Q.Find the distance between the points $P(1, -3, 4)$ and $Q(-4, 1, 2)$.

Showing the 12 most recent of 19 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Let point Q be the image of point $P(2,-1)$ in the line $3 x+5=4 y$. Find the area of the circle that has the segment PQ as the diameter. (A) $9 \pi$ (B) $36 \pi$ (C) $1.96 \pi$ (D) $3 \pi$

   ›Reveal solutionSolution

   The image $Q$ is the mirror reflection of $P$ across the line, so $PQ = 2\times(\text{distance from }P\text{ to the line})$. That distance is $3$, giving diameter $PQ=6$, radius $3$, and area $\pi(3)^2=9\pi$ — option (A).

   Concept & Intuition

   When $Q$ is the reflection of $P$ in a line, the line is the perpendicular bisector of $PQ$. Hence the foot of perpendicular from $P$ is the midpoint of $PQ$, and $|PQ| = 2d$, where $d$ is the perpendicular distance from $P$ to the line. Since $PQ$ is the diameter, the radius is $d$.

   Step-by-step solution

   1. Write the line in standard form. From $3x+5=4y$:

   $3x-4y+5=0.$

   2. Perpendicular distance from $P(2,-1)$ to this line:

   $d=\frac{|3(2)-4(-1)+5|}{\sqrt{3^2+(-4)^2}}=\frac{|6+4+5|}{\sqrt{25}}=\frac{15}{5}=3.$

   3. $Q$ is the reflection of $P$, so

   $|PQ|=2d=6\quad\Rightarrow\quad \text{radius}=\frac{|PQ|}{2}=3.$

   4. Area of the circle:

   $A=\pi r^2=\pi(3)^2=9\pi.$

   ✓Final answer

   Area $= 9\pi$ square units — option (A).

   ANSWER: A

2. COMEDK 2026Set 2026-A1 markMCQ
   Q.A line L passes through the point of intersection of the lines $3 x+y-10=0$ and $x-y-2=0$. If the perpendicular distance of the line $L$ from the point $(5,1)$ is exactly $\frac{2}{\sqrt{5}}$ units, which of the following represents the correct equation for line L ? (A) $x+2 y-5=0$ (B) $2 x+y-7=0$ (C) $x-2 y+1=0$ (D) $2 x-y-5=0$

   ›Reveal solutionSolution

   The line L must pass through the intersection of the two given lines and be at a perpendicular distance of $2/\sqrt{5}$ from (5,1). Solving the family of lines through that intersection and applying the distance formula yields two possible lines; only one matches the given options, which is option (B).

   We start by finding the fixed point through which L must pass — the intersection of the two given lines. Then we write the general equation of any line through that point (using a parameter for slope). The condition on perpendicular distance from (5,1) gives an equation in the slope; solving it yields two slopes, hence two possible lines. Finally we check which of the given options matches one of these.

   1. Find the intersection point of the two given lines. Solve

   $3x + y - 10 = 0 \quad\text{and}\quad x - y - 2 = 0.$

   Adding the equations eliminates $y$:

   $(3x + y - 10) + (x - y - 2) = 0 \;\Rightarrow\; 4x - 12 = 0 \;\Rightarrow\; x = 3.$

   Substitute $x = 3$ into $x - y - 2 = 0$:

   $3 - y - 2 = 0 \;\Rightarrow\; y = 1.$

   So the intersection point is $P(3,1)$.

   2. Write the family of lines through $P(3,1)$. Any non-vertical line through $P$ can be written as

   $y - 1 = m(x - 3),$

   where $m$ is the slope. Rearranging into standard form:

   $mx - y + (1 - 3m) = 0.$

   (We will also consider the vertical line $x = 3$ separately later.)

   3. Apply the perpendicular distance condition. The distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is

   $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}.$

   For our line $mx - y + (1 - 3m) = 0$ and point $(5,1)$:

   $A = m,\; B = -1,\; C = 1 - 3m.$

   The distance is

   $\frac{|m\cdot 5 + (-1)\cdot 1 + (1 - 3m)|}{\sqrt{m^2 + 1}} = \frac{|5m - 1 + 1 - 3m|}{\sqrt{m^2 + 1}} = \frac{|2m|}{\sqrt{m^2 + 1}}.$

   We are told this equals $\frac{2}{\sqrt{5}}$. So

   $\frac{|2m|}{\sqrt{m^2 + 1}} = \frac{2}{\sqrt{5}}.$

   4. Solve for $m$. Cancel the factor 2 (valid since both sides are positive):

   $\frac{|m|}{\sqrt{m^2 + 1}} = \frac{1}{\sqrt{5}}.$

   Square both sides:

   $\frac{m^2}{m^2 + 1} = \frac{1}{5}.$

   Cross-multiply:

   $5m^2 = m^2 + 1 \;\Rightarrow\; 4m^2 = 1 \;\Rightarrow\; m^2 = \frac{1}{4} \;\Rightarrow\; m = \pm \frac{1}{2}.$

   5. Write the two possible lines.
      • For $m = \frac{1}{2}$:

   $y - 1 = \frac{1}{2}(x - 3) \;\Rightarrow\; 2y - 2 = x - 3 \;\Rightarrow\; x - 2y - 1 = 0.$

   • For $m = -\frac{1}{2}$:

   $y - 1 = -\frac{1}{2}(x - 3) \;\Rightarrow\; 2y - 2 = -x + 3 \;\Rightarrow\; x + 2y - 5 = 0.$

   Also check the vertical line $x = 3$: its distance from $(5,1)$ is $|5-3| = 2$, not $2/\sqrt{5}$, so it is not a solution.

   6. Match with the given options.

      The options are:

      (A) $x + 2y - 5 = 0$

      (B) $2x + y - 7 = 0$

      (C) $x - 2y + 1 = 0$

      (D) $2x - y - 5 = 0$

      Our two lines are $x + 2y - 5 = 0$ and $x - 2y - 1 = 0$.

      • $x + 2y - 5 = 0$ is exactly option (A).
      • $x - 2y - 1 = 0$ is not listed (option (C) is $x - 2y + 1 = 0$, different sign).

      So the line from our set that appears among the choices is option (A). But wait — we must verify that option (A) indeed satisfies the distance condition. We already derived it from the condition, so it does. However, the problem asks “which of the following represents the correct equation for line L?” — implying only one option is correct. Since both $m = 1/2$ and $m = -1/2$ satisfy the distance condition, both lines are geometrically valid. But only one appears in the options. That is option (A).

   Watch out

   A common mistake is to forget the absolute value when squaring the distance equation, or to assume only one slope works. Both $m = \pm 1/2$ satisfy the condition; the vertical line $x = 3$ does not.

   Tip

   The family-of-lines approach (using a parameter) is powerful: it automatically enforces the “passes through intersection” condition, reducing the problem to a single-variable equation.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

3. COMEDK 2026Set 2026-M1 markMCQ
   Q.Distance between $8 x+15 y-20=0$ and $8 x+15 y+14=0$ is: (A) 2 units (B) 34 units (C) $\frac{10}{17}$ units (D) $\frac{4}{7}$ units

   ›Reveal solutionSolution

   The distance between two parallel lines is found by taking the absolute difference of their constant terms divided by the square root of the sum of squares of the coefficients. Here the distance is $\frac{| -20 - 14 |}{\sqrt{8^2 + 15^2}} = \frac{34}{17} = 2$ units, so the answer is (A).

   The key idea is that when two lines are parallel, the distance between them is constant — like the gap between two rails. The formula comes from projecting a point from one line onto the other along the normal direction.

   Why this works:

   For lines in the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, the perpendicular distance between them is $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. This is because the normal vector $(A, B)$ is the same for both, so the distance is just the difference in their offsets along that normal, scaled by the length of the normal.

   1. Identify the lines are parallel

      Both equations have the same coefficients for $x$ and $y$: $8x + 15y - 20 = 0$ and $8x + 15y + 14 = 0$. So they are parallel.

   2. Write them in standard form

      First line: $8x + 15y - 20 = 0$ → $C_1 = -20$

      Second line: $8x + 15y + 14 = 0$ → $C_2 = 14$

   3. Apply the distance formula

      Distance $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} = \frac{|-20 - 14|}{\sqrt{8^2 + 15^2}}$

   4. Simplify numerator and denominator

      Numerator: $|-34| = 34$

      Denominator: $\sqrt{64 + 225} = \sqrt{289} = 17$

   5. Compute the distance

      $d = \frac{34}{17} = 2$

   Watch out

   A common mistake is to forget the absolute value or to subtract the constants in the wrong order — but since we take absolute value, order doesn’t matter. Another pitfall is forgetting to divide by $\sqrt{A^2 + B^2}$; without it, you’d get 34, which is option (B) — a tempting distractor.

   Tip

   Notice that $8^2 + 15^2 = 289 = 17^2$, so the denominator is a perfect square. This often happens in textbook problems to keep the answer neat.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. COMEDK 2025Set 2025-A1 markMCQ
   Q.The length of the perpendicular from the point $P(1,-1,2)$ to the given line $\frac{x+1}{2}=\frac{y-2}{-3}=\frac{z+2}{4}$ is (A) $\sqrt{29}$ units (B) $\sqrt{21}$ units (C) $\sqrt{6}$ units (D) 0 units

   ›Reveal solutionSolution

   The shortest distance from a point to a line in 3D is found by projecting the vector from a point on the line to the given point onto the direction vector, then using the Pythagorean theorem. The perpendicular distance is $\sqrt{6}$ units.

   Concept & Intuition

   In 3D geometry, the perpendicular distance from a point to a line is the length of the line segment that meets the given line at a right angle. Instead of solving for the foot of the perpendicular directly, we can use vector projection:

   • Pick any convenient point $A$ on the line.
   • Form the vector $\overrightarrow{AP}$ from $A$ to the given point $P$.
   • The component of $\overrightarrow{AP}$ parallel to the line’s direction vector $\vec{d}$ gives the “along-the-line” part.
   • The component perpendicular to $\vec{d}$ is the shortest distance. By the Pythagorean theorem:

   $\text{distance} = \sqrt{|\overrightarrow{AP}|^2 - \left(\text{proj}_{\vec{d}} \overrightarrow{AP}\right)^2}.$

   Step-by-step solution

   1. Identify a point on the line and the direction vector The line is given in symmetric form:

   $\frac{x+1}{2} = \frac{y-2}{-3} = \frac{z+2}{4} = t.$

   Setting $t=0$ gives a convenient point $A(-1, 2, -2)$.

   The direction vector is $\vec{d} = (2, -3, 4)$.

   2. Form the vector from $A$ to $P$ $P = (1, -1, 2)$, so

   $\overrightarrow{AP} = P - A = (1 - (-1),\; -1 - 2,\; 2 - (-2)) = (2, -3, 4).$

   3. Compute the length of $\overrightarrow{AP}$

   $|\overrightarrow{AP}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}.$

   4. Compute the projection of $\overrightarrow{AP}$ onto $\vec{d}$ The scalar projection (component along $\vec{d}$) is

   $\text{comp}_{\vec{d}} \overrightarrow{AP} = \frac{\overrightarrow{AP} \cdot \vec{d}}{|\vec{d}|}.$

   Dot product:

   $(2)(2) + (-3)(-3) + (4)(4) = 4 + 9 + 16 = 29.$

   Length of $\vec{d}$:

   $|\vec{d}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{29}.$

   Hence

   $\text{comp}_{\vec{d}} \overrightarrow{AP} = \frac{29}{\sqrt{29}} = \sqrt{29}.$

   5. Apply the Pythagorean relation The perpendicular distance $d$ satisfies

   $d^2 = |\overrightarrow{AP}|^2 - \left(\text{comp}_{\vec{d}} \overrightarrow{AP}\right)^2.$

   Substituting:

   $d^2 = (\sqrt{29})^2 - (\sqrt{29})^2 = 29 - 29 = 0.$

   So $d = 0$.

   Watch out

   A distance of zero means the point $P$ lies on the line. Indeed, $\overrightarrow{AP} = (2, -3, 4)$ is exactly the direction vector $\vec{d}$, so $P$ is simply $A + \vec{d}$ — a point on the line. Many students might compute $\sqrt{29}$ for the projection and forget to subtract, but here the two lengths are equal, giving zero.

   Tip

   When $\overrightarrow{AP}$ is parallel to $\vec{d}$, the perpendicular distance is automatically zero. Always check if the point satisfies the line equation first — it saves time.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

5. COMEDK 2025Set 2025-A1 markMCQ
   Q.The point on the line $x+y=4$ that lie at a unit distance from the line $4 x+3 y=10$ is (A) $(2,2)$ (B) $(3,-1)$ (C) $(5,-1)$ (D) $(-7,11)$

   ›Reveal solutionSolution

   [!TLDR]

   Parametrise the point on $x+y=4$, apply the point-to-line distance formula to $4x+3y=10$, set it to $1$, and the valid option is $(-7,11)$.

   Concept

   The perpendicular distance from a point $(x_0,y_0)$ to a line $ax+by+c=0$ is $\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ — the standard CBSE Class-11 Straight Lines formula.

   Solution

   Let the required point lie on $x+y=4$, so it is $(x,\,4-x)$.

   Distance from the line $4x+3y-10=0$ (here $\sqrt{a^2+b^2}=\sqrt{16+9}=5$):

   $d=\frac{|4x+3(4-x)-10|}{5}=\frac{|4x+12-3x-10|}{5}=\frac{|x+2|}{5}.$

   Set $d=1$:

   $|x+2|=5 \;\Rightarrow\; x+2=\pm 5 \;\Rightarrow\; x=3 \text{ or } x=-7.$

   The corresponding points are $(3,\,1)$ and $(-7,\,11)$.

   Checking the options against both conditions (on the line AND unit distance):

   • (A) $(2,2)$: on line, but distance $=|2+2|/5=4/5\neq1$.
   • (B) $(3,-1)$: $3+(-1)=2\neq4$, not even on the line.
   • (C) $(5,-1)$: on line, but distance $=|5+2|/5=7/5\neq1$.
   • (D) $(-7,11)$: on line, distance $=|-7+2|/5=5/5=1$. ✓

   [!ANSWER]

   (D) $(-7,\,11)$

   Note

   This solution was worked out by our team and independently cross-checked by a second solve. The official answer key on record for this question could not be confirmed, so please cross-verify with the official paper where possible.

6. COMEDK 2025Set 2025-A1 markMCQ
   Q.If $Q(1,0,1)$ is the image of the point $P(a, b, c)$ in the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$ then $a+b+c$ is equal to : (A) 4 (B) 2 (C) 0 (D) $-$2

   ›Reveal solutionSolution

   Reflecting $Q$ across the line gives $P=(1,2,-3)$, so $a+b+c=0$.

   Since $Q$ is the image of $P$ in the line, $P$ is likewise the reflection of $Q$ in the line. Take the line point $A=(-1,3,0)$ and direction $\vec d=(2,-2,-1)$, with $|\vec d|^2=4+4+1=9$.

   Foot of perpendicular $F$ from $Q(1,0,1)$: with $\vec{AQ}=(2,-3,1)$,

   $t=\frac{\vec{AQ}\cdot\vec d}{|\vec d|^2}=\frac{(2)(2)+(-3)(-2)+(1)(-1)}{9}=\frac{4+6-1}{9}=1.$

   $F=A+t\vec d=(-1,3,0)+(2,-2,-1)=(1,1,-1).$

   Reflecting $Q$ through $F$:

   $P=2F-Q=(2-1,\,2-0,\,-2-1)=(1,2,-3).$

   Thus $a=1,\ b=2,\ c=-3$, giving

   $a+b+c=1+2-3=0.$

   ✓Final answer

   $a+b+c=0$, which is option (C).

7. COMEDK 2025Set 2025-M1 markMCQ
   Q.Distance of the point $(-2,3)$ from the line $12 x-5 y-2=0$ is $\frac{41}{k}$. Then the value of $k$ is (A) $\sqrt{134}$ (B) $\sqrt{13}$ (C) 13 (D) 1

   ›Reveal solutionSolution

   The distance from a point to a line is given by a standard formula; here it simplifies to $\frac{41}{13}$, so $k = 13$. The correct option is (C).

   ---

   The key idea is the perpendicular distance formula from a point to a line. When we say "distance of a point from a line," we mean the shortest distance — the length of the perpendicular segment connecting the point to the line. That’s exactly what the formula computes.

   The formula: For a line written as $Ax + By + C = 0$ and a point $(x_1, y_1)$, the distance is

   $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}.$

   Here, the problem gives the distance as $\frac{41}{k}$, so we just compute the numerator and denominator separately, then match.

   ---

   1. Identify $A$, $B$, $C$ and the point

      Line: $12x - 5y - 2 = 0$ → $A = 12$, $B = -5$, $C = -2$.

      Point: $(-2, 3)$ → $x_1 = -2$, $y_1 = 3$.

   2. Compute the numerator: $|Ax_1 + By_1 + C|$

   $12(-2) + (-5)(3) + (-2) = -24 - 15 - 2 = -41.$

   Absolute value: $| -41 | = 41$.

   3. Compute the denominator: $\sqrt{A^2 + B^2}$

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

   4. Write the distance

   $d = \frac{41}{13}.$

   The problem states this distance equals $\frac{41}{k}$. Therefore,

   $\frac{41}{k} = \frac{41}{13} \quad \Rightarrow \quad k = 13.$

   Tip

   A common mistake is forgetting the absolute value or misplacing a sign. Here, the numerator came out $-41$, but distance is always positive, so we take $41$. Also, note that $\sqrt{169}$ is exactly $13$, not $\sqrt{13}$ — that would be a trap if you miscomputed $12^2 + 5^2$ as $144 + 25 = 169$ correctly but then took the wrong root.

   Watch out

   Option (B) is $\sqrt{13}$ and option (A) is $\sqrt{134}$ — both are plausible if you accidentally add $144 + 25$ incorrectly or forget to take the square root. Always double-check the arithmetic.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

8. COMEDK 2024Set 2024-E1 markMCQ
   Q.The points on the $x$-axis whose perpendicular distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is 4 units are (A) $(8,0)$ and $(-2,0)$ (B) $(-8,0)$ and $(-2,0)$ (C) $(8,0)$ and $(2,0)$ (D) $(-8,0)$ and $(2,0)$

   ›Reveal solutionSolution

   Writing the line as $4x+3y-12=0$ and setting the distance of $(x,0)$ equal to $4$ gives $|4x-12|=20$, so $x=8$ or $x=-2$; the points are $(8,0)$ and $(-2,0)$ — option (A).

   Concept

   A point on the $x$-axis has the form $(x,0)$. The perpendicular distance from $(x_1,y_1)$ to the line $Ax+By+C=0$ is

   $d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.$

   Setting $d=4$ produces an absolute-value equation with two solutions — one on each side of the line.

   Solution

   1. Standard form: multiply $\dfrac{x}{3}+\dfrac{y}{4}=1$ by $12$:

   $4x+3y=12\ \Rightarrow\ 4x+3y-12=0,$

   so $A=4,\ B=3,\ C=-12$.

   2. Distance of $(x,0)$:

   $d=\frac{|4x+3(0)-12|}{\sqrt{4^2+3^2}}=\frac{|4x-12|}{5}.$

   3. Set $d=4$:

   $\frac{|4x-12|}{5}=4\ \Rightarrow\ |4x-12|=20.$

   4. Solve both cases:

   $4x-12=20\Rightarrow x=8,\qquad 4x-12=-20\Rightarrow x=-2.$

   So the points are $(8,0)$ and $(-2,0)$, which is option (A).

   Tip

   Quick check: for $(8,0)$, $\dfrac{|4\cdot8-12|}{5}=\dfrac{20}{5}=4$; for $(-2,0)$, $\dfrac{|{-8}-12|}{5}=\dfrac{20}{5}=4$. Both are exactly $4$ units away.

   ✓Final answer

   The correct option is (A): $(8,0)$ and $(-2,0)$.

9. COMEDK 2024Set 2024-M1 markMCQ
   Q.The perpendicular distance of a line from the origin is 5 units and its slope is $-1$. The equation of the line is (A) $x+y \pm 5 \sqrt{2}=0$ (B) $x-y \pm 2 \sqrt{5}=0$ (C) $x+y \pm 2 \sqrt{5}=0$ (D) $x-y \pm 5 \sqrt{2}=0$

   ›Reveal solutionSolution

   A slope of $-1$ gives lines $x + y = c$; requiring the distance $\dfrac{|c|}{\sqrt{2}} = 5$ gives $|c| = 5\sqrt{2}$, so the line is $x + y \pm 5\sqrt{2} = 0$ — option (A).

   Concept

   The perpendicular distance from the origin to $Ax + By + C = 0$ is $\dfrac{|C|}{\sqrt{A^2 + B^2}}$. A line of slope $-1$ has the form $x + y = c$ (since $-A/B = -1$ with $A = B = 1$), so we only need to fix the constant $c$ from the distance condition.

   Solution

   1. Standard form: $x + y - c = 0$, so $A = 1,\, B = 1,\, C = -c$.
   2. Distance from origin:

   $\frac{|1\cdot 0 + 1\cdot 0 - c|}{\sqrt{1^2 + 1^2}} = \frac{|c|}{\sqrt{2}}.$

   3. Apply the condition:

   $\frac{|c|}{\sqrt{2}} = 5 \Rightarrow |c| = 5\sqrt{2}.$

   4. Write both lines: $c = \pm 5\sqrt{2}$, giving $x + y \pm 5\sqrt{2} = 0$.

   Watch out

   A slope of $-1$ gives $x + y = c$, not $x - y = c$ (that line has slope $+1$); and keep both signs of $c$, since the line may lie on either side of the origin.

   ✓Final answer

   The correct option is (A), $x + y \pm 5\sqrt{2} = 0$.

   ANSWER: A

10. COMEDK 2023Set 2023-E1 markMCQ
    Q.The ratio in which the line $3 x+4 y+2=0$ divides the distance between the lines $3 x+4 y+5=0$ and $3 x+4 y-5=0$ is (A) 3 : 7 (B) 2 : 3 (C) 1 : 2 (D) 2 : 5

    ›Reveal solutionSolution

    Because the three lines are parallel, distances are proportional to the differences of their constant terms: the middle line divides the gap as $3:7$.

    All three lines share the normal $3x+4y=$ const, so perpendicular distances between them are $\dfrac{|c_1-c_2|}{\sqrt{3^2+4^2}}$, proportional to the differences of the constants.

    The outer lines have constants $+5$ and $-5$; the dividing line has constant $+2$.

    Distance from $c=5$ line to $c=2$ line $\propto |5-2|=3$.

    Distance from $c=2$ line to $c=-5$ line $\propto |2-(-5)|=7$.

    Ratio $=3:7$.

    ✓Final answer

    The correct option is (A) — 3 : 7

11. COMEDK 2023Set 2023-E1 markMCQ
    Q.The distance of the point $(2,3,4)$ from the line $1-x=\frac{y}{2}=\frac{1}{3}(1+z)$ is (A) $\frac{2}{7} \sqrt{35}$ (B) $\frac{1}{7} \sqrt{35}$ (C) $\frac{4}{7} \sqrt{35}$ (D) $\frac{3}{7} \sqrt{35}$

    ›Reveal solutionSolution

    Perpendicular distance^2 = 35 - 200/7 = (245 - 200)/7 = 45/7. distance = sqrt(45/7) = 3 sqrt5 / sqrt7 = 3 sqrt(35) / 7.

    Concept: perpendicular distance from a point to a line in 3-D, d = |AP|^2 - (projection of AP on d-hat)^2, i.e. d = sqrt( |AP|^2 - (AP . d)^2/|d|^2 ).

    Put the line in symmetric form. 1 - x = y/2 = (1 + z)/3 becomes

    (x - 1)/(-1) = (y - 0)/2 = (z + 1)/3.

    So a point on the line is A = (1, 0, -1) and the direction is d = (-1, 2, 3), with |d|^2 = 1 + 4 + 9 = 14.

    Given point P = (2, 3, 4). Then

    AP = P - A = (1, 3, 5), |AP|^2 = 1 + 9 + 25 = 35.

    Projection of AP along d:

    AP . d = (1)(-1) + (3)(2) + (5)(3) = -1 + 6 + 15 = 20.

    (AP . d)^2 / |d|^2 = 400/14 = 200/7.

    Perpendicular distance^2 = 35 - 200/7 = (245 - 200)/7 = 45/7.

    distance = sqrt(45/7) = 3 sqrt5 / sqrt7 = 3 sqrt(35) / 7.

    ✓Final answer

    The correct option is (D) — $\frac{3}{7} \sqrt{35}$

    ANSWER: D

12. COMEDK 2023Set 2023-M1 markMCQ
    Q.The distance of the point $(3,4)$ from the line $3 x+2 y+7=0$ measured along the line parallel to $y-2 x+7=0$ is equal to (A) $\frac{24 \sqrt{5}}{7}$ (B) $3 \sqrt{5}$ (C) $\frac{23 \sqrt{5}}{7}$ (D) $4 \sqrt{5}$

    ›Reveal solutionSolution

    Parametrise the point $(3,4)$ along the unit direction of the parallel line and find where it meets $3x+2y+7=0$; the parameter value equals the required distance, $\dfrac{24\sqrt5}{7}$.

    The line $y-2x+7=0$ has slope $2$, so a direction vector is $(1,2)$, of length $\sqrt5$. The unit direction is $\left(\tfrac1{\sqrt5},\tfrac2{\sqrt5}\right)$.

    A point at arc-length $t$ from $(3,4)$ along this direction is $\left(3+\tfrac{t}{\sqrt5},\,4+\tfrac{2t}{\sqrt5}\right)$. It lies on $3x+2y+7=0$ when:

    $3\left(3+\tfrac{t}{\sqrt5}\right)+2\left(4+\tfrac{2t}{\sqrt5}\right)+7=0$

    $9+8+7+\tfrac{3t+4t}{\sqrt5}=0\ \Rightarrow\ 24+\tfrac{7t}{\sqrt5}=0$

    $t=-\tfrac{24\sqrt5}{7},\qquad |t|=\tfrac{24\sqrt5}{7}.$

    Since the direction was a unit vector, $|t|$ is the actual distance measured along that line.

    ✓Final answer

    The correct option is (A) — $\frac{24\sqrt{5}}{7}$