Question

Find the angle between the x-axis the line joining the points (3,-1) and (4,-2)

Answer 1

Hint: Use the fact that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Substitute the value of ${{x}_{1}},{{x}_{2}},{{y}_{1}},{{y}_{2}}$ in each case and hence find the slopes of the lines which is also the tangent of the angle which the line makes with the positive x-axis when going anticlockwise from the x-axis. The value of m gives the slope of the line and then equate it to the tangent of the angle which the line makes with the positive x-axis when going anticlockwise from the x-axis as follows
\[m=\tan \theta \]
(Where \[\theta \] is the angle that the line makes with the positive x-axis when going anticlockwise from the x-axis)
Complete step-by-step answer:
[i] We have $A\equiv \left( 3,-1 \right)$ and $B\equiv \left( 4,-2 \right)$
We know that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Here ${{x}_{1}}=3,{{x}_{2}}=4,{{y}_{1}}=-1$ and ${{y}_{2}}=-2$
Hence, we have
$m=\dfrac{-2-(-1)}{4-3}=\dfrac{-1}{1}=-1$
Hence the slope of the line is$-1$ .
Now. As mentioned in the hint, we can equate this value of slope that is ‘m’ to tangent of the angle as follows
\[\begin{align}
  & m=\tan \theta \\
 & -1=\tan \theta \\
 & \theta ={{\tan }^{-1}}(-1) \\
 & \theta ={{135}^{\circ }} \\
\end{align}\]
(As tan inverse of -1 is \[{{135}^{\circ }}\])
Hence the angle that the line makes with the positive x-axis when going anticlockwise from the x-axis is \[{{135}^{\circ }}\] .
Note: Alternative solution:
Alternatively, assume that the equation of the line is $y=mx+c$. Since the line passes through the points, the points satisfy the equation of the line. Hence form two linear equations in two variables m and c. Solve for m and c. The value of m gives the slope of the line and then equate it to the tangent of the angle which the line makes with the positive x-axis when going anticlockwise from the x-axis as follows
\[m=\tan \theta \]
(Where \[\theta \] is the angle that the line makes with the positive x-axis when going anticlockwise from the x-axis)