Question

Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right)$ be two points on a non-vertical line $l$ whose inclination is $\theta$. Then,
I. Slope of the line $l$ is $m=\frac{y_2-y_1}{x_2-x_1}$, whether $\theta$ is acute or obtuse.
II. Slope of the line $l$ is $m=\frac{y_2-y_1}{x_2-x_1}$, when $\theta$ is acute only.

• A. I and II are true
• B. I is true and II is false
• C. I is false and II is true
• D. I and II are false

Answer 1

Answer: (B)

Case I When angle $\theta$ is acute.

From figure, $\angle M P Q=\theta$
Therefore, slope of line $I=m=\tan \theta ....$(i)
But in $\triangle M P Q$, we have
$\tan \theta=\frac{M Q}{M P}=\frac{y_2-y_1}{x_2-x_1} ....$(ii)
From Eqs. (i) and (ii), we have
$m=\frac{y_2-y_1}{x_2-x_1}$
Case II When angle $\theta$ is obtuse.

From figure, we have
$\angle M P Q=180^{\circ}-\theta$
Therefore, $\theta=180^{\circ}-\angle M P Q$
Now, slope of the line $l$ is
$m =\tan \theta$
$ =\tan \left(180^{\circ}-\angle M P Q\right)$
$ =-\tan \angle M P Q$
$=-\frac{M Q}{M P} $
$=-\frac{y_2-y_1}{x_1-x_2}$
$=\frac{y_2-y_1}{x_2-x_1}$
Consequently, we see that in both the cases, the slope $m$ of the line / through the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is given by
$m=\frac{y_2-y_1}{x_2-x_1} \text {. }$