General Equation of a Line — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics General Equation of a Line MCQs with step-by-step solutions (20 questions). Part of Straight Lines. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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 $m = -A/B = -5/(-2) = \dfrac{5}{2}$.
Q2 — 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 $7$.)
Q3 — 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 $x = -\dfrac{9}{4}$.
Q4 — 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 $k = 4$.
Q5 — 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 $k = 3$.
Q6 — General Equation of a Line · medium · numerical
When $4x + 3y - 24 = 0$ is written in intercept form $\dfrac{x}{a} + \dfrac{y}{b} = 1$, the values of $a$ and $b$ are:
A. $a = 6,\ b = 8$ ✓ Correct
B. $a = 4,\ b = 3$
C. $a = 8,\ b = 6$
D. $a = 6,\ b = -8$
Solution: $4x + 3y = 24$ gives $\dfrac{x}{6} + \dfrac{y}{8} = 1$, so $a = 6$ and $b = 8$.
Q7 — General Equation of a Line · medium · numerical
When $x + y = 4$ is reduced to the normal form $x\cos\omega + y\sin\omega = p$, the values of $\omega$ and $p$ are:
A. $\omega = 30^\circ,\ p = 2\sqrt{2}$
B. $\omega = 60^\circ,\ p = 2$
C. $\omega = 45^\circ,\ p = 4$
D. $\omega = 45^\circ,\ p = 2\sqrt{2}$ ✓ Correct
Solution: Divide by $\sqrt{1^2 + 1^2} = \sqrt{2}$: $\cos\omega = \sin\omega = \dfrac{1}{\sqrt{2}}$, so $\omega = 45^\circ$ and $p = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}$.
Q8 — General Equation of a Line · medium · numerical
When $\sqrt{3}x + y = 8$ is reduced to the normal form $x\cos\omega + y\sin\omega = p$, the values of $\omega$ and $p$ are:
A. $\omega = 30^\circ,\ p = 8$
B. $\omega = 30^\circ,\ p = 4$ ✓ Correct
C. $\omega = 60^\circ,\ p = 2$
D. $\omega = 60^\circ,\ p = 4$
Solution: Divide by $\sqrt{3 + 1} = 2$: $\dfrac{\sqrt{3}}{2}x + \dfrac{1}{2}y = 4$, so $\cos\omega = \dfrac{\sqrt{3}}{2}$, $\sin\omega = \dfrac{1}{2}$, giving $\omega = 30^\circ$, $p = 4$.
Q9 — General Equation of a Line · medium · numerical
For the line $x - \sqrt{3}y + 8 = 0$, the normal form $x\cos\omega + y\sin\omega = p$ has:
A. $\omega = 120^\circ,\ p = 8$
B. $\omega = 120^\circ,\ p = 4$ ✓ Correct
C. $\omega = 60^\circ,\ p = 4$
D. $\omega = 150^\circ,\ p = 4$
Solution: Rewrite as $-x + \sqrt{3}y = 8$ and divide by $\sqrt{1 + 3} = 2$: $\cos\omega = -\dfrac{1}{2}$, $\sin\omega = \dfrac{\sqrt{3}}{2}$, so $\omega = 120^\circ$ and $p = 4$.
Q10 — General Equation of a Line · medium · numerical
The equation of the line through $(2, 3)$ parallel to the line $4x - 5y + 6 = 0$ is:
A. $4x - 5y - 7 = 0$
B. $5x + 4y - 22 = 0$
C. $4x - 5y + 7 = 0$ ✓ Correct
D. $4x + 5y - 23 = 0$
Solution: A parallel line is $4x - 5y + c = 0$; through $(2, 3)$: $8 - 15 + c = 0$, so $c = 7$.
Q11 — General Equation of a Line · medium · numerical
The equation of the line through $(-1, 2)$ perpendicular to the line $2x + 3y - 4 = 0$ is:
A. $3x - 2y - 7 = 0$
B. $2x + 3y - 4 = 0$
C. $3x + 2y - 1 = 0$
D. $3x - 2y + 7 = 0$ ✓ Correct
Solution: A perpendicular line is $3x - 2y + c = 0$; through $(-1, 2)$: $-3 - 4 + c = 0$, so $c = 7$.
Q12 — General Equation of a Line · medium · numerical
The point of intersection of the lines $3x + y - 10 = 0$ and $x - 2y + 6 = 0$ is:
A. $(2, -4)$
B. $(2, 4)$ ✓ Correct
C. $(1, 7)$
D. $(4, 2)$
Solution: From the first line $y = 10 - 3x$; substituting: $x - 2(10 - 3x) + 6 = 0$ gives $7x = 14$, so $x = 2$, $y = 4$.
Q13 — General Equation of a Line · medium · numerical
If the line $kx + 3y - 12 = 0$ passes through the point $(3, 2)$, then $k$ equals:
A. $6$
B. $3$
C. $2$ ✓ Correct
D. $-2$
Solution: Substitute the point: $3k + 6 - 12 = 0$, so $k = 2$.
Q14 — General Equation of a Line · medium · numerical
If the lines $4x + 6y - 10 = 0$ and $2x + 3y + k = 0$ are coincident, then $k$ equals:
A. $5$
B. $10$
C. $-10$
D. $-5$ ✓ Correct
Solution: Coincident lines have all coefficients proportional: $\dfrac{4}{2} = \dfrac{6}{3} = \dfrac{-10}{k}$, so $k = -5$.
Q15 — General Equation of a Line · hard · numerical
The area of the triangle formed by the line $5x + 12y - 60 = 0$ with the coordinate axes is:
A. $15$
B. $60$
C. $25$
D. $30$ ✓ Correct
Solution: The intercepts are $x = 12$ and $y = 5$, so the area is $\dfrac{1}{2}(12)(5) = 30$.
Q16 — General Equation of a Line · hard · numerical
The equation of the line passing through the intersection of $x + 2y - 5 = 0$ and $3x - y - 1 = 0$ and also through the point $(4, -1)$ is:
A. $x + y + 3 = 0$
B. $3x - y - 1 = 0$
C. $x + y - 3 = 0$ ✓ Correct
D. $x - y + 1 = 0$
Solution: The two lines meet at $(1, 2)$; the line through $(1, 2)$ and $(4, -1)$ has slope $\dfrac{-1 - 2}{4 - 1} = -1$, giving $x + y - 3 = 0$.
Q17 — General Equation of a Line · hard · numerical
If the line $3x + 2y - k = 0$ (with $k > 0$) forms a triangle of area $12$ with the coordinate axes, then $k$ equals:
A. $24$
B. $6\sqrt{2}$
C. $6$
D. $12$ ✓ Correct
Solution: The intercepts are $\dfrac{k}{3}$ and $\dfrac{k}{2}$, so area $= \dfrac{1}{2}\cdot\dfrac{k}{3}\cdot\dfrac{k}{2} = \dfrac{k^2}{12} = 12$, giving $k^2 = 144$, $k = 12$.
Q18 — General Equation of a Line · hard · numerical
The equation of the line through the point of intersection of $x + y - 6 = 0$ and $x - y - 2 = 0$ that is perpendicular to $5x + 12y - 7 = 0$ is:
A. $12x - 5y - 38 = 0$ ✓ Correct
B. $12x + 5y - 58 = 0$
C. $5x + 12y - 44 = 0$
D. $12x - 5y + 38 = 0$
Solution: The two lines meet at $(4, 2)$; a perpendicular to $5x + 12y - 7 = 0$ is $12x - 5y + c = 0$, and $48 - 10 + c = 0$ gives $c = -38$.
Q19 — General Equation of a Line · hard · numerical
If the line $ax + by - 12 = 0$ is parallel to $2x + 5y = 0$ and passes through $(-4, 4)$, then:
A. $a = 4,\ b = 10$
B. $a = 2,\ b = -5$
C. $a = 5,\ b = 2$
D. $a = 2,\ b = 5$ ✓ Correct
Solution: Parallel to $2x + 5y = 0$ means $a = 2t$, $b = 5t$; substituting $(-4, 4)$: $-8t + 20t = 12$, so $t = 1$, giving $a = 2$, $b = 5$.
Q20 — General Equation of a Line · hard · numerical
A line passes through the intersection of $2x + y - 10 = 0$ and $x - y + 1 = 0$ and is parallel to $2x + 3y = 0$. The area of the triangle it forms with the coordinate axes is:
A. $\dfrac{27}{2}$
B. $54$
C. $27$ ✓ Correct
D. $18$
Solution: The given lines meet at $(3, 4)$, so the required line is $2x + 3y = 18$, with intercepts $9$ and $6$; area $= \dfrac{1}{2}(9)(6) = 27$.