {mx - y - 2m + 3 = 0} \right)$ is (7/5) so we have,
$ \Rightarrow \dfrac{7}{5} = \dfrac{{\left| {3m - 2 - 2m + 3} \right|}}{{\sqrt {{m^2} + {{\left( { - 1} \right)}^2}} }}$
Now simplify this we have,
$ \Rightarrow \dfrac{7}{5} = \dfrac{{\left| {m + 1} \right|}}{{\sqrt {{m^2} + 1} }}$
Now squaring on both sides we have,
\[ \Rightarrow {\left( {\dfrac{7}{5}} \right)^2} = {\left( {\dfrac{{\left| {m + 1} \right|}}{{\sqrt {{m^2} + 1} }}} \right)^2}\]
$ \Rightarrow \dfrac{{49}}{{25}} = \dfrac{{{m^2} + 1 + 2m}}{{{m^2} + 1}}$
$ \Rightarrow 49{m^2} + 49 = 25{m^2} + 50m + 25$
$ \Rightarrow 24{m^2} - 50m + 24 = 0$
Divide by 2 throughout we have,
$ \Rightarrow 12{m^2} - 25m + 12 = 0$
Now factorize the above equation we have,
$ \Rightarrow 12{m^2} - 16m - 9m + 12 = 0$
$ \Rightarrow 4m\left( {3m - 4} \right) - 3\left( {3m - 4} \right) = 0$
$ \Rightarrow \left( {4m - 3} \right)\left( {3m - 4} \right) = 0$
$ \Rightarrow m = \dfrac{3}{4},\dfrac{4}{3}$
Now from equation (4) we have,
When m = (3/4)
$ \Rightarrow \dfrac{3}{4} \times x - y - 2 \times \dfrac{3}{4} + 3 = 0$
Multiply by 4 throughout we have,
$ \Rightarrow 3x - 4y - 6 + 12 = 0$
$ \Rightarrow 3x - 4y + 6 = 0$
When m = (4/3)
$ \Rightarrow \dfrac{4}{3} \times x - y - 2 \times \dfrac{4}{3} + 3 = 0$
Multiply by 3 throughout we have,
$ \Rightarrow 4x - 3y - 8 + 9 = 0$
$ \Rightarrow 4x - 3y + 1 = 0$
So this is the required equation of line.
Hence option (A) is the correct answer.

Note:There can be a second method to solve the linear equation in two variables to find the points of intersection. In the second method that is the method of elimination we simply make the coefficients of any one variable same for both the equations and then by application of some basic arithmetic operations like addition/subtraction we eliminate this variable and thus an equation is obtained in a single variable only, solve it to get the value of the variable. The value of the other previously eliminated variable can be obtained by simply equating any one of the two equations given.