Q.Find the distance between the following pairs of points:

Showing the 12 most recent of 46 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The position vectors of the points A, B and C with respect to the origin O are $\sqrt{3}\bar{i} + \bar{j}$, $\bar{i} + \sqrt{3}\bar{j}$ and $\beta\bar{i} + (1-\beta)\bar{j}$ respectively. If the distance of 'C' from the bisector of the acute angle between OA and OB is $\dfrac{3}{\sqrt{2}}$, then the sum of the possible values of $\beta$ is (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Uses the fact that the internal bisector of the angle between two vectors is along the sum of their unit vectors, then applies the point-to-line distance formula.

   Concept and Intuition

   When two vectors from the origin make an angle with each other, the direction that bisects that angle is obtained by adding their unit vectors — this is because the parallelogram built from two equal-length (unit) vectors always has its diagonal bisecting the angle between them.

   Step-by-Step Solution

   1. $\vec{OA}=\sqrt3\bar i+\bar j$, $|\vec{OA}|=\sqrt{3+1}=2$. Unit vector $=\left(\dfrac{\sqrt3}{2},\dfrac12\right)$, which is at angle $30^\circ$ from the x-axis.
   2. $\vec{OB}=\bar i+\sqrt3\bar j$, $|\vec{OB}|=2$. Unit vector $=\left(\dfrac12,\dfrac{\sqrt3}{2}\right)$, at angle $60^\circ$.
   3. The angle $AOB=60^\circ-30^\circ=30^\circ$, which is itself acute, so the required bisector bisects this $30^\circ$ angle, landing at $45^\circ$.
   4. Bisector direction $=$ sum of the two unit vectors $=\left(\dfrac{\sqrt3+1}{2},\dfrac{1+\sqrt3}{2}\right)\propto(1,1)$. So the bisector line through $O$ is $y=x$, i.e. $x-y=0$.
   5. $C=(\beta,\,1-\beta)$. Distance from $C$ to $x-y=0$ is $\dfrac{|\beta-(1-\beta)|}{\sqrt{1^2+(-1)^2}}=\dfrac{|2\beta-1|}{\sqrt2}$.
   6. Given this equals $\dfrac{3}{\sqrt2}$: $|2\beta-1|=3 \Rightarrow 2\beta-1=3$ or $2\beta-1=-3$, giving $\beta=2$ or $\beta=-1$.
   7. Sum of possible values of $\beta = 2+(-1)=1$.

   Common Mistakes

   • Forgetting to normalize $OA,OB$ before adding (the bisector rule only works for unit vectors).
   • Missing the negative root of $|2\beta-1|=3$ and reporting only one value of $\beta$.

   ✓Final answer

   The correct option is (A) — $1$.

   ANSWER: A

2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The perpendicular distance from the point $(1, 1, -2)$ on to the line $\dfrac{x - 1}{-1} = \dfrac{y}{2} = \dfrac{z}{1}$ is (A) $1$ (B) $2\sqrt{3}$ (C) $\sqrt{5}$ (D) $\dfrac{1}{\sqrt{2}}$

   ›Reveal solutionSolution

   Using the cross-product formula for point-to-line distance in 3D, the perpendicular distance from $(1,1,-2)$ to the given line is $\sqrt5$.

   Concept and Intuition

   For a line through point $A$ with direction vector $\vec d$, the perpendicular distance from an external point $P$ is $\dfrac{|\vec{AP}\times\vec d|}{|\vec d|}$ — the area of the parallelogram spanned by $\vec{AP}$ and $\vec d$, divided by the base length $|\vec d|$, gives the height, which is exactly this perpendicular distance.

   Step-by-Step Solution

   1. The line $\dfrac{x-1}{-1}=\dfrac{y}{2}=\dfrac{z}{1}$ passes through $A=(1,0,0)$ with direction $\vec d=(-1,2,1)$.
   2. $P=(1,1,-2)$, so $\vec{AP}=P-A=(0,1,-2)$.
   3. $\vec{AP}\times\vec d = \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&1&-2\\-1&2&1\end{vmatrix}$ $= \mathbf i(1\cdot1-(-2)\cdot2) - \mathbf j(0\cdot1-(-2)(-1)) + \mathbf k(0\cdot2-1\cdot(-1))$ $= \mathbf i(1+4) - \mathbf j(0-2) + \mathbf k(0+1) = (5,2,1)$.
   4. $|\vec{AP}\times\vec d| = \sqrt{25+4+1}=\sqrt{30}$.
   5. $|\vec d|=\sqrt{1+4+1}=\sqrt6$.
   6. Distance $=\dfrac{\sqrt{30}}{\sqrt6}=\sqrt5$.

   Common Mistakes

   • Taking $\vec{PA}$ vs $\vec{AP}$ inconsistently (doesn't affect magnitude, but sign errors elsewhere can creep in).
   • Arithmetic slip in the $3\times3$ determinant expansion.

   ✓Final answer

   The correct option is (C) — $\sqrt{5}$.

   ANSWER: C

3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If the distance of a variable point $P$ from a fixed line $2x - y + 1 = 0$ is twice the distance of $P$ from another fixed line $2x + y - 2 = 0$, then a point on the locus of $P$ is (A) $\left(\dfrac{1}{4}, \dfrac{3}{4}\right)$ (B) $(2, 3)$ (C) $(1, 1)$ (D) $\left(\dfrac{3}{2}, \dfrac{1}{4}\right)$

   ›Reveal solutionSolution

   Translate the distance ratio into an absolute-value equation and split into two linear loci; check which option satisfies one of them. Answer: $(1,1)$.

   Concept and Intuition

   The distance from a point to a line $ax+by+c=0$ is $\dfrac{|ax+by+c|}{\sqrt{a^2+b^2}}$. Both given lines have the same $\sqrt{a^2+b^2}=\sqrt{4+1}=\sqrt5$, so the ratio condition simplifies to a direct relation between the linear expressions themselves (no square roots survive), producing a pair of straight lines (the locus of $P$).

   Step-by-Step Solution

   1. Distance to $2x-y+1=0$ is $\dfrac{|2x-y+1|}{\sqrt5}$; distance to $2x+y-2=0$ is $\dfrac{|2x+y-2|}{\sqrt5}$.
   2. Given condition: $\dfrac{|2x-y+1|}{\sqrt5} = 2\cdot\dfrac{|2x+y-2|}{\sqrt5} \Rightarrow |2x-y+1| = 2|2x+y-2|$.
   3. Case (same sign): $2x-y+1 = 2(2x+y-2) = 4x+2y-4 \Rightarrow -2x-3y+5=0 \Rightarrow 2x+3y=5$.
   4. Case (opposite sign): $2x-y+1 = -2(2x+y-2) = -4x-2y+4 \Rightarrow 6x+y-3=0 \Rightarrow 6x+y=3$.
   5. Check each option against these two lines: $(1,1)$ gives $2(1)+3(1)=5$ ✓ (satisfies the first line exactly). None of the other listed points satisfy either line.

   Common Mistakes

   • Forgetting the locus is the union of TWO lines (both sign cases), and checking only one.
   • Dropping the absolute value and losing one of the two branches.

   ✓Final answer

   The correct option is (C) — $(1,1)$.

   ANSWER: C

4. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The product of the lengths of the perpendiculars drawn from the point $(1,2)$ to the pair of lines $2x^2 - 3xy - 2y^2 = 0$ is (A) $\dfrac{12}{5}$ (B) $2$ (C) $\dfrac{4}{5}$ (D) $6$

   ›Reveal solutionSolution

   Factor the homogeneous pair of lines, then multiply the two perpendicular distances from the given point. Answer: $\dfrac{12}{5}$.

   Concept and Intuition

   A homogeneous second-degree equation $ax^2+2hxy+by^2=0$ always represents a pair of straight lines through the origin. Once factored into two linear factors, the perpendicular distance from any point to each line can be found by the standard point-to-line distance formula, and multiplied.

   Step-by-Step Solution

   1. Factor $2x^2-3xy-2y^2$: try $(2x+y)(x-2y) = 2x^2-4xy+xy-2y^2 = 2x^2-3xy-2y^2$ — this matches.
   2. So the pair of lines is $2x+y=0$ and $x-2y=0$.
   3. Distance from $(1,2)$ to $2x+y=0$: $\dfrac{|2(1)+2|}{\sqrt{2^2+1^2}} = \dfrac{4}{\sqrt5}$.
   4. Distance from $(1,2)$ to $x-2y=0$: $\dfrac{|1-2(2)|}{\sqrt{1^2+2^2}} = \dfrac{3}{\sqrt5}$.
   5. Product $= \dfrac{4}{\sqrt5}\times\dfrac{3}{\sqrt5} = \dfrac{12}{5}$.

   (Cross-check via the standard formula: product of perpendiculars from $(x_1,y_1)$ to $ax^2+2hxy+by^2=0$ is $\dfrac{|ax_1^2+2hx_1y_1+by_1^2|}{\sqrt{(a-b)^2+4h^2}} = \dfrac{|2(1)+2(-\frac32)(2)+(-2)(4)|}{\sqrt{16+9}} = \dfrac{|2-6-8|}{5}=\dfrac{12}{5}$ — confirms the result.)

   Common Mistakes

   • Errors while factoring the quadratic (sign of the middle term especially).
   • Forgetting to divide by $\sqrt{a^2+b^2}$ (the norm of the line's coefficients) in the distance formula.

   ✓Final answer

   The correct option is (A) — $\dfrac{12}{5}$.

   ANSWER: A

5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If a non-horizontal line L is passing through the point $(4, -2)$ and the distance of L from the origin is 2 units, then the equation of the line L is (A) $4x + 3y - 10 = 0$ (B) $x + y - 2\sqrt{2} = 0$ (C) $3x + 4y - 4 = 0$ (D) $2x + 3y - 2 = 0$

   ›Reveal solutionSolution

   Set up the line through $(4,-2)$ with unknown slope, use the point-to-line distance formula with the origin, solve for slope (rejecting the horizontal solution), and get $4x+3y-10=0$ — option (A).

   Concept and Intuition

   A line through a fixed point with unknown slope $m$ can be written as $y-y_0=m(x-x_0)$. Imposing a distance condition from a fixed point (here, the origin) using the standard point-to-line distance formula

   $d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

   gives an equation in $m$ alone, which we solve. A "non-horizontal" line simply excludes the trivial solution $m=0$ (a line parallel to the x-axis), which often appears as an extraneous root in these setups since horizontal/vertical lines can independently satisfy distance conditions.

   Step-by-Step Solution

   1. Line through $(4,-2)$ with slope $m$: $y+2=m(x-4)$, i.e. $mx-y-4m-2=0$.
   2. Distance from origin $(0,0)$ to this line:

   $d = \frac{|m(0)-0-4m-2|}{\sqrt{m^2+1}} = \frac{|-4m-2|}{\sqrt{m^2+1}} = \frac{|4m+2|}{\sqrt{m^2+1}}.$

   3. Set this equal to 2: $\dfrac{|4m+2|}{\sqrt{m^2+1}}=2 \implies |4m+2| = 2\sqrt{m^2+1}$.
   4. Square both sides: $(4m+2)^2 = 4(m^2+1) \implies 16m^2+16m+4 = 4m^2+4$.
   5. Simplify: $12m^2+16m = 0 \implies 4m(3m+4)=0 \implies m=0$ or $m=-\dfrac43$.
   6. Since the line must be non-horizontal, reject $m=0$; take $m=-\dfrac43$.
   7. Substitute back: $y+2 = -\dfrac43(x-4) \implies 3y+6=-4(x-4)=-4x+16 \implies 4x+3y+6-16=0 \implies 4x+3y-10=0$.
   8. Verify: passes through $(4,-2)$: $4(4)+3(-2)-10=16-6-10=0$ ✓. Distance from origin: $\dfrac{|{-10}|}{\sqrt{16+9}}=\dfrac{10}{5}=2$ ✓.

   Common Mistakes

   • Keeping the $m=0$ solution (the horizontal line $y=-2$, which also has distance 2 from the origin) despite the problem explicitly excluding horizontal lines.
   • Sign errors when clearing the absolute value before squaring — squaring both sides sidesteps this but only if done carefully on the full expression.

   ✓Final answer

   The correct option is (A) — $4x+3y-10=0$.

   ANSWER: A

6. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The shortest distance between the lines $\bar r = \bar a + t\bar b$ and $\bar r = \bar c + s\bar d$ when $\bar a = \bar i - 2\bar j + 2\bar k$, $\bar b = 3\bar i - 2\bar j - 2\bar k$, $\bar c = 6\bar i + 2\bar j + 2\bar k$ and $\bar d = -4\bar i - \bar k$ is (A) 9 (B) $\dfrac{6\sqrt3}{\sqrt7}$ (C) $\dfrac{\sqrt7}{2\sqrt3}$ (D) $\dfrac{5}{\sqrt3}$

   ›Reveal solutionSolution

   The shortest distance between two skew lines is the projection of the vector joining a point on each line onto the common perpendicular direction $\bar b\times\bar d$; it evaluates to $\dfrac{6\sqrt3}{\sqrt7}$.

   Concept and Intuition

   Two skew lines each carry a direction vector; their cross product $\bar b\times\bar d$ points along the unique direction perpendicular to both — the direction of the shortest connecting segment. Projecting the vector between any two points on the lines onto this direction gives the shortest distance, since all other components of that connecting vector lie within the plane spanned by $\bar b,\bar d$ and contribute nothing to the perpendicular separation.

   Step-by-Step Solution

   1. $\bar a=(1,-2,2),\ \bar b=(3,-2,-2),\ \bar c=(6,2,2),\ \bar d=(-4,0,-1)$.
   2. $\bar c-\bar a=(6-1,\,2-(-2),\,2-2)=(5,4,0)$.
   3. $\bar b\times\bar d = \left((-2)(-1)-(-2)(0),\ (-2)(-4)-(3)(-1),\ (3)(0)-(-2)(-4)\right)=(2,\ 8+3,\ 0-8)=(2,11,-8)$.
   4. $|\bar b\times\bar d|=\sqrt{2^2+11^2+8^2}=\sqrt{4+121+64}=\sqrt{189}=3\sqrt{21}$.
   5. $(\bar c-\bar a)\cdot(\bar b\times\bar d)=5(2)+4(11)+0(-8)=10+44=54$.
   6. Shortest distance $=\dfrac{|54|}{3\sqrt{21}}=\dfrac{18}{\sqrt{21}}=\dfrac{18\sqrt{21}}{21}=\dfrac{6\sqrt{21}}{7}$.
   7. Rewrite: $\dfrac{6\sqrt{21}}{7}=6\sqrt{\dfrac{21}{49}}=6\sqrt{\dfrac{3}{7}}=\dfrac{6\sqrt3}{\sqrt7}$, matching the given option form.

   Common Mistakes

   • Computing $\bar b\times\bar d$ with a sign slip on the middle (j) component.
   • Forgetting to take the absolute value / not simplifying the surd fully to compare with the answer choices.

   ✓Final answer

   The correct option is (B) $\dfrac{6\sqrt3}{\sqrt7}$.

   ANSWER: B

7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The lines $L_1 \equiv 2x + y + 1 = 0$ and $L_2 \equiv x - 2y + 4 = 0$ intersect at A. Let P be a point at a distance of 5 units from $L_1 = 0$ and $\alpha$ units from $L_2 = 0$. If M, N are the feet of the perpendiculars from P on to the lines $L_1 = 0$ and $L_2 = 0$ respectively and the area of the quadrilateral AMPN is 25 sq. units, then point P lies on the line (A) $x + 3y - 3 = 0$ (B) $3x - y + 12 = 0$ (C) $x + 3y = 0$ (D) $3x - y = 0$

   ›Reveal solutionSolution

   The equal-area condition forces P to be equidistant from the two perpendicular lines, so P lies on their angle bisector: $x+3y-3=0$.

   Concept and Intuition

   Since $L_1\perp L_2$, the quadrilateral $AMPN$ (A the intersection point, M and N the feet of perpendiculars from P onto $L_1$ and $L_2$) has three right angles (at A, M, N) and is therefore a rectangle. In a rectangle, area = product of adjacent sides, and opposite sides are equal — this directly links the given area to the two perpendicular distances from P.

   Step-by-Step Solution

   1. Check perpendicularity: $L_1: 2x+y+1=0$ (normal $(2,1)$), $L_2: x-2y+4=0$ (normal $(1,-2)$). Dot product $2(1)+1(-2)=0$, so $L_1\perp L_2$.
   2. AMPN is a rectangle: $PM$ (distance from P to $L_1$) $=5$, and $PN$ (distance from P to $L_2$) $=\alpha$; opposite sides give $AM=PN=\alpha$ and $AN=PM=5$.
   3. Area of rectangle $= AM\cdot AN = \alpha\times 5 = 25 \Rightarrow \alpha=5$.
   4. So P is at distance 5 from both lines — P lies on an angle bisector of $L_1,L_2$: $\dfrac{2x+y+1}{\sqrt{5}} = \pm\dfrac{x-2y+4}{\sqrt{5}}$ (normal magnitudes both $\sqrt5$).
   5. Taking "+": $2x+y+1 = x-2y+4 \Rightarrow x+3y-3=0$ — this matches option (A) exactly.

   Common Mistakes

   • Missing that $L_1\perp L_2$ (needed to identify AMPN as a rectangle rather than a general quadrilateral).
   • Not recognizing that the derived $\alpha=5$ means P is equidistant from both lines, hence lies on a bisector.

   ✓Final answer

   The correct option is (A) — $x + 3y - 3 = 0$.

   ANSWER: A

8. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $P(\alpha, \alpha + 1)$ is the foot of the perpendicular drawn from the origin to the line L and the $x$-intercept of L is $\left(-\frac{5}{2}\right)$, then the sum of the squares of the distances from the origin to all such possible points P, is (A) $\dfrac{45}{8}$ (B) $\dfrac{15}{2}$ (C) $\dfrac{12}{5}$ (D) $\dfrac{10}{7}$

   ›Reveal solutionSolution

   Two values of $\alpha$ satisfy the intercept condition; summing $OP^2$ for both gives $\frac{45}{8}$.

   Concept and Intuition

   If P is the foot of the perpendicular from the origin O to a line L, then OP is perpendicular to L and OP's length equals the distance from O to L. The line L can be written directly using P: since L is perpendicular to OP at P, its equation is $x\cdot\alpha + y\cdot(\alpha+1) = \alpha^2+(\alpha+1)^2$ (a line through P normal to the direction $(\alpha,\alpha+1)$, with the constant chosen so P lies on it).

   Step-by-Step Solution

   1. Since $P(\alpha,\alpha+1)$ is the foot of perpendicular from O to L, the equation of L is $\alpha x + (\alpha+1)y = \alpha^2+(\alpha+1)^2$ (perpendicular to $OP$, passing through $P$).
   2. x-intercept (set $y=0$): $x = \dfrac{\alpha^2+(\alpha+1)^2}{\alpha} = -\dfrac{5}{2}$.
   3. So $\alpha^2+(\alpha+1)^2 = -\dfrac{5\alpha}{2}$. Expand: $2\alpha^2+2\alpha+1 = -\dfrac{5\alpha}{2}$, multiply by 2: $4\alpha^2+4\alpha+2=-5\alpha \Rightarrow 4\alpha^2+9\alpha+2=0$.
   4. Solve: $\alpha = \dfrac{-9\pm\sqrt{81-32}}{8}=\dfrac{-9\pm7}{8}$, giving $\alpha=-\dfrac14$ or $\alpha=-2$.
   5. $OP^2 = \alpha^2+(\alpha+1)^2$ for each: at $\alpha=-\frac14$: $\frac{1}{16}+\frac{9}{16}=\frac{10}{16}=\frac58$. At $\alpha=-2$: $4+1=5$.
   6. Sum $= \dfrac58+5 = \dfrac{45}{8}$.

   Common Mistakes

   • Forgetting there are two valid values of $\alpha$ (the question explicitly asks for the sum over "all such possible points P").
   • Sign errors when expanding $(\alpha+1)^2$ or clearing the fraction in the intercept equation.

   ✓Final answer

   The correct option is (A) — $\dfrac{45}{8}$.

   ANSWER: A

9. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If M is the foot of the perpendicular drawn from $P(1,2,-1)$ to the plane passing through the point $A(3,-2,1)$ and perpendicular to the vector $4\bar{i} + 7\bar{j} - 4\bar{k}$, then the length of PM is (A) $\dfrac{32}{9}$ (B) $\dfrac{28}{9}$ (C) $\dfrac{26}{5}$ (D) $\dfrac{22}{5}$

   ›Reveal solutionSolution

   $PM$ is simply the perpendicular distance from $P$ to the given plane, computed via the point-to-plane distance formula: $\dfrac{28}{9}$.

   Concept and Intuition

   The foot of the perpendicular from a point to a plane is the closest point on the plane to that point, and the distance from the point to that foot equals the perpendicular distance from the point to the plane — no need to actually find $M$'s coordinates.

   Step-by-Step Solution

   1. The plane passes through $A(3,-2,1)$ and has normal vector $\bar n=(4,7,-4)$. Its equation: $4(x-3)+7(y+2)-4(z-1)=0$.
   2. Expand: $4x-12+7y+14-4z+4=0 \Rightarrow 4x+7y-4z+6=0$.
   3. Distance from $P(1,2,-1)$ to this plane: $d=\dfrac{|4(1)+7(2)-4(-1)+6|}{\sqrt{4^2+7^2+(-4)^2}}=\dfrac{|4+14+4+6|}{\sqrt{16+49+16}}=\dfrac{28}{\sqrt{81}}=\dfrac{28}{9}$.
   4. Since $M$ is the foot of the perpendicular from $P$, $PM$ equals exactly this distance.

   Common Mistakes

   • Actually computing $M$'s coordinates by parametrizing the perpendicular line — correct but unnecessary extra work versus the direct distance formula.
   • Sign/arithmetic slip in expanding the plane equation.

   ✓Final answer

   The correct option is (B) — $\dfrac{28}{9}$.

   ANSWER: B

10. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.The perpendicular distance from origin to the tangent drawn at the point $P\left(\dfrac{\pi}{4}\right)$ to the circle $x^2 + y^2 - 4x - 4y + 6 = 0$ is (A) $4$ (B) $3\sqrt{2}$ (C) $6$ (D) $5\sqrt{2}$

    ›Reveal solutionSolution

    This tests converting a parametric angle on a circle to Cartesian coordinates, writing the tangent at that point using the standard "replace $x^2\to xx_1$" tangent rule, and computing a perpendicular distance from the origin. The distance comes out to $3\sqrt2$.

    Concept and Intuition

    Any circle $(x-h)^2+(y-k)^2=r^2$ can be parametrised as $(h+r\cos\theta,\,k+r\sin\theta)$, with $\theta$ measured from the centre. Given "the point $P(\pi/4)$" means we plug $\theta=\pi/4$ into this parametrisation to get actual coordinates. Then the tangent to a circle $x^2+y^2+2gx+2fy+c=0$ at a point $(x_1,y_1)$ on the circle is obtained by the standard substitution rule ($x^2\to xx_1$, $y^2\to yy_1$, $x\to\frac{x+x_1}2$, $y\to\frac{y+y_1}2$), giving $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.

    Step-by-Step Solution

    1. Circle: $x^2+y^2-4x-4y+6=0$. Complete the square: $(x-2)^2+(y-2)^2=4+4-6=2$. Centre $(2,2)$, radius $r=\sqrt2$.
    2. Parametric form: $P(\theta)=(2+\sqrt2\cos\theta,\,2+\sqrt2\sin\theta)$.
    3. At $\theta=\pi/4$: $\cos\theta=\sin\theta=\dfrac{\sqrt2}{2}$, so

    $P=\left(2+\sqrt2\cdot\frac{\sqrt2}2,\;2+\sqrt2\cdot\frac{\sqrt2}2\right)=(2+1,\,2+1)=(3,3).$

    4. Tangent to $x^2+y^2-4x-4y+6=0$ at $(x_1,y_1)=(3,3)$: using $g=-2,f=-2,c=6$,

    $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$

    $3x+3y-2(x+3)-2(y+3)+6=0$

    $3x+3y-2x-6-2y-6+6=0 \;\Rightarrow\; x+y-6=0.$

    5. Perpendicular distance from origin $(0,0)$ to $x+y-6=0$:

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

    Common Mistakes

    • Misreading $P(\pi/4)$ as a point with $x$- or $y$-coordinate equal to $\pi/4$ instead of a parametric-angle point on the circle.
    • Forgetting to complete the square correctly to identify the centre and radius (sign errors on $g,f$).
    • Errors applying the tangent substitution rule, e.g. forgetting the $\frac{x+x_1}2$ becomes just $(x+x_1)$ after clearing the factor of 2 that's already built into $2g$.

    ✓Final answer

    The correct option is (B) — $3\sqrt{2}$.

    ANSWER: B

11. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If P is a variable point which is at a distance of 2 units from the line $2x - 3y + 1 = 0$ and $\sqrt{13}$ units from the point $(5, 6)$, then the equation of the locus of P is (A) $4x^2 + 12xy - 5y^2 - 44x - 42y + 245 = 0$ (B) $12xy - 5y^2 - 44x - 42y + 243 = 0$ (C) $8x^2 + 12xy - 5y^2 - 44x - 42y + 243 = 0$ (D) $12xy - 13y^2 - 44x - 42y + 245 = 0$

    ›Reveal solutionSolution

    Combine the squared line-distance condition and the squared point-distance condition, eliminating their common quadratic part, to get the locus of P.

    Concept and Intuition

    The perpendicular distance from a point to a line, when squared, clears the square root: $\left(\dfrac{2x-3y+1}{\sqrt{13}}\right)^2=4\Rightarrow(2x-3y+1)^2=52$. This is a genuine second-degree relation in $x,y$ (not literally two parallel lines once left un-square-rooted). Combined with the circle condition from the fixed-distance-from-a-point requirement, subtracting a suitable multiple cancels the matching $x^2$ terms and produces the single combined-locus equation that the options are shaped like.

    Step-by-Step Solution

    1. Line condition: $\dfrac{|2x-3y+1|}{\sqrt{2^2+(-3)^2}}=2\Rightarrow(2x-3y+1)^2=4\times13=52$. Expanding: $4x^2+9y^2+1-12xy+4x-6y=52\Rightarrow 4x^2-12xy+9y^2+4x-6y-51=0$ … (i)
    2. Point condition: $(x-5)^2+(y-6)^2=13\Rightarrow x^2-10x+25+y^2-12y+36=13\Rightarrow x^2+y^2-10x-12y+48=0$ … (ii)
    3. Multiply (ii) by 4: $4x^2+4y^2-40x-48y+192=0$.
    4. Subtract this from (i): $-12xy+5y^2+44x+42y-243=0$.
    5. Multiply by $-1$: $12xy-5y^2-44x-42y+243=0$.

    Common Mistakes

    • Forgetting the $\sqrt{13}$ normalizing factor when squaring the perpendicular distance (using $52$ instead of, say, $4$).
    • Sign slip while subtracting the two expanded equations.

    ✓Final answer

    The correct option is (B) — $12xy-5y^2-44x-42y+243=0$.

    ANSWER: B

12. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.For a positive real number p, if the perpendicular distance from a point $-\bar{i}+p\bar{j}-3\bar{k}$ to the plane $\bar{r}\cdot(2\bar{i}-3\bar{j}+6\bar{k})=7$ is 6 units, then $p =$ (A) $\dfrac{4}{5}$ (B) $\dfrac{5}{6}$ (C) $6$ (D) $5$

    ›Reveal solutionSolution

    Applying the point-to-plane distance formula and solving for the positive root gives $p=5$.

    Concept and Intuition

    The perpendicular distance from a point $(x_0,y_0,z_0)$ to the plane $ax+by+cz=d$ is $\dfrac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}$ — a direct plug-in once the plane is written in Cartesian form.

    Step-by-Step Solution

    1. The plane $\bar r\cdot(2\bar i-3\bar j+6\bar k)=7$ is $2x-3y+6z=7$, with normal magnitude $\sqrt{4+9+36}=\sqrt{49}=7$.
    2. Point: $(-1,\ p,\ -3)$.
    3. Distance $=\dfrac{|2(-1)-3(p)+6(-3)-7|}{7} = \dfrac{|-2-3p-18-7|}{7}=\dfrac{|-27-3p|}{7}$.
    4. Set equal to 6: $|-27-3p|=42$. Since $p>0$, $-27-3p<0$, so $27+3p=42 \Rightarrow 3p=15 \Rightarrow p=5$.
    5. (The other algebraic root, $p=-23$, is rejected since $p$ must be a positive real number.)

    Common Mistakes

    • Dropping the absolute value and getting a sign wrong, which can produce a spurious negative or fractional $p$.
    • Forgetting to discard the negative root when the problem specifies $p$ is a positive real number.

    ✓Final answer

    The correct option is (D) — $5$.

    ANSWER: D