Distance of a Point from a Line — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Distance of a Point from 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 — 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: $d = \dfrac{|3(2) - 4(1) + 8|}{\sqrt{3^2 + 4^2}} = \dfrac{10}{5} = 2$.
Q2 — 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: $d = \dfrac{|5(0) + 12(0) - 39|}{\sqrt{25 + 144}} = \dfrac{39}{13} = 3$.
Q3 — 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: $d = \dfrac{|1 + 2 + 5|}{\sqrt{1 + 1}} = \dfrac{8}{\sqrt{2}} = 4\sqrt{2}$.
Q4 — 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: $d = \dfrac{|c_1 - c_2|}{\sqrt{A^2 + B^2}} = \dfrac{|-9 - 11|}{5} = \dfrac{20}{5} = 4$.
Q5 — 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: $d = \dfrac{|12(5) - 5(1) + 10|}{\sqrt{144 + 25}} = \dfrac{65}{13} = 5$.
Q6 — Distance of a Point from a Line · medium · numerical
The distance between the parallel lines $3x - 4y + 5 = 0$ and $6x - 8y - 15 = 0$ is:
A. $\dfrac{5}{2}$ ✓ Correct
B. $4$
C. $\dfrac{5}{4}$
D. $\dfrac{25}{2}$
Solution: Write the second line as $3x - 4y - \dfrac{15}{2} = 0$; then $d = \dfrac{\left|5 + \frac{15}{2}\right|}{5} = \dfrac{25/2}{5} = \dfrac{5}{2}$.
Q7 — Distance of a Point from a Line · medium · numerical
If the perpendicular distance of the point $(5, 2)$ from the line $4x - 3y + k = 0$ is $4$, the positive value of $k$ is:
A. $20$
B. $34$
C. $-34$
D. $6$ ✓ Correct
Solution: $\dfrac{|20 - 6 + k|}{5} = 4$ gives $|k + 14| = 20$, so $k = 6$ or $k = -34$; the positive value is $6$.
Q8 — Distance of a Point from a Line · medium · numerical
The points on the $x$-axis whose perpendicular distance from the line $4x + 3y - 12 = 0$ is $4$ are:
A. $(8, 0)$ and $(-2, 0)$ ✓ Correct
B. $(-8, 0)$ and $(2, 0)$
C. $(8, 0)$ and $(2, 0)$
D. $(3, 0)$ and $(-3, 0)$
Solution: For $(a, 0)$: $\dfrac{|4a - 12|}{5} = 4$, so $4a - 12 = \pm 20$, giving $a = 8$ or $a = -2$.
Q9 — Distance of a Point from a Line · medium · numerical
The distance between the parallel lines $x - y + 9 = 0$ and $x - y + 3 = 0$ is:
A. $6$
B. $3$
C. $3\sqrt{2}$ ✓ Correct
D. $2\sqrt{3}$
Solution: $d = \dfrac{|9 - 3|}{\sqrt{1 + 1}} = \dfrac{6}{\sqrt{2}} = 3\sqrt{2}$.
Q10 — Distance of a Point from a Line · medium · numerical
The point on the line $4x + 3y - 21 = 0$ that lies in the first quadrant and is equidistant from the two coordinate axes is:
A. $(-3, 3)$
B. $(3, -3)$
C. $(3, 3)$ ✓ Correct
D. $\left(\dfrac{21}{4}, 0\right)$
Solution: Equidistant from both axes in the first quadrant means $y = x$; then $4x + 3x = 21$, so $x = 3$, giving $(3, 3)$.
Q11 — Distance of a Point from a Line · medium · numerical
The line $8x + 15y - 120 = 0$ meets the coordinate axes at $A$ and $B$. The length of the perpendicular from the origin to $AB$ is:
A. $8$
B. $\dfrac{120}{17}$ ✓ Correct
C. $\dfrac{60}{17}$
D. $\dfrac{120}{23}$
Solution: $d = \dfrac{|-120|}{\sqrt{64 + 225}} = \dfrac{120}{17}$; check by area: $A(15, 0)$, $B(0, 8)$, $AB = 17$ and $\dfrac{1}{2}(17)\cdot\dfrac{120}{17} = 60 = \dfrac{1}{2}(15)(8)$.
Q12 — Distance of a Point from a Line · medium · numerical
The perpendicular distance of the point $(-2, 3)$ from the line $5x + 12y + 65 = 0$ is:
A. $7$ ✓ Correct
B. $91$
C. $\dfrac{19}{13}$
D. $5$
Solution: $d = \dfrac{|5(-2) + 12(3) + 65|}{\sqrt{25 + 144}} = \dfrac{91}{13} = 7$.
Q13 — Distance of a Point from a Line · medium · numerical
The points on the $y$-axis whose perpendicular distance from the line $12x + 5y - 10 = 0$ is $5$ are:
A. $(0, -15)$ and $(0, 11)$
B. $(0, 15)$ and $(0, -11)$ ✓ Correct
C. $(0, 13)$ and $(0, -13)$
D. $(0, 15)$ and $(0, 11)$
Solution: For $(0, b)$: $\dfrac{|5b - 10|}{13} = 5$, so $5b - 10 = \pm 65$, giving $b = 15$ or $b = -11$.
Q14 — Distance of a Point from a Line · medium · numerical
The point on the line $x - y - 1 = 0$ that is equidistant from the points $A(3, 0)$ and $B(11, 2)$ is:
A. $(6, 5)$ ✓ Correct
B. $(7, 6)$
C. $(6, -5)$
D. $(4, 3)$
Solution: The perpendicular bisector of $AB$ is $4x + y - 29 = 0$; with $y = x - 1$ this gives $5x = 30$, so the point is $(6, 5)$, where both distances equal $\sqrt{34}$.
Q15 — Distance of a Point from a Line · hard · numerical
The point on the line $x + y = 8$ nearest to the origin is:
A. $(2, 6)$
B. $(4, -4)$
C. $(0, 8)$
D. $(4, 4)$ ✓ Correct
Solution: The nearest point is the foot of the perpendicular from the origin, which lies along $y = x$; then $x + x = 8$ gives $(4, 4)$.
Q16 — Distance of a Point from a Line · hard · numerical
The centroid of the triangle with vertices $(0, 1)$, $(2, 9)$ and $(4, 5)$ is at a distance $d$ from the line $3x + 4y + 4 = 0$. Then $d$ equals:
A. $\dfrac{22}{5}$
B. $6$ ✓ Correct
C. $30$
D. $5$
Solution: The centroid is $\left(\dfrac{0+2+4}{3}, \dfrac{1+9+5}{3}\right) = (2, 5)$, so $d = \dfrac{|3(2) + 4(5) + 4|}{5} = \dfrac{30}{5} = 6$.
Q17 — Distance of a Point from a Line · hard · numerical
A line parallel to $5x - 12y + 2 = 0$ and at a distance $1$ from it is $5x - 12y + c = 0$. The greater value of $c$ is:
A. $15$ ✓ Correct
B. $11$
C. $-11$
D. $13$
Solution: $\dfrac{|c - 2|}{\sqrt{25 + 144}} = 1$ gives $|c - 2| = 13$, so $c = 15$ or $c = -11$; the greater value is $15$.
Q18 — Distance of a Point from a Line · hard · numerical
If the distance of the point $(k, 2)$ from the line $12x - 5y - 3 = 0$ is $2$, the positive value of $k$ is:
A. $-\dfrac{13}{12}$
B. $\dfrac{13}{4}$ ✓ Correct
C. $\dfrac{13}{2}$
D. $\dfrac{13}{6}$
Solution: $\dfrac{|12k - 10 - 3|}{13} = 2$ gives $|12k - 13| = 26$, so $12k = 39$ or $12k = -13$; the positive value is $k = \dfrac{13}{4}$.
Q19 — Distance of a Point from a Line · hard · numerical
If the distance between the parallel lines $5x + 12y - 8 = 0$ and $10x + 24y + k = 0$ is $1$, the positive value of $k$ is:
A. $-42$
B. $42$
C. $10$ ✓ Correct
D. $5$
Solution: Write the second line as $5x + 12y + \dfrac{k}{2} = 0$; then $\dfrac{\left|\frac{k}{2} + 8\right|}{13} = 1$ gives $k = 10$ or $k = -42$.
Q20 — Distance of a Point from a Line · hard · numerical
The distance of the point of intersection of the lines $2x - y - 7 = 0$ and $x + 3y - 14 = 0$ from the line $5x - 12y + 24 = 0$ is:
A. $\dfrac{35}{13}$
B. $\dfrac{85}{13}$
C. $13$
D. $1$ ✓ Correct
Solution: The lines meet at $(5, 3)$, and $d = \dfrac{|5(5) - 12(3) + 24|}{13} = \dfrac{13}{13} = 1$.