Slope of a Line — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Slope of a Line MCQs with step-by-step solutions (14 questions). Part of Straight Lines. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Slope of a Line · easy · numerical
The inclination of a line whose slope is $\sqrt{3}$ is:
A. $30^\circ$
B. $45^\circ$
C. $120^\circ$
D. $60^\circ$ ✓ Correct
Solution: $\tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ$.
Q2 — Slope of a Line · easy · numerical
The slope of the line passing through the points $(-2, 5)$ and $(4, -7)$ is:
A. $-\dfrac{1}{2}$
B. $-2$ ✓ Correct
C. $\dfrac{1}{2}$
D. $2$
Solution: $m = \dfrac{-7 - 5}{4 - (-2)} = \dfrac{-12}{6} = -2$.
Q3 — Slope of a Line · easy · numerical
The slope of the line passing through the points $(-5, -4)$ and $(3, 2)$ is:
A. $-\dfrac{3}{4}$
B. $\dfrac{4}{3}$
C. $\dfrac{3}{4}$ ✓ Correct
D. $-\dfrac{4}{3}$
Solution: $m = \dfrac{2 - (-4)}{3 - (-5)} = \dfrac{6}{8} = \dfrac{3}{4}$.
Q4 — Slope of a Line · medium · numerical
If the line through the points $(-4, y)$ and $(2, 8)$ has slope $\dfrac{3}{2}$, then $y$ equals:
A. $17$
B. $-9$
C. $-1$ ✓ Correct
D. $1$
Solution: $\dfrac{8 - y}{2 - (-4)} = \dfrac{3}{2} \Rightarrow 8 - y = 9 \Rightarrow y = -1$.
Q5 — Slope of a Line · medium · numerical
If the line through $(3, k)$ and $(7, 5)$ is parallel to the line through $(-1, 2)$ and $(1, 8)$, then $k$ equals:
A. $-1$
B. $17$
C. $7$
D. $-7$ ✓ Correct
Solution: The second line has slope $\dfrac{8 - 2}{1 - (-1)} = 3$; equal slopes give $\dfrac{5 - k}{7 - 3} = 3 \Rightarrow 5 - k = 12 \Rightarrow k = -7$.
Q6 — Slope of a Line · medium · numerical
If the line through $(k, 5)$ and $(4, -1)$ is perpendicular to the line through $(-6, -1)$ and $(0, 2)$, then $k$ equals:
A. $16$
B. $1$ ✓ Correct
C. $-1$
D. $7$
Solution: The second line has slope $\dfrac{2 - (-1)}{0 - (-6)} = \dfrac{1}{2}$, so the first must have slope $-2$: $\dfrac{-1 - 5}{4 - k} = -2 \Rightarrow 4 - k = 3 \Rightarrow k = 1$.
Q7 — Slope of a Line · medium · numerical
The acute angle between two lines whose slopes are $5$ and $\dfrac{2}{3}$ is:
A. $90^\circ$
B. $45^\circ$ ✓ Correct
C. $30^\circ$
D. $60^\circ$
Solution: $\tan\theta = \left|\dfrac{5 - \frac{2}{3}}{1 + 5 \cdot \frac{2}{3}}\right| = \dfrac{13/3}{13/3} = 1 \Rightarrow \theta = 45^\circ$.
Q8 — Slope of a Line · medium · numerical
The acute angle between two lines whose slopes are $\sqrt{3}$ and $\dfrac{1}{\sqrt{3}}$ is:
A. $30^\circ$ ✓ Correct
B. $60^\circ$
C. $15^\circ$
D. $45^\circ$
Solution: $\tan\theta = \left|\dfrac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}}}\right| = \dfrac{2/\sqrt{3}}{2} = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ$.
Q9 — Slope of a Line · medium · numerical
The acute angle between two lines whose slopes are $\sqrt{3}$ and $-\sqrt{3}$ is:
A. $30^\circ$
B. $60^\circ$ ✓ Correct
C. $45^\circ$
D. $90^\circ$
Solution: $\tan\theta = \left|\dfrac{-\sqrt{3} - \sqrt{3}}{1 + (\sqrt{3})(-\sqrt{3})}\right| = \left|\dfrac{-2\sqrt{3}}{-2}\right| = \sqrt{3} \Rightarrow \theta = 60^\circ$.
Q10 — Slope of a Line · medium · numerical
If the points $(x, 4)$, $(2, -6)$ and $(5, 9)$ are collinear, then $x$ equals:
A. $0$
B. $-4$
C. $4$ ✓ Correct
D. $6$
Solution: The slope through $(2, -6)$ and $(5, 9)$ is $\dfrac{15}{3} = 5$; equating $\dfrac{4 - (-6)}{x - 2} = 5$ gives $x - 2 = 2$, i.e. $x = 4$.
Q11 — Slope of a Line · medium · numerical
A straight line makes an angle of $45^\circ$ with a line of slope $4$. If the slope of the straight line is positive, it is equal to:
A. $\dfrac{5}{3}$
B. $-\dfrac{3}{5}$
C. $\dfrac{3}{5}$ ✓ Correct
D. $-\dfrac{5}{3}$
Solution: $\dfrac{m - 4}{1 + 4m} = \pm 1$ gives $m = -\dfrac{5}{3}$ or $m = \dfrac{3}{5}$; the positive slope is $\dfrac{3}{5}$.
Q12 — Slope of a Line · medium · numerical
The inclination of the line passing through the points $(-6, 2)$ and $(-1, -3)$ is:
A. $120^\circ$
B. $135^\circ$ ✓ Correct
C. $45^\circ$
D. $150^\circ$
Solution: $m = \dfrac{-3 - 2}{-1 - (-6)} = \dfrac{-5}{5} = -1 = \tan 135^\circ$, so the inclination is $135^\circ$.
Q13 — Slope of a Line · hard · numerical
The line joining $A(-2, -3)$ and $B(1, 0)$ is rotated about $A$ through $15^\circ$ in the anticlockwise direction. The slope of the line in its new position is:
A. $\dfrac{1}{\sqrt{3}}$
B. $1$
C. $2 + \sqrt{3}$
D. $\sqrt{3}$ ✓ Correct
Solution: Slope of $AB = \dfrac{0 - (-3)}{1 - (-2)} = 1$, so its inclination is $45^\circ$; after the rotation the inclination is $60^\circ$ and the slope is $\tan 60^\circ = \sqrt{3}$.
Q14 — Slope of a Line · hard · numerical
The angle between two lines is $45^\circ$ and the slope of one line is $6$ times the slope of the other. The sum of all possible positive values of the smaller slope is:
A. $\dfrac{5}{12}$
B. $\dfrac{5}{6}$ ✓ Correct
C. $\dfrac{1}{3}$
D. $\dfrac{1}{2}$
Solution: With slopes $m$ and $6m$: $\left|\dfrac{6m - m}{1 + 6m^2}\right| = 1$; for $m > 0$ this gives $6m^2 - 5m + 1 = 0$, so $m = \dfrac{1}{2}$ or $m = \dfrac{1}{3}$, whose sum is $\dfrac{5}{6}$.