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Straight Lines — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Straight Lines MCQs with step-by-step solutions covering Introduction & Basics, Slope of a Line, Various Forms of the Equation of a Line, General Equation of a Line, Distance of a Point from a Line, High-Yield JEE Extensions. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Introduction & Basics · easy · numerical
The distance between the points $(3, -2)$ and $(-5, 4)$ is:
A. $2\sqrt{7}$
B. $6\sqrt{2}$
C. $10$ ✓ Correct
D. $8$
Solution: $d = \sqrt{(-5-3)^2 + (4+2)^2} = \sqrt{64+36} = 10$.
Q2 — Introduction & Basics · easy · numerical
The distance between the points $(6, 8)$ and $(2, 5)$ is:
A. $7$
B. $5$ ✓ Correct
C. $\sqrt{7}$
D. $25$
Solution: $d = \sqrt{(6-2)^2 + (8-5)^2} = \sqrt{16+9} = 5$.
Q3 — Introduction & Basics · easy · numerical
The midpoint of the segment joining $(-6, 3)$ and $(2, -9)$ is:
A. $(-2, -3)$ ✓ Correct
B. $(-2, 3)$
C. $(-4, -6)$
D. $(2, -3)$
Solution: Midpoint $= \left(\dfrac{-6+2}{2}, \dfrac{3-9}{2}\right) = (-2, -3)$.
Q4 — Introduction & Basics · easy · numerical
The point which divides the segment joining $A(-1, 7)$ and $B(4, -3)$ internally in the ratio $2 : 3$ is:
A. $(1, -3)$
B. $(1, 3)$ ✓ Correct
C. $(2, 1)$
D. $(3, 1)$
Solution: By the section formula, $P = \left(\dfrac{2(4)+3(-1)}{5}, \dfrac{2(-3)+3(7)}{5}\right) = (1, 3)$.
Q5 — Introduction & Basics · easy · numerical
The centroid of the triangle with vertices $(5, 2)$, $(-2, 6)$ and $(6, 7)$ is:
A. $(3, 5)$ ✓ Correct
B. $(3, -5)$
C. $(5, 3)$
D. $(9, 15)$
Solution: Centroid $= \left(\dfrac{5-2+6}{3}, \dfrac{2+6+7}{3}\right) = (3, 5)$.
Q6 — Slope of a Line · easy · numerical
The slope of a line whose inclination is $135^\circ$ is:
A. $-\sqrt{3}$
B. $\sqrt{3}$
C. $-1$ ✓ Correct
D. $1$
Solution: $m = \tan 135^\circ = -\tan 45^\circ = -1$.
Q7 — Slope of a Line · easy · numerical
The slope of a line whose inclination is $30^\circ$ is:
A. $-\dfrac{1}{\sqrt{3}}$
B. $\sqrt{3}$
C. $\dfrac{1}{\sqrt{3}}$ ✓ Correct
D. $\dfrac{1}{2}$
Solution: $m = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$.
Q8 — 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: $\tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ$.
Q9 — 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: $m = \dfrac{-7 - 5}{4 - (-2)} = \dfrac{-12}{6} = -2$.
Q10 — 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: $m = \dfrac{2 - (-4)}{3 - (-5)} = \dfrac{6}{8} = \dfrac{3}{4}$.
Q11 — Various Forms of the Equation of a Line · easy · numerical
The equation of the vertical line passing through the point $(-7, 9)$ is:
A. $x = -7$ ✓ Correct
B. $y = 9$
C. $y = -7$
D. $x = 9$
Solution: A vertical line has equation $x = a$; through $(-7, 9)$ it is $x = -7$.
Q12 — Various Forms of the Equation of a Line · easy · numerical
The equation of the line with slope $-3$ and y-intercept $5$ is:
A. $y = 3x + 5$
B. $y = 5x - 3$
C. $y = -3x - 5$
D. $y = -3x + 5$ ✓ Correct
Solution: Slope-intercept form: $y = mx + c = -3x + 5$.
Q13 — Various Forms of the Equation of a Line · easy · numerical
The equation of the line passing through $(-4, 3)$ with slope $2$ is:
A. $y - 3 = 2(x - 4)$
B. $y - 3 = 2(x + 4)$ ✓ Correct
C. $y + 3 = 2(x - 4)$
D. $y + 4 = 2(x - 3)$
Solution: Point-slope form: $y - y_1 = m(x - x_1)$, i.e. $y - 3 = 2(x + 4)$.
Q14 — Various Forms of the Equation of a Line · easy · numerical
The equation of the line which cuts off intercepts $4$ and $7$ on the x-axis and y-axis respectively is:
A. $\dfrac{x}{4} + \dfrac{y}{7} = 1$ ✓ Correct
B. $\dfrac{x}{7} + \dfrac{y}{4} = 1$
C. $\dfrac{x}{4} - \dfrac{y}{7} = 1$
D. $4x + 7y = 1$
Solution: Intercept form: $\dfrac{x}{a} + \dfrac{y}{b} = 1$ with $a = 4$ and $b = 7$.
Q15 — Various Forms of the Equation of a Line · easy · numerical
The equation of the line passing through the points $(3, -1)$ and $(-3, -5)$ is:
A. $3x - 2y = 9$
B. $2x - 3y = 3$
C. $2x - 3y = 9$ ✓ Correct
D. $2x + 3y = 9$
Solution: Slope $= \dfrac{-5 - (-1)}{-3 - 3} = \dfrac{2}{3}$; then $y + 1 = \dfrac{2}{3}(x - 3)$ gives $2x - 3y = 9$ (both points satisfy it).
Q16 — General Equation of a Line · easy · numerical
The slope of the line $5x - 2y + 7 = 0$ is:
A. $\dfrac{5}{2}$ ✓ Correct
B. $-\dfrac{5}{2}$
C. $\dfrac{2}{5}$
D. $-\dfrac{2}{5}$
Solution: For $Ax + By + C = 0$, slope $m = -A/B = -5/(-2) = \dfrac{5}{2}$.
Q17 — General Equation of a Line · easy · numerical
The $y$-intercept of the line $3x + 7y - 21 = 0$ is:
A. $7$
B. $21$
C. $-3$
D. $3$ ✓ Correct
Solution: Put $x = 0$: $7y = 21$, so $y = 3$. (Putting $y = 0$ instead gives the $x$-intercept $7$.)
Q18 — General Equation of a Line · easy · numerical
The $x$-intercept of the line $4x - 6y + 9 = 0$ is:
A. $\dfrac{9}{4}$
B. $-\dfrac{3}{2}$
C. $-\dfrac{9}{4}$ ✓ Correct
D. $\dfrac{3}{2}$
Solution: Put $y = 0$: $4x + 9 = 0$, so $x = -\dfrac{9}{4}$.
Q19 — General Equation of a Line · easy · numerical
If the lines $3x + ky - 6 = 0$ and $6x + 8y + 5 = 0$ are parallel, then $k$ equals:
A. $4$ ✓ Correct
B. $16$
C. $8$
D. $-4$
Solution: Parallel lines have proportional coefficients of $x$ and $y$: $\dfrac{3}{6} = \dfrac{k}{8}$, so $k = 4$.
Q20 — General Equation of a Line · easy · numerical
If the lines $2x + 3y - 1 = 0$ and $kx - 2y + 5 = 0$ are perpendicular, then $k$ equals:
A. $3$ ✓ Correct
B. $6$
C. $\dfrac{4}{3}$
D. $-3$
Solution: Perpendicularity requires $A_1A_2 + B_1B_2 = 0$: $2k + 3(-2) = 0$, so $k = 3$.
Q21 — Distance of a Point from a Line · easy · numerical
The perpendicular distance of the point $(2, 1)$ from the line $3x - 4y + 8 = 0$ is:
A. $\dfrac{2}{5}$
B. $10$
C. $\dfrac{6}{5}$
D. $2$ ✓ Correct
Solution: $d = \dfrac{|3(2) - 4(1) + 8|}{\sqrt{3^2 + 4^2}} = \dfrac{10}{5} = 2$.
Q22 — Distance of a Point from a Line · easy · numerical
The distance of the origin from the line $5x + 12y - 39 = 0$ is:
A. $3$ ✓ Correct
B. $\dfrac{39}{17}$
C. $6$
D. $39$
Solution: $d = \dfrac{|5(0) + 12(0) - 39|}{\sqrt{25 + 144}} = \dfrac{39}{13} = 3$.
Q23 — Distance of a Point from a Line · easy · numerical
The distance of the point $(1, -2)$ from the line $x - y + 5 = 0$ is:
A. $4$
B. $8$
C. $4\sqrt{2}$ ✓ Correct
D. $2\sqrt{2}$
Solution: $d = \dfrac{|1 + 2 + 5|}{\sqrt{1 + 1}} = \dfrac{8}{\sqrt{2}} = 4\sqrt{2}$.
Q24 — Distance of a Point from a Line · easy · numerical
The distance between the parallel lines $3x + 4y - 9 = 0$ and $3x + 4y + 11 = 0$ is:
A. $20$
B. $5$
C. $2$
D. $4$ ✓ Correct
Solution: $d = \dfrac{|c_1 - c_2|}{\sqrt{A^2 + B^2}} = \dfrac{|-9 - 11|}{5} = \dfrac{20}{5} = 4$.
Q25 — Distance of a Point from a Line · easy · numerical
The perpendicular distance of the point $(5, 1)$ from the line $12x - 5y + 10 = 0$ is:
A. $65$
B. $5$ ✓ Correct
C. $6$
D. $\dfrac{45}{13}$
Solution: $d = \dfrac{|12(5) - 5(1) + 10|}{\sqrt{144 + 25}} = \dfrac{65}{13} = 5$.
Q26 — High-Yield JEE Extensions · easy · numerical
If the pair of lines $kx^2 + 7xy - 4y^2 = 0$ are perpendicular to each other, then $k$ equals:
A. $0$
B. $4$ ✓ Correct
C. $7$
D. $-4$
Solution: A homogeneous pair $ax^2 + 2hxy + by^2 = 0$ is perpendicular when $a + b = 0$: here $k - 4 = 0$, so $k = 4$.
Q27 — High-Yield JEE Extensions · easy · numerical
The angle between the pair of lines $x^2 + 4xy + y^2 = 0$ is:
A. $60^\circ$ ✓ Correct
B. $90^\circ$
C. $45^\circ$
D. $30^\circ$
Solution: $\tan\theta = \dfrac{2\sqrt{h^2 - ab}}{|a + b|} = \dfrac{2\sqrt{4 - 1}}{2} = \sqrt{3}$, so $\theta = 60^\circ$.
Q28 — High-Yield JEE Extensions · easy · numerical
The separate equations of the lines given by $2x^2 - 5xy + 2y^2 = 0$ are:
A. $x - y = 0$ and $2x - y = 0$
B. $2x + y = 0$ and $x + 2y = 0$
C. $2x - y = 0$ and $x - 2y = 0$ ✓ Correct
D. $2x - y = 0$ and $x + 2y = 0$
Solution: $2x^2 - 5xy + 2y^2 = (2x - y)(x - 2y)$, so the lines are $2x - y = 0$ and $x - 2y = 0$.
Q29 — High-Yield JEE Extensions · easy · numerical
If the line $(3x + y - 5) + \lambda(x - 3y + 5) = 0$ passes through the point $(0, 2)$, then $\lambda$ equals:
A. $-3$ ✓ Correct
B. $3$
C. $-\dfrac{1}{3}$
D. $\dfrac{1}{3}$
Solution: Substitute $(0, 2)$: $(0 + 2 - 5) + \lambda(0 - 6 + 5) = -3 - \lambda = 0$, so $\lambda = -3$.
Q30 — High-Yield JEE Extensions · easy · numerical
For every value of $\lambda$, the line $(x + y - 2) + \lambda(x - y) = 0$ passes through the fixed point:
A. $(-1, 1)$
B. $(1, 1)$ ✓ Correct
C. $(2, 0)$
D. $(1, -1)$
Solution: The fixed point satisfies both $x + y - 2 = 0$ and $x - y = 0$, giving $(1, 1)$.