Q.What are the points on the y-axis whose distance from the line $\dfrac{x}{3} + \dfrac{y}{4} = 1$ is $4$ units.

Showing the 12 most recent of 18 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The shortest distance from the point $(-10, 10, -10)$ to the $z$-axis, is (A) $2\sqrt{10}$ (B) $10\sqrt{3}$ (C) $10$ (D) $3\sqrt{10}$ (E) $10\sqrt{2}$

   ›Reveal solutionSolution

   Distance from a point to the $z$-axis is $\sqrt{x^2+y^2}$.

   For $(-10,10,-10)$, the shortest distance to the $z$-axis is $\sqrt{(-10)^2+(10)^2}=\sqrt{200}=10\sqrt2$.

   ✓Final answer

   The correct option is (E).

2. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The shortest distance between the straight line $\vec{r}=3\hat{i}+12\hat{j}+5\hat{k}+\lambda(2\hat{i}),\ \lambda\in\mathbb{R}$ and the $x$-axis is (A) $7$ (B) $12$ (C) $3\sqrt{5}$ (D) $5\sqrt{3}$ (E) $13$

   ›Reveal solutionSolution

   The line is parallel to the $x$-axis; the shortest distance is $\sqrt{12^2+5^2}=13$.

   The line $\vec r=3\hat i+12\hat j+5\hat k+\lambda(2\hat i)$ has direction $(1,0,0)$, the same as the $x$-axis, so the two lines are parallel.

   The shortest distance is the perpendicular distance from any point on the line, say $(3,12,5)$, to the $x$-axis. Distance from the $x$-axis ignores the $x$-coordinate:

   $d=\sqrt{12^{2}+5^{2}}=\sqrt{144+25}=\sqrt{169}=13.$

   ✓Final answer

   The correct option is (E).

3. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.The nearest point on the line $x+2y=5$ from the point $P(7,9)$ is equal to (A) $(6,1)$ (B) $(7,6)$ (C) $(2,3)$ (D) $(8,3)$ (E) $(3,1)$

   ›Reveal solutionSolution

   The nearest point is the foot of the perpendicular from $P$ to the line. Parametrize along the normal direction $(1,2)$: point $(7+t,9+2t)$ on the line gives $t=-4$, so the foot is $(3,1)$.

   The nearest point on $x+2y=5$ from $P(7,9)$ is the foot of the perpendicular. The line's normal direction is $(1,2)$, so points on the perpendicular through $P$ are

   $(7+t,\;9+2t).$

   Require this to lie on the line:

   $(7+t)+2(9+2t)=5\;\Rightarrow\;7+t+18+4t=5\;\Rightarrow\;25+5t=5\;\Rightarrow\;t=-4.$

   Foot of perpendicular:

   $(7-4,\;9-8)=(3,1).$

   ✓Final answer

   The correct option is (E).

4. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.The distance of the point $P(1,-3)$ from the line $2y-3x=4$ is (A) $\sqrt{13}$ units (B) $13$ units (C) $7$ units (D) $\sqrt{7}$ units (E) $2\sqrt{13}$ units

   ›Reveal solutionSolution

   Applying the point–line distance formula gives $\frac{13}{\sqrt{13}}=\sqrt{13}$.

   Write the line $2y-3x=4$ as $-3x+2y-4=0$. Distance from $P(1,-3)$:

   $d=\frac{|-3(1)+2(-3)-4|}{\sqrt{(-3)^2+2^2}}=\frac{|-3-6-4|}{\sqrt{13}}=\frac{13}{\sqrt{13}}=\sqrt{13}\ \text{units}.$

   ✓Final answer

   The correct option is (A).

5. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The perpendicular distance between the lines $3x + 4y - 6 = 0$ and $6x + 8y + 18 = 0$ is (A) 15 (B) 12 (C) 9 (D) 3 (E) 0

   ›Reveal solutionSolution

   Make the two lines have identical $x,y$ coefficients, then apply the parallel-line distance formula.

   Divide $6x+8y+18=0$ by $2$: $3x+4y+9=0$. This is parallel to $3x+4y-6=0$.

   Distance $=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}=\frac{|-6-9|}{\sqrt{9+16}}=\frac{15}{5}=3$.

   ✓Final answer

   The correct option is (D).

6. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The shortest distance between the lines $\vec{r} = \hat{i} + \hat{j} + 3\hat{k} + \lambda(2\hat{i} + 2\hat{j} + \hat{k})$ and $\vec{r} = (2\mu + 1)\hat{i} + (2\mu - 1)\hat{j} + (\mu + 1)\hat{k}$, where $\lambda$ and $\mu$ are parameters, is (A) 1 (B) 6 (C) 3 (D) 4 (E) 2

   ›Reveal solutionSolution

   Recognize the lines as parallel, then apply the parallel-line distance formula.

   Line 1: point $(1,1,3)$, direction $(2,2,1)$. Line 2: $(2\mu+1,2\mu-1,\mu+1)$ has point $(1,-1,1)$ at $\mu=0$ and direction $(2,2,1)$ — same direction, so parallel.

   Joining vector $=(1-1,-1-1,1-3)=(0,-2,-2)$.

   $(0,-2,-2)\times(2,2,1)=(2,-4,4)$, magnitude $\sqrt{4+16+16}=6$.

   Distance $=\frac{6}{|(2,2,1)|}=\frac{6}{3}=2$.

   ✓Final answer

   The correct option is (E).

7. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.Let $ABC$ be an equilateral triangle. If the coordinates of $A$ are $(-2, 2)$ and the side $BC$ is along the line $x + y = 6$, then the length of the side of the triangle is (A) $2\sqrt{3}$ (B) $3\sqrt{2}$ (C) $4\sqrt{6}$ (D) $6\sqrt{6}$ (E) $2\sqrt{6}$

   ›Reveal solutionSolution

   The side length is $2\sqrt{6}$.

   Concept and Intuition

   In an equilateral triangle the perpendicular distance from a vertex to the opposite side is the altitude $h=\tfrac{\sqrt3}{2}s$. Compute that distance from $A$ to line $BC$, then recover $s$.

   Step-by-Step Solution

   1. Perpendicular distance from $A(-2,2)$ to $x+y=6$: $h=\dfrac{|-2+2-6|}{\sqrt2}=\dfrac{6}{\sqrt2}=3\sqrt2$.
   2. For an equilateral triangle $h=\dfrac{\sqrt3}{2}s$, so $s=\dfrac{2h}{\sqrt3}=\dfrac{2\cdot 3\sqrt2}{\sqrt3}=\dfrac{6\sqrt2}{\sqrt3}$.
   3. $s=6\sqrt{2/3}=\dfrac{6\sqrt6}{3}=2\sqrt6$.

   Common Mistakes

   • Using $h=s$ instead of $h=\tfrac{\sqrt3}{2}s$.
   • Dropping the $\sqrt2$ in the point–line distance formula.

   ✓Final answer

   The correct option is (E) — $2\sqrt6$.

   ANSWER: E

8. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The line $x + y = 2$ touches a circle. If the centre of the circle is at $(-4,0)$, then the radius of the circle is (A) $2\sqrt{2}$ (B) $\frac{3\sqrt{2}}{2}$ (C) $\sqrt{2}$ (D) $\frac{\sqrt{2}}{2}$ (E) $3\sqrt{2}$

   ›Reveal solutionSolution

   A tangent line's distance from the centre equals the radius; compute the perpendicular distance from $(-4,0)$ to $x+y-2=0$.

   The line $x + y = 2$, i.e. $x + y - 2 = 0$, touches the circle, so the radius equals the perpendicular distance from the centre $(-4,0)$:

   $r = \frac{|(-4) + 0 - 2|}{\sqrt{1^2 + 1^2}} = \frac{|-6|}{\sqrt2} = \frac{6}{\sqrt2} = 3\sqrt2.$

   ✓Final answer

   The correct option is (E).

9. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.If the distance of the line $4x - 3y + k = 0$ from the point $(1, 2)$ is $5$ units, then the values of $k$ are (A) $27,-23$ (B) $-27,23$ (C) $29,-24$ (D) $-29,24$ (E) $-28,-25$

   ›Reveal solutionSolution

   Distance formula: $\frac{|4(1)-3(2)+k|}{5}=5\Rightarrow|k-2|=25\Rightarrow k=27,\,-23$.

   Apply the point-line distance. $d=\frac{|4x_0-3y_0+k|}{\sqrt{4^{2}+(-3)^{2}}}=\frac{|4(1)-3(2)+k|}{5}=\frac{|k-2|}{5}$.

   Set equal to 5. $\frac{|k-2|}{5}=5\Rightarrow|k-2|=25\Rightarrow k-2=\pm25\Rightarrow k=27\text{ or }k=-23$.

   ✓Final answer

   The correct option is (A).

10. KEAM 2025Set eng-2025-04264 marksMCQ
    Q.Let $\overrightarrow{OP} = 2\hat{j}$ be the position vector a point P. Let $\vec{r} = \hat{j} + \lambda(\hat{i} + \hat{j})$ be a straight line. The distance of the point P from the line is (A) $\dfrac{\sqrt{2}}{2}$ (B) $\dfrac{\sqrt{3}}{3}$ (C) $\dfrac{\sqrt{6}}{3}$ (D) $\dfrac{\sqrt{2}}{3}$ (E) $\dfrac{\sqrt{2}}{4}$

    ›Reveal solutionSolution

    Point $P=(0,2)$; line passes through $(0,1)$ with direction $(1,1)$. Distance $=\dfrac{|(P-A)\times \hat d|}{1}=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$.

    Here $\overrightarrow{OP}=2\hat j$ gives $P=(0,2)$. The line $\vec r=\hat j+\lambda(\hat i+\hat j)$ passes through $A=(0,1)$ with direction $\vec d=(1,1)$.

    Vector $\overrightarrow{AP}=(0,1)$. The 2-D cross magnitude with $\vec d$: $|0\cdot1-1\cdot1|=1$. Divide by $|\vec d|=\sqrt2$:

    $\text{distance}=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}.$

    ✓Final answer

    The correct option is (A).

11. KEAM 2025Set eng-2025-04284 marksMCQ
    Q.The point with integral coordinates on the line $x + y = 1$, that lie at a distance 2 units from the line $5x + 12y = 0$, is (A) $(-7, 8)$ (B) $(-1, 2)$ (C) $(-12, 13)$ (D) $(-2, 3)$ (E) $(-3, 4)$

    ›Reveal solutionSolution

    Parametrize the point on $x+y=1$ and impose the distance condition; the integer solution is $(-2,3)$.

    A point on $x+y=1$ is $(a,1-a)$. Its distance from $5x+12y=0$ is

    $\frac{|5a+12(1-a)|}{\sqrt{5^2+12^2}}=\frac{|12-7a|}{13}=2.$

    Thus $|12-7a|=26$, giving $12-7a=26\Rightarrow a=-2$ (integer), or $12-7a=-26\Rightarrow a=\tfrac{38}{7}$ (not an integer).

    So $a=-2$ and the point is $(-2,\,1-(-2))=(-2,3)$.

    ✓Final answer

    The correct option is (D).

12. KEAM 2025Set eng-2025-04284 marksMCQ
    Q.The shortest distance between the point $(2, 3, 4)$ and the line $\dfrac{x - 4}{-2} = \dfrac{y - 4}{2} = \dfrac{z - 6}{1}$ is (A) $12$ (B) $9$ (C) $3$ (D) $\sqrt{5}$ (E) $\sqrt{3}$

    ›Reveal solutionSolution

    Use the point-to-line formula $\dfrac{|\vec{AP}\times\vec d|}{|\vec d|}$ with $A=(4,4,6)$, $\vec d=(-2,2,1)$.

    The line passes through $A=(4,4,6)$ with direction $\vec d=(-2,2,1)$, $|\vec d|=3$. For $P=(2,3,4)$:

    $\vec{AP}=P-A=(-2,-1,-2).$

    Compute the cross product:

    $\vec{AP}\times\vec d=\begin{vmatrix} i & j & k\\ -2 & -1 & -2\\ -2 & 2 & 1\end{vmatrix}=(3,\,6,\,-6),$

    with magnitude $\sqrt{9+36+36}=9$.

    The shortest distance is

    $\frac{|\vec{AP}\times\vec d|}{|\vec d|}=\frac{9}{3}=3.$

    ✓Final answer

    The correct option is (C).