Question

he equation of the line passing through $(4,-6)$ and makes an angle $45{}^\circ $ with the positive x-axis, is
A. $x-y-10=0$
B. $x-3y-22=0$
C. $x-2y-16=0$
D. None of these

Answer 1

Hint: In this question, we are to find the equation of the line with an angle of inclination of $45{}^\circ $ with the positive x-axis. To find this, we know that the slope of the line is calculated by $\tan \theta $ where $\theta $ is the angle of inclination of the line. By substituting these values, we can form the required equation of the line.



Formula Used:The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
The slope of a line with the equation in the form of $ax+by+c=0$ is $m=\dfrac{-a}{b}$
The slope of a line with an inclination of angle $\theta $ is $m=\tan \theta $
The equation of the line (point-slope form) is
$y-{{y}_{1}}=m(x-{{x}_{1}})$



Complete step by step solution:Given that,
A-line is passing through the point $(4,-6)$ and the makes an inclination $\theta =45{}^\circ $.
Then, the required line’s slope is
$\begin{align}
  & m=\tan \theta \\
 & \text{ }=\tan 45{}^\circ \\
 & \text{ }=1 \\
\end{align}$
Thus, the equation of the line with the slope $m=1$ and passing through the point $(4,-6)$ is
$\begin{align}
  & y-{{y}_{1}}=m(x-{{x}_{1}}) \\
 & \Rightarrow y+6=1(x-4) \\
 & \Rightarrow x-y-4-6=0 \\
 & \text{ }\therefore x-y-10=0 \\
\end{align}$



Option ‘A’ is correct



Note: Here we may go wrong with the slope of the line. Here an angle of inclination of the line is given. So, by the definition of the slope, we can calculate the slope using this inclination angle. Then, the equation of the required line is formed by using the point-slope form.