Question

The equations of the lines which pass through the points of intersection of the lines \[4x-3y-1=0\] and \[2x-5y+3=0\] and are equally inclined to the axes are
A. $y\pm x=0$
B. $y-1=\pm 1(x-1)$
C. $x-1=\pm 2(y-1)$
D. None of these

Answer 1

Hint: In this question, we are to find the equations of the lines that pass through the point of intersection of the given lines. Since those lines are equally inclined with the axes, their slopes are $m=\pm 1$. By using this we can calculate the required equations.

Formula Used:The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
We can write the above equation in the point-slope form as
$y-{{y}_{1}}=m(x-{{x}_{1}})$
The slope of the line with the equation $ax+by+c=0$ is $m=\frac{-a}{b}$
The point of intersection of two non-parallel lines ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ is
\[\left( \frac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}},\frac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)\]



Complete step by step solution:Given that,
The required equations pass through the point of intersection of the lines
\[\begin{align}
  & 4x-3y-1=0\text{ }...(1) \\
 & 2x-5y+3=0\text{ }...(2) \\
\end{align}\]
From (2),
\[\begin{align}
  & 2x-5y+3=0 \\
 & \Rightarrow 2x=5y-3\text{ }...(3) \\
\end{align}\]
Substituting (3) in (1), we get
$\begin{align}
  & 4x-3y-1=0 \\
 & \Rightarrow 2(2x)-3y-1=0 \\
 & \Rightarrow 2(5y-3)-3y-1=0 \\
 & \Rightarrow 10y-6-3y-1=0 \\
 & \Rightarrow 7y=7 \\
 & \text{ }\therefore y=1 \\
\end{align}$
Then,
$\begin{align}
  & 2x=5y-3 \\
 & \Rightarrow 2x=5(1)-3 \\
 & \Rightarrow 2x=2 \\
 & \text{ }\therefore x=1 \\
\end{align}$
Thus, the point of intersection of the given lines is $({{x}_{1}},{{y}_{1}})=(1,1)$
Since it is given that the lines are equally inclined to the axes, the slopes are $m=\pm 1$.
Then, by the point-slope form formula,
$\begin{align}
  & y-{{y}_{1}}=m(x-{{x}_{1}}) \\
 & y-1=\pm 1(x-1) \\
\end{align}$
Therefore, the required equations are $y-1=\pm 1(x-1)$.



Option ‘B’ is correct

Note: Here we may go wrong with the slope. Since it is mentioned that the angle of inclination is equal for both the lines, the angles are $45{}^\circ $ and $135{}^\circ $. So, the slopes are $\tan 45{}^\circ =1$ and $\tan 135{}^\circ =-1$ respectively. By this, the required equations are calculated. For calculating the point of intersection, the direct formula is also used as another way to find it.