Various Forms of the Equation of a Line — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Various Forms of the 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 — 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 $x = -7$.
Q2 — 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: $y = mx + c = -3x + 5$.
Q3 — 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. $y - 3 = 2(x + 4)$.
Q4 — 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 $b = 7$.
Q5 — 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).
Q6 — Various Forms of the Equation of a Line · medium · numerical
The line through $(2, -4)$ with slope $2$ cuts intercepts on the x-axis and y-axis respectively equal to:
A. $4$ and $-8$ ✓ Correct
B. $8$ and $-4$
C. $4$ and $8$
D. $-4$ and $8$
Solution: $y + 4 = 2(x - 2) \Rightarrow y = 2x - 8$, so the x-intercept is $4$ and the y-intercept is $-8$.
Q7 — Various Forms of the Equation of a Line · medium · numerical
The equation of the line passing through $(-3, 5)$ and having equal (non-zero) intercepts on the axes is:
A. $y - x = 8$
B. $x + y = -2$
C. $x + y = 8$
D. $x + y = 2$ ✓ Correct
Solution: Equal intercepts give $x + y = a$; substituting $(-3, 5)$: $a = -3 + 5 = 2$, so the line is $x + y = 2$.
Q8 — Various Forms of the Equation of a Line · medium · numerical
A line passes through $(2, 6)$ and its intercepts on the x-axis and y-axis are in the ratio $2 : 3$. Its equation is:
A. $3x + 2y = 18$ ✓ Correct
B. $3x + 2y = 12$
C. $2x + 3y = 22$
D. $2x + 3y = 18$
Solution: With intercepts $2t$ and $3t$: $\dfrac{x}{2t} + \dfrac{y}{3t} = 1 \Rightarrow 3x + 2y = 6t$; through $(2, 6)$: $6t = 6 + 12 = 18$, so $3x + 2y = 18$.
Q9 — Various Forms of the Equation of a Line · medium · numerical
The portion of a line intercepted between the coordinate axes is bisected at the point $(-5, 3)$. The equation of the line is:
A. $3x + 5y = 30$
B. $3x - 5y = 30$
C. $3x - 5y + 30 = 0$ ✓ Correct
D. $5x - 3y + 30 = 0$
Solution: The line meets the axes at $(-10, 0)$ and $(0, 6)$ (their midpoint is $(-5, 3)$), so $\dfrac{x}{-10} + \dfrac{y}{6} = 1$, i.e. $3x - 5y + 30 = 0$.
Q10 — Various Forms of the Equation of a Line · medium · numerical
The equation of the line on which the perpendicular from the origin has length $7$ and this perpendicular makes an angle of $30^\circ$ with the positive x-axis is:
A. $x + \sqrt{3}y = 14$
B. $\sqrt{3}x + y = 7$
C. $\sqrt{3}x + y = 14$ ✓ Correct
D. $\sqrt{3}x - y = 14$
Solution: Normal form: $x\cos 30^\circ + y\sin 30^\circ = 7 \Rightarrow \dfrac{\sqrt{3}}{2}x + \dfrac{1}{2}y = 7$, i.e. $\sqrt{3}x + y = 14$.
Q11 — Various Forms of the Equation of a Line · medium · numerical
The equation of the line at a perpendicular distance of $3$ units from the origin, the perpendicular making an angle of $60^\circ$ with the positive x-axis, is:
A. $x - \sqrt{3}y = 6$
B. $x + \sqrt{3}y = 6$ ✓ Correct
C. $\sqrt{3}x + y = 6$
D. $x + \sqrt{3}y = 3$
Solution: $x\cos 60^\circ + y\sin 60^\circ = 3 \Rightarrow \dfrac{1}{2}x + \dfrac{\sqrt{3}}{2}y = 3$, i.e. $x + \sqrt{3}y = 6$.
Q12 — Various Forms of the Equation of a Line · medium · numerical
For the line $x + y = 4$, the length $p$ of the perpendicular from the origin and the angle $\omega$ this perpendicular makes with the positive x-axis are:
A. $p = 4,\ \omega = 45^\circ$
B. $p = 2\sqrt{2},\ \omega = 135^\circ$
C. $p = 2,\ \omega = 45^\circ$
D. $p = 2\sqrt{2},\ \omega = 45^\circ$ ✓ Correct
Solution: Dividing by $\sqrt{1^2 + 1^2} = \sqrt{2}$: $\dfrac{x}{\sqrt{2}} + \dfrac{y}{\sqrt{2}} = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}$, so $\cos\omega = \sin\omega = \dfrac{1}{\sqrt{2}}$, giving $\omega = 45^\circ$ and $p = 2\sqrt{2}$.
Q13 — Various Forms of the Equation of a Line · medium · numerical
The area (in square units) of the triangle formed by the line $3x + 4y = 12$ with the coordinate axes is:
A. $24$
B. $7$
C. $12$
D. $6$ ✓ Correct
Solution: The intercepts are $4$ and $3$, so the area is $\dfrac{1}{2} \times 4 \times 3 = 6$.
Q14 — Various Forms of the Equation of a Line · medium · numerical
The equation of the line on which the perpendicular from the origin has length $5$ and makes an angle of $120^\circ$ with the positive x-axis is:
A. $x + \sqrt{3}y = 10$
B. $x - \sqrt{3}y = 10$
C. $\sqrt{3}y - x = 10$ ✓ Correct
D. $\sqrt{3}x - y = 10$
Solution: $x\cos 120^\circ + y\sin 120^\circ = 5 \Rightarrow -\dfrac{1}{2}x + \dfrac{\sqrt{3}}{2}y = 5$, i.e. $\sqrt{3}y - x = 10$.
Q15 — Various Forms of the Equation of a Line · hard · numerical
A line passes through $(2, 2)$ and the sum of its intercepts on the axes is $9$. If its x-intercept is smaller than its y-intercept, the equation of the line is:
A. $x + 2y = 6$
B. $x + y = 9$
C. $2x + y = 9$
D. $2x + y = 6$ ✓ Correct
Solution: With $\dfrac{x}{a} + \dfrac{y}{9 - a} = 1$ through $(2, 2)$: $a^2 - 9a + 18 = 0 \Rightarrow a = 3$ or $a = 6$; the smaller x-intercept gives $a = 3$, $b = 6$, i.e. $\dfrac{x}{3} + \dfrac{y}{6} = 1$, or $2x + y = 6$.
Q16 — Various Forms of the Equation of a Line · hard · numerical
A line meets the x-axis at $A$ and the y-axis at $B$. The point $(-2, 4)$ divides $AB$ in the ratio $1 : 2$ (measured from $A$ towards $B$). The equation of the line is:
A. $4x + y + 12 = 0$
B. $x - 4y + 12 = 0$
C. $4x - y = 12$
D. $4x - y + 12 = 0$ ✓ Correct
Solution: With $A(a, 0)$ and $B(0, b)$, the dividing point is $\left(\dfrac{2a}{3}, \dfrac{b}{3}\right) = (-2, 4)$, so $a = -3$, $b = 12$ and $\dfrac{x}{-3} + \dfrac{y}{12} = 1$, i.e. $4x - y + 12 = 0$.
Q17 — Various Forms of the Equation of a Line · hard · numerical
The line $4x + 5y + c = 0$ forms with the coordinate axes a triangle of area $10$ square units. The positive value of $c$ is:
A. $20$ ✓ Correct
B. $5$
C. $10$
D. $40$
Solution: The intercepts are $-\dfrac{c}{4}$ and $-\dfrac{c}{5}$, so the area is $\dfrac{1}{2} \cdot \dfrac{c^2}{20} = \dfrac{c^2}{40} = 10 \Rightarrow c^2 = 400 \Rightarrow c = 20$.
Q18 — Various Forms of the Equation of a Line · hard · numerical
When the line $\sqrt{3}x - y + 8 = 0$ is written in the normal form $x\cos\omega + y\sin\omega = p$, the values of $p$ and $\omega$ are:
A. $p = 8,\ \omega = 150^\circ$
B. $p = 4,\ \omega = 60^\circ$
C. $p = 4,\ \omega = 30^\circ$
D. $p = 4,\ \omega = 150^\circ$ ✓ Correct
Solution: Rewriting with a positive right side: $-\sqrt{3}x + y = 8$; dividing by $\sqrt{3 + 1} = 2$ gives $-\dfrac{\sqrt{3}}{2}x + \dfrac{1}{2}y = 4$, so $\cos\omega = -\dfrac{\sqrt{3}}{2}$, $\sin\omega = \dfrac{1}{2}$, i.e. $\omega = 150^\circ$ and $p = 4$.
Q19 — Various Forms of the Equation of a Line · hard · numerical
A line through $(2, 1)$ forms with the coordinate axes a triangle of area $4$ square units in the first quadrant. The equation of the line is:
A. $x + 2y = 8$
B. $2x + y = 4$
C. $x + y = 3$
D. $x + 2y = 4$ ✓ Correct
Solution: With $\dfrac{x}{a} + \dfrac{y}{b} = 1$: $ab = 8$ and $\dfrac{2}{a} + \dfrac{1}{b} = 1$ give $b^2 - 4b + 4 = 0 \Rightarrow b = 2$, $a = 4$, so the line is $\dfrac{x}{4} + \dfrac{y}{2} = 1$, i.e. $x + 2y = 4$.
Q20 — Various Forms of the Equation of a Line · hard · numerical
The area (in square units) of the triangle formed by the coordinate axes and the line passing through the points $(-6, 8)$ and $(9, -2)$ is:
A. $6$
B. $12$ ✓ Correct
C. $10$
D. $24$
Solution: The line has slope $\dfrac{-2 - 8}{9 - (-6)} = -\dfrac{2}{3}$ and equation $2x + 3y = 12$, with intercepts $6$ and $4$; area $= \dfrac{1}{2} \times 6 \times 4 = 12$.