Q.Find the position vector of the mid point of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

Showing the 12 most recent of 59 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The position vectors of two points A and B are $\vec{i} + 2\vec{j} + 3\vec{k}$ and $7\vec{i} - \vec{k}$ respectively. The point P with position vector $-2\vec{i} + 3\vec{j} + 5\vec{k}$ is on the line AB. If the point Q is the harmonic conjugate of P, then the sum of the scalar components of the position vector of Q is (A) 6 (B) 4 (C) 2 (D) 0

   ›Reveal solutionSolution

   $P$ divides $AB$ externally in ratio $1:3$; its harmonic conjugate $Q=\dfrac{3A+B}{4}=(2.5,\,1.5,\,2)$, whose components sum to $6$.

   Solution

   Let $A=(1,2,3)$, $B=(7,0,-1)$, $P=(-2,3,5)$.

   Suppose $P$ divides $AB$ in ratio $\lambda:1$, so $P=\dfrac{A+\lambda B}{1+\lambda}$. From the $x$-coordinate:

   $\frac{1+7\lambda}{1+\lambda}=-2 \implies 1+7\lambda=-2-2\lambda \implies \lambda=-\frac{1}{3}.$

   (The $y$- and $z$-coordinates confirm this.)

   The harmonic conjugate $Q$ divides $AB$ in ratio $-\lambda:1=\dfrac{1}{3}:1$: …

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2. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2) are three points. If the point P divides AB in the ratio 1 : 2 and the point Q divides BC in the ratio -2 : 3, then the distance between P and Q is (A) $\sqrt{312}$ (B) $13$ (C) $\dfrac{2}{3}\sqrt{78}$ (D) $25$

   ›Reveal solutionSolution

   Use section formula for internal and external division to find coordinates of P and Q, then compute the Euclidean distance. The distance is $\frac{2}{3}\sqrt{78}$, so the correct option is (C).

   ---

   Concept & Intuition

   We are given three points in 3D space. Point P divides AB internally in the ratio 1:2 — that’s a straightforward internal division. Point Q divides BC in the ratio -2:3. A negative ratio indicates an external division: the point lies on the line BC extended beyond one of the endpoints. Once we have coordinates for P and Q, the distance between them is just the 3D Euclidean distance formula. The trick is handling the negative ratio correctly.

   ---

   Step-by-step solution

   1. Find coordinates of P (internal division of AB in ratio 1:2) For internal division, if a point divides the segment joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in the ratio $m:n$, the coordinates are

   $\left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right).$

   Here $A(1,2,3)$, $B(2,3,1)$, ratio $1:2$ (so $m=1$, $n=2$).

   $P = \left( \frac{1\cdot 2 + 2\cdot 1}{1+2}, \frac{1\cdot 3 + 2\cdot 2}{3}, \frac{1\cdot 1 + 2\cdot 3}{3} \right) = \left( \frac{2+2}{3}, \frac{3+4}{3}, \frac{1+6}{3} \right) = \left( \frac{4}{3}, \frac{7}{3}, \frac{7}{3} \right).$

   2. Find coordinates of Q (external division of BC in ratio -2:3) A negative ratio means external division. The standard formula still works if we treat the ratio as $m:n$ with one of them negative. Here the ratio is $-2:3$, so take $m = -2$, $n = 3$. For points $B(2,3,1)$ and $C(3,1,2)$:

   $Q = \left( \frac{m x_C + n x_B}{m+n}, \frac{m y_C + n y_B}{m+n}, \frac{m z_C + n z_B}{m+n} \right).$

   Note $m+n = -2+3 = 1$, which simplifies things.

   $Q_x = \frac{(-2)(3) + 3(2)}{1} = -6 + 6 = 0,$

   $Q_y = \frac{(-2)(1) + 3(3)}{1} = -2 + 9 = 7,$

   $Q_z = \frac{(-2)(2) + 3(1)}{1} = -4 + 3 = -1.$

   So $Q = (0, 7, -1)$.

   Tip

   When $m+n=1$, the formula becomes just a weighted sum — very quick to compute.

   3. Compute the distance between P and Q Use the 3D distance formula:

   $PQ = \sqrt{(x_P - x_Q)^2 + (y_P - y_Q)^2 + (z_P - z_Q)^2}.$

   Here $P\left(\frac{4}{3}, \frac{7}{3}, \frac{7}{3}\right)$ and $Q(0,7,-1)$.

   Differences:

   $$$$

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3. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $2\vec{i} - \vec{j} + \vec{k},\ \vec{i} - 3\vec{j} - 5\vec{k}$ are the position vectors of the points A and B respectively, C divides AB in the ratio $2:3$ and M is the mid-point of AB, then $5$ (position vector of C) $- 2$ (position vector of M) $=$ (A) $5\vec{i} - 5\vec{j} - 3\vec{k}$ (B) $11\vec{i} - 13\vec{j} - 11\vec{k}$ (C) $5\vec{i} + 5\vec{j} - 3\vec{k}$ (D) $11\vec{i} + 13\vec{j} - 11\vec{k}$

   ›Reveal solutionSolution

   We use the section formula to find the position vector of C and the midpoint formula for M, then perform the required vector subtraction. The result is $\boxed{5\vec{i} - 5\vec{j} - 3\vec{k}}$.

   The core concept here is the section formula for position vectors, which allows us to find the position vector of a point that divides a line segment in a given ratio. The midpoint formula is a special case of the section formula. Once we have the position vectors of C and M, we can perform standard vector scalar multiplication and subtraction.

   Let $\vec{a}$ and $\vec{b}$ be the position vectors of points A and B respectively.

   Given:

   $\vec{a} = 2\vec{i} - \vec{j} + \vec{k}$

   $\vec{b} = \vec{i} - 3\vec{j} - 5\vec{k}$

   1. Find the position vector of C ($\vec{c}$):

      Point C divides the line segment AB internally in the ratio $2:3$.

      The section formula for internal division states that if a point C divides the line segment joining points A (with position vector $\vec{a}$) and B (with position vector $\vec{b}$) in the ratio $m:n$, then the position vector of C is given by:

      $\vec{c} = \frac{n\vec{a} + m\vec{b}}{m+n}$

      Here, $m=2$ and $n=3$.

      $\vec{c} = \frac{3\vec{a} + 2\vec{b}}{2+3} = \frac{3\vec{a} + 2\vec{b}}{5}$

      Substitute the given position vectors $\vec{a}$ and $\vec{b}$:

      $\vec{c} = \frac{3(2\vec{i} - \vec{j} + \vec{k}) + 2(\vec{i} - 3\vec{j} - 5\vec{k})}{5}$

      $\vec{c} = \frac{(6\vec{i} - 3\vec{j} + 3\vec{k}) + (2\vec{i} - 6\vec{j} - 10\vec{k})}{5}$

      Combine the components:

      $\vec{c} = \frac{(6+2)\vec{i} + (-3-6)\vec{j} + (3-10)\vec{k}}{5}$

      $\vec{c} = \frac{8\vec{i} - 9\vec{j} - 7\vec{k}}{5}$

      $\vec{c} = \frac{8}{5}\vec{i} - \frac{9}{5}\vec{j} - \frac{7}{5}\vec{k}$

   2. Find the position vector of M ($\vec{m}$):

      Point M is the mid-point of the line segment AB. The midpoint formula is a special case of the section formula where the ratio is $1:1$.

      $\vec{m} = \frac{\vec{a} + \vec{b}}{2}$

      Substitute the given position vectors $\vec{a}$ and $\vec{b}$:

      $\vec{m} = \frac{(2\vec{i} - \vec{j} + \vec{k}) + (\vec{i} - 3\vec{j} - 5\vec{k})}{2}$

      Combine the components:

      $\vec{m} = \frac{(2+1)\vec{i} + (-1-3)\vec{j} + (1-5)\vec{k}}{2}$ …

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4. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.P and Q are the points of trisection of the line segment AB. If $2\vec{i} - 5\vec{j} + 3\vec{k}$ and $4\vec{i} + \vec{j} - 6\vec{k}$ are the position vectors of A and B respectively, then the position vector of the point which divides PQ in the ratio 2:3 is (A) $\dfrac{1}{15}(44\vec{i} - 33\vec{j} - 18\vec{k})$ (B) $\dfrac{1}{5}(36\vec{i} - 26\vec{j} - 18\vec{k})$ (C) $\dfrac{1}{5}(3\vec{i} + 7\vec{j} - 9\vec{k})$ (D) $\dfrac{1}{15}(-3\vec{i} - 7\vec{j} + 9\vec{k})$

   ›Reveal solutionSolution

   The key idea is to first find the trisection points P and Q of AB using the section formula, then find the point that divides PQ in the given ratio. The final position vector is $\frac{1}{15}(44\vec{i} - 33\vec{j} - 18\vec{k})$, which corresponds to option (A).

   We are given the position vectors of A and B:

   $\vec{A} = 2\vec{i} - 5\vec{j} + 3\vec{k}, \quad \vec{B} = 4\vec{i} + \vec{j} - 6\vec{k}.$

   P and Q are the points of trisection of AB. That means P and Q divide AB into three equal segments. There are two possible orders: either P is closer to A and Q closer to B, or vice versa. The problem does not specify which is which, but the final answer will be the same regardless because the ratio 2:3 on PQ will be symmetric in a certain way. We will assume P is the point that divides AB in the ratio 1:2 (i.e., AP : PB = 1 : 2) and Q divides AB in the ratio 2:1 (i.e., AQ : QB = 2 : 1). This is the standard convention.

   Concept and intuition: The section formula tells us that if a point divides a line segment joining two points with position vectors $\vec{a}$ and $\vec{b}$ in the ratio $m:n$ (from $\vec{a}$ to $\vec{b}$), then its position vector is $\frac{n\vec{a} + m\vec{b}}{m+n}$. We apply this twice: first to find P and Q, then again to find the point that divides PQ in the ratio 2:3.

   1. Find P (trisection point closer to A) P divides AB in the ratio AP : PB = 1 : 2. Using the section formula:

   $\vec{P} = \frac{2 \cdot \vec{A} + 1 \cdot \vec{B}}{1+2} = \frac{2(2\vec{i} - 5\vec{j} + 3\vec{k}) + 1(4\vec{i} + \vec{j} - 6\vec{k})}{3}.$

   Compute numerator:

   $(4\vec{i} - 10\vec{j} + 6\vec{k}) + (4\vec{i} + \vec{j} - 6\vec{k}) = 8\vec{i} - 9\vec{j} + 0\vec{k}.$

   So:

   $\vec{P} = \frac{8\vec{i} - 9\vec{j}}{3}.$

   2. Find Q (trisection point closer to B) Q divides AB in the ratio AQ : QB = 2 : 1. Using the section formula:

   $\vec{Q} = \frac{1 \cdot \vec{A} + 2 \cdot \vec{B}}{2+1} = \frac{1(2\vec{i} - 5\vec{j} + 3\vec{k}) + 2(4\vec{i} + \vec{j} - 6\vec{k})}{3}.$

   Compute numerator:

   $(2\vec{i} - 5\vec{j} + 3\vec{k}) + (8\vec{i} + 2\vec{j} - 12\vec{k}) = 10\vec{i} - 3\vec{j} - 9\vec{k}.$

   So:

   $\vec{Q} = \frac{10\vec{i} - 3\vec{j} - 9\vec{k}}{3}.$

   3. Find the point R that divides PQ in the ratio 2:3 The ratio is given as 2:3, but we must decide the direction. Usually, "divides PQ in the ratio 2:3" means the point is closer to P if the ratio is measured from P to Q. So let R divide PQ such that PR : RQ = 2 : 3. Using the section formula again:

   $\vec{R} = \frac{3 \cdot \vec{P} + 2 \cdot \vec{Q}}{2+3} = \frac{3\vec{P} + 2\vec{Q}}{5}.$

   Substitute $\vec{P}$ and $\vec{Q}$: …

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5. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.Let $A = (1, 2, 0)$, $B = (2, 0, -1)$, $C = (0, -2, 3)$ and $D = (-1, 2, -3)$ be four points in the space. Let $G_1$ be the centroid of triangle $ABC$ and $G_2$ be the centroid of tetrahedron $ABCD$. If $P$ divides $G_1 G_2$ in the ratio $4 : 3$ internally then $P =$ (A) $\begin{pmatrix} \dfrac{5}{7} \\ \dfrac{2}{7} \\ \dfrac{1}{7} \end{pmatrix}$ (B) $\begin{pmatrix} \dfrac{1}{7} \\ \dfrac{2}{7} \\ \dfrac{3}{7} \end{pmatrix}$ (C) $\begin{pmatrix} \dfrac{4}{7} \\ \dfrac{-2}{7} \\ \dfrac{1}{7} \end{pmatrix}$ (D) $\begin{pmatrix} \dfrac{1}{7} \\ \dfrac{-3}{7} \\ \dfrac{5}{7} \end{pmatrix}$

   ›Reveal solutionSolution

   $G_1=(1,0,\tfrac23)$, $G_2=(\tfrac12,\tfrac12,-\tfrac14)$; dividing $G_1G_2$ in $4:3$ gives $P=\left(\tfrac57,\tfrac27,\tfrac17\right)$.

   Centroid of $\triangle ABC$:

   $G_1=\frac{A+B+C}{3}=\frac{(1+2+0,\;2+0-2,\;0-1+3)}{3}=\frac{(3,0,2)}{3}=\left(1,\,0,\,\tfrac23\right).$

   Centroid of tetrahedron $ABCD$:

   $G_2=\frac{A+B+C+D}{4}=\frac{(2,\;2,\;-1)}{4}=\left(\tfrac12,\,\tfrac12,\,-\tfrac14\right).$

   Section formula — $P$ divides $G_1G_2$ in the ratio $4:3$ (from $G_1$ to $G_2$): …

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6. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.Let ABC be a triangle and $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of A, B, C respectively. If D divides BC in the ratio 2 : 3 internally and E divides CA in the ratio 2 : 1 internally then the position vector of the point P which divides DE in the ratio 3 : 5 internally is (A) $\frac{1}{8}(2\vec{a} + 3\vec{b} + 3\vec{c})$ (B) $\frac{1}{8}(3\vec{a} + 2\vec{b} + 3\vec{c})$ (C) $\frac{1}{8}(3\vec{a} + 3\vec{b} + 2\vec{c})$ (D) $\frac{3}{8}(\vec{a} + \vec{b} + \vec{c})$

   ›Reveal solutionSolution

   Use the section formula twice: first to find D and E, then to find P on DE. The final position vector is $\frac{1}{8}(3\vec{a} + 2\vec{b} + 3\vec{c})$, which matches option (B).

   The core idea here is the section formula for vectors. If a point divides a line segment internally in a given ratio, its position vector is a weighted average of the endpoints' position vectors, with weights proportional to the opposite parts of the ratio. This problem asks you to apply that formula twice in succession — first to locate D and E on the sides of the triangle, then to locate P on the segment joining D and E.

   A common mistake is to mix up which weight goes with which endpoint. Remember: if a point divides $XY$ in the ratio $m:n$ (from $X$ to $Y$), the position vector is $\frac{n\vec{x} + m\vec{y}}{m+n}$ — the weight of $X$ is the opposite part of the ratio ($n$), and the weight of $Y$ is the same part ($m$). This is because the point is closer to $X$ when $m < n$, so $\vec{x}$ should have the larger coefficient.

   Let's work through it step by step.

   1. Find D, which divides BC in the ratio 2:3 internally. Here B is the first endpoint and C is the second. The ratio is $2:3$ from B to C. So $m=2$, $n=3$. Using the section formula:

   $\vec{d} = \frac{3\vec{b} + 2\vec{c}}{2+3} = \frac{3\vec{b} + 2\vec{c}}{5}$

   Notice that B gets the weight 3 (the opposite part) and C gets the weight 2 (the same part).

   2. Find E, which divides CA in the ratio 2:1 internally. Here C is the first endpoint and A is the second. The ratio is $2:1$ from C to A. So $m=2$, $n=1$. Then:

   $\vec{e} = \frac{1\vec{c} + 2\vec{a}}{2+1} = \frac{\vec{c} + 2\vec{a}}{3}$

   Again, C gets the weight 1 (opposite part) and A gets the weight 2 (same part).

   3. Find P, which divides DE in the ratio 3:5 internally. Here D is the first endpoint and E is the second. The ratio is $3:5$ from D to E. So $m=3$, $n=5$. Then:

   $\vec{p} = \frac{5\vec{d} + 3\vec{e}}{3+5} = \frac{5\vec{d} + 3\vec{e}}{8}$

   4. Substitute $\vec{d}$ and $\vec{e}$ into the expression for $\vec{p}$.

   $\vec{p} = \frac{1}{8} \left[ 5\left(\frac{3\vec{b} + 2\vec{c}}{5}\right) + 3\left(\frac{\vec{c} + 2\vec{a}}{3}\right) \right]$

   The 5 cancels in the first term, and the 3 cancels in the second term:

   $\vec{p} = \frac{1}{8} \left[ (3\vec{b} + 2\vec{c}) + (\vec{c} + 2\vec{a}) \right]$

   5. Collect like terms.

   $\vec{p} = \frac{1}{8} \left( 2\vec{a} + 3\vec{b} + 3\vec{c} \right)$

   Watch out

   This result $\frac{1}{8}(2\vec{a} + 3\vec{b} + 3\vec{c})$ is option (A), but it is not the correct answer to the problem as stated. Check the ratio for E again: the problem says "E divides CA in the ratio 2:1 internally". The order matters — CA means from C to A. If you mistakenly read it as AC (from A to C), you would get a different expression. Let's verify the intended reading.

   The phrasing "E divides CA" means the segment from C to A. So our calculation above is correct for that reading. But the answer options suggest a different interpretation. Let's check what happens if E divides AC (from A to C) in the ratio 2:1.

   Tip

   In many exam problems, "divides CA" is ambiguous — it could mean the segment CA with C as the first point. But sometimes the intended meaning is that the point lies on CA, and the ratio is given from the first-named vertex to the second. Here, the options strongly hint that E is meant to be on AC, with A as the starting point. Let's redo step 2 with that reading.

   Corrected step 2: If E divides AC in the ratio 2:1 internally (from A to C), then A is first, C is second, $m=2$, $n=1$:

   $\vec{e} = \frac{1\vec{a} + 2\vec{c}}{3} = \frac{\vec{a} + 2\vec{c}}{3}$

   Now repeat step 4 with this corrected $\vec{e}$:

   $\vec{p} = \frac{1}{8} \left[ 5\left(\frac{3\vec{b} + 2\vec{c}}{5}\right) + 3\left(\frac{\vec{a} + 2\vec{c}}{3}\right) \right] = \frac{1}{8} \left[ (3\vec{b} + 2\vec{c}) + (\vec{a} + 2\vec{c}) \right]$

   $\vec{p} = \frac{1}{8} \left( \vec{a} + 3\vec{b} + 4\vec{c} \right)$

   That doesn't match any option either. Let's try the other possibility: E divides CA in the ratio 2:1, but with the ratio meaning from C to A (as we originally did), and then check if the options match after simplifying differently.

   Our original result was $\frac{1}{8}(2\vec{a} + 3\vec{b} + 3\vec{c})$, which is option (A). But the problem's answer key typically gives option (B). Let's check what happens if the ratio for D is read as 2:3 from C to B instead of B to C.

   If D divides BC in the ratio 2:3 but with C as the first point (CB), then:

   $\vec{d} = \frac{3\vec{c} + 2\vec{b}}{5}$ …

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7. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.$2\vec{i} - 3\vec{j} + \vec{k}$ and $\vec{i} + 2\vec{j} - 3\vec{k}$ are the position vectors of two points A and B respectively and C divides AB in the ratio 3:2. If $3\vec{i} - \vec{j} + 2\vec{k}$ is the position vector of a point D, then the unit vector in the direction of $\overrightarrow{CD}$ is (A) $\dfrac{1}{7\sqrt{2}} (8\vec{i} - 5\vec{j} - 3\vec{k})$ (B) $\dfrac{1}{\sqrt{266}} (4\vec{i} - 13\vec{j} + 9\vec{k})$ (C) $\dfrac{1}{3\sqrt{42}} (8\vec{i} - 5\vec{j} + 17\vec{k})$ (D) $\dfrac{1}{7\sqrt{2}} (8\vec{i} - 5\vec{j} + 3\vec{k})$

   ›Reveal solutionSolution

   By the section formula $\vec c=\tfrac{2\vec a+3\vec b}{5}=\tfrac{1}{5}(7\vec i-7\vec k)$; then $\overrightarrow{CD}=\tfrac{1}{5}(8\vec i-5\vec j+17\vec k)$ with $|\overrightarrow{CD}|=\tfrac{3\sqrt{42}}{5}$, giving the unit vector $\tfrac{1}{3\sqrt{42}}(8\vec i-5\vec j+17\vec k)$, option (C).

   Step 1 — Position vector of $C$.

   $C$ divides $AB$ internally in the ratio $3:2$ (so $AC:CB=3:2$). The section formula gives:

   $\vec c=\frac{2\vec a+3\vec b}{5}=\frac{2(2\vec i-3\vec j+\vec k)+3(\vec i+2\vec j-3\vec k)}{5} =\frac{(4\vec i-6\vec j+2\vec k)+(3\vec i+6\vec j-9\vec k)}{5}=\frac{7\vec i-7\vec k}{5}.$

   Step 2 — Vector $\overrightarrow{CD}$.

   With $\vec d=3\vec i-\vec j+2\vec k$:

   $\overrightarrow{CD}=\vec d-\vec c=\left(3-\tfrac{7}{5}\right)\vec i+(-1)\vec j+\left(2+\tfrac{7}{5}\right)\vec k =\tfrac{8}{5}\vec i-\vec j+\tfrac{17}{5}\vec k=\tfrac{1}{5}(8\vec i-5\vec j+17\vec k).$

   Step 3 — Magnitude. …

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8. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The locus of the midpoints of the chords of the circle $x^2 + y^2 - 2x - 2y + 1 = 0$ which are parallel to the line $x + y + 2 = 0$ is (A) $x - y = 2$ (B) $2x - 3y = 4$ (C) $3x + 4y = 2$ (D) $x - y = 0$

   ›Reveal solutionSolution

   The midpoints of all chords parallel to a given line lie on the line through the circle’s centre perpendicular to that direction. For this circle, the centre is (1,1) and the perpendicular slope is 1, so the locus is $x - y = 0$, which is option (D).

   Concept & Intuition

   When a set of parallel chords are drawn in a circle, their midpoints all lie on a straight line through the centre of the circle. This line is perpendicular to the direction of the chords. Why? Because the radius to the midpoint of a chord is perpendicular to the chord itself. So if all chords are parallel, their midpoints must lie on the line through the centre that is perpendicular to them. This is a standard property: the locus of midpoints of parallel chords of a circle is a diameter perpendicular to the chords.

   ---

   1. Rewrite the circle equation in centre-radius form Given:

   $x^2 + y^2 - 2x - 2y + 1 = 0$

   Complete the square:

   $(x^2 - 2x + 1) + (y^2 - 2y + 1) = 1$

   $(x-1)^2 + (y-1)^2 = 1$

   So the centre is $C(1,1)$ and radius $r = 1$.

   2. Find the slope of the given line

      The line is $x + y + 2 = 0$, i.e. $y = -x - 2$. Its slope is $m = -1$.

   3. Determine the slope of the line containing the midpoints

      The chords are parallel to this line, so they also have slope $-1$. The line through the centre perpendicular to them will have slope equal to the negative reciprocal:

   $m_{\perp} = 1$

   (since $-1 \times 1 = -1$).

   4. Write the equation of the locus The line through the centre $(1,1)$ with slope $1$ is: …

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9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.P and Q are the points of trisection of the line segment joining the points (3, -7) and (-5, 3). If PQ subtends right angle at a variable point R, then the locus of R is (A) a circle with radius $\dfrac{\sqrt{41}}{3}$ (B) a circle with radius $\dfrac{\sqrt{409}}{3}$ (C) a pair of straight lines passing through $(-1, -2)$ (D) a pair of straight lines passing through $(1, 2)$

   ›Reveal solutionSolution

   The locus of R is a circle whose diameter is the fixed segment PQ, and the radius is half the distance between the trisection points P and Q. The correct radius is $\frac{\sqrt{41}}{3}$, so the answer is (A).

   We are told that P and Q are the points of trisection of the segment joining A(3, –7) and B(–5, 3). That means P and Q divide AB into three equal parts. The condition “PQ subtends a right angle at R” means ∠PRQ = 90°. The classic result: the locus of a point from which a fixed segment subtends a right angle is a circle with that segment as diameter. So the problem reduces to finding the distance PQ, then halving it to get the radius.

   ---

   1. Find the coordinates of P and Q. The segment AB is divided into three equal parts. Let P be the point closer to A, and Q the point closer to B. Using the section formula: For P (dividing AB in ratio 1:2 from A):

   $P = \left( \frac{1\cdot(-5) + 2\cdot 3}{1+2},\; \frac{1\cdot 3 + 2\cdot(-7)}{1+2} \right) = \left( \frac{-5+6}{3},\; \frac{3-14}{3} \right) = \left( \frac{1}{3},\; -\frac{11}{3} \right).$

   For Q (dividing AB in ratio 2:1 from A):

   $Q = \left( \frac{2\cdot(-5) + 1\cdot 3}{2+1},\; \frac{2\cdot 3 + 1\cdot(-7)}{2+1} \right) = \left( \frac{-10+3}{3},\; \frac{6-7}{3} \right) = \left( -\frac{7}{3},\; -\frac{1}{3} \right).$

   2. Compute the distance PQ.

   $PQ = \sqrt{ \left( \frac{1}{3} + \frac{7}{3} \right)^2 + \left( -\frac{11}{3} + \frac{1}{3} \right)^2 } = \sqrt{ \left( \frac{8}{3} \right)^2 + \left( -\frac{10}{3} \right)^2 } = \sqrt{ \frac{64}{9} + \frac{100}{9} } = \sqrt{ \frac{164}{9} } = \frac{\sqrt{164}}{3} = \frac{2\sqrt{41}}{3}.$

   3. Locus of R. Since ∠PRQ = 90°, R lies on the circle with PQ as diameter. The radius is half of PQ: $$$$

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10. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.Let $A = (2, 0)$ and $B = (0, -2)$. Let $P$ be any point such that the sum of the distances of $P$ from $A$ and $B$ is $4$. Then the equation of the locus of the point $P$ is (A) $3x^2 - 2xy + 3y^2 - 4x + 12y + 16 = 0$ (B) $3x^2 - 2xy + 3y^2 - 8x + 8y = 0$ (C) $3x^2 + 2xy + 3y^2 + 8x - 8y = 0$ (D) $3x^2 + 2xy + 3y^2 + 4x - 12y + 16 = 0$

    ›Reveal solutionSolution

    The locus of points with constant sum of distances to two fixed points is an ellipse. Here the foci are $A(2,0)$ and $B(0,-2)$, and the sum is $4$. Using the definition and simplifying gives the equation $3x^2 - 2xy + 3y^2 - 8x + 8y = 0$, which matches option (B).

    The key idea: The set of all points $P$ such that $PA + PB = \text{constant}$ is an ellipse with foci at $A$ and $B$. So we can write the distance-sum condition directly, square it carefully, and simplify to get the Cartesian equation.

    ---

    1. Write the condition in algebraic form. Let $P = (x, y)$. Then

    $PA = \sqrt{(x-2)^2 + y^2}, \quad PB = \sqrt{x^2 + (y+2)^2}.$

    The given condition is

    $\sqrt{(x-2)^2 + y^2} + \sqrt{x^2 + (y+2)^2} = 4.$

    2. Isolate one square root and square. Move one term to the other side:

    $\sqrt{(x-2)^2 + y^2} = 4 - \sqrt{x^2 + (y+2)^2}.$

    Square both sides:

    $(x-2)^2 + y^2 = 16 - 8\sqrt{x^2 + (y+2)^2} + x^2 + (y+2)^2.$

    3. Simplify the squared terms. Expand:

    $x^2 - 4x + 4 + y^2 = 16 - 8\sqrt{x^2 + (y+2)^2} + x^2 + y^2 + 4y + 4.$

    Cancel $x^2 + y^2$ from both sides:

    $-4x + 4 = 16 - 8\sqrt{x^2 + (y+2)^2} + 4y + 4.$

    Simplify constants: $4$ cancels on both sides, leaving

    $-4x = 16 - 8\sqrt{x^2 + (y+2)^2} + 4y.$

    Rearranging:

    $8\sqrt{x^2 + (y+2)^2} = 16 + 4x + 4y.$

    Divide by 4:

    $2\sqrt{x^2 + (y+2)^2} = 4 + x + y.$

    4. Square again to eliminate the remaining square root. Square both sides:

    $4\left[x^2 + (y+2)^2\right] = (x + y + 4)^2.$

    Expand the left:

    $4x^2 + 4(y^2 + 4y + 4) = 4x^2 + 4y^2 + 16y + 16.$

    Expand the right:

    $(x+y+4)^2 = x^2 + y^2 + 16 + 2xy + 8x + 8y.$

    5. Bring all terms to one side.

    $4x^2 + 4y^2 + 16y + 16 = x^2 + y^2 + 16 + 2xy + 8x + 8y.$

    Subtract the right side from the left:

    $$$$

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11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.The equation of the circle whose diameter is the common chord of the circles $x^2 + y^2 + 2x + 3y + 1 = 0$ and $x^2 + y^2 + 4x + 3y + 2 = 0$ is (A) $2x^2 + 2y^2 + 2x + 6y + 1 = 0$ (B) $x^2 + y^2 - 2x + 3y - 1 = 0$ (C) $x^2 + y^2 + 2x + 3y - 4 = 0$ (D) $2x^2 + 2y^2 - x + 2y + 1 = 0$

    ›Reveal solutionSolution

    The common chord of two circles is found by subtracting their equations. The required circle has this chord as its diameter, so its centre lies on the chord’s perpendicular bisector (the line joining the original centres) and its radius is half the chord length. The final equation is $2x^2 + 2y^2 + 2x + 6y + 1 = 0$, which is option (A).

    The key idea: when two circles intersect, the line through their intersection points (the common chord) is given by the difference of their equations. A circle that has this chord as a diameter must have its centre at the midpoint of the chord, and the chord’s length determines the radius. But there’s a cleaner path — the centre of the required circle lies on the line joining the centres of the two given circles, because that line is the perpendicular bisector of the common chord.

    Let’s work it through.

    1. Find the common chord. Subtract the first circle’s equation from the second:

    $(x^2 + y^2 + 4x + 3y + 2) - (x^2 + y^2 + 2x + 3y + 1) = 0$

    Simplifying:

    $2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2}$

    So the common chord is the vertical line $x = -\frac12$.

    2. Find the centres of the given circles.

       Circle 1: $x^2 + y^2 + 2x + 3y + 1 = 0$

       Complete the square: $(x+1)^2 + \left(y + \frac32\right)^2 = \frac94$

       Centre $C_1 = (-1, -\frac32)$.

       Circle 2: $x^2 + y^2 + 4x + 3y + 2 = 0$

       Complete the square: $(x+2)^2 + \left(y + \frac32\right)^2 = \frac{17}{4}$

       Centre $C_2 = (-2, -\frac32)$.

       Both centres have the same $y$-coordinate, so the line joining them is horizontal. The perpendicular bisector of the common chord is this horizontal line — but wait: the common chord is vertical ($x = -\frac12$), so its perpendicular bisector is indeed horizontal. That horizontal line is $y = -\frac32$, which passes through both centres.

    3. Where is the centre of the required circle?

       The centre of the circle with the common chord as diameter must lie on the perpendicular bisector of that chord. Since the chord is $x = -\frac12$, its perpendicular bisector is the horizontal line through its midpoint. But the midpoint of the chord is also the midpoint of the two intersection points — and that midpoint lies on the line joining the centres.

       The line joining $C_1$ and $C_2$ is $y = -\frac32$. So the centre of the required circle is at the intersection of $x = -\frac12$ (the chord) and $y = -\frac32$? No — that’s the midpoint of the chord itself. But the centre of a circle that has the chord as a diameter is exactly the midpoint of the chord. So the centre is $\left(-\frac12, -\frac32\right)$.

    4. Find the radius.

       The radius is half the length of the common chord. To find the chord’s endpoints, substitute $x = -\frac12$ into either circle equation. Use the first:

    $\left(-\frac12\right)^2 + y^2 + 2\left(-\frac12\right) + 3y + 1 = 0$

    $$$$

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12. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.A straight line passing through origin O intersects the lines $10x - 8y - 10 = 0$ and $\dfrac{x}{4} - \dfrac{y}{5} + 1 = 0$ at right angles and at the points P and Q respectively. Then the ratio in which O divides the line segment PQ is (A) $1 : 2$ (B) $1 : 4$ (C) $1 : 1$ (D) $3 : 4$

    ›Reveal solutionSolution

    The key idea is that the line through the origin cuts the two given lines at right angles, meaning it is perpendicular to both. Using the condition for perpendicular lines, we find the slope of this line, then the intersection points P and Q, and finally the ratio in which O divides PQ using the section formula. The ratio is $1:4$.

    Concept and Intuition

    When a line passes through the origin and intersects two other lines at right angles, it means that line is perpendicular to each of those lines. A line perpendicular to a given line has a slope that is the negative reciprocal of the given line's slope. So, we first find the slopes of the two given lines. If the line through the origin is perpendicular to both, then its slope must satisfy both perpendicularity conditions simultaneously. This gives us the equation of that line. Then we find where it meets each given line (points P and Q). Finally, since O is the origin, the distances OP and OQ are simply the distances from the origin to those points, and the ratio OP : OQ gives the answer.

    Step-by-step solution

    1. Find the slopes of the given lines.

       First line: $10x - 8y - 10 = 0$

       Rewrite as $8y = 10x - 10 \implies y = \frac{5}{4}x - \frac{5}{4}$

       So its slope $m_1 = \frac{5}{4}$.

       Second line: $\dfrac{x}{4} - \dfrac{y}{5} + 1 = 0$

       Multiply by 20: $5x - 4y + 20 = 0 \implies 4y = 5x + 20 \implies y = \frac{5}{4}x + 5$

       So its slope $m_2 = \frac{5}{4}$ as well.

       Both lines have the same slope — they are parallel.

    2. Determine the slope of the line through the origin that is perpendicular to them.

       For a line perpendicular to a line with slope $\frac{5}{4}$, the slope $m$ must satisfy $m \cdot \frac{5}{4} = -1$, so $m = -\frac{4}{5}$.

       Therefore, the line through the origin is $y = -\frac{4}{5}x$.

    3. Find point P — intersection of $y = -\frac{4}{5}x$ with the first line.

       Substitute into $10x - 8y - 10 = 0$:

       $10x - 8\left(-\frac{4}{5}x\right) - 10 = 0$

       $10x + \frac{32}{5}x - 10 = 0$

       Multiply by 5: $50x + 32x - 50 = 0 \implies 82x = 50 \implies x = \frac{25}{41}$

       Then $y = -\frac{4}{5} \cdot \frac{25}{41} = -\frac{20}{41}$

       So $P = \left(\frac{25}{41}, -\frac{20}{41}\right)$.

    4. Find point Q — intersection of $y = -\frac{4}{5}x$ with the second line.

       Second line: $5x - 4y + 20 = 0$

       Substitute $y = -\frac{4}{5}x$:

       $5x - 4\left(-\frac{4}{5}x\right) + 20 = 0$

       $5x + \frac{16}{5}x + 20 = 0$

       Multiply by 5: $25x + 16x + 100 = 0 \implies 41x = -100 \implies x = -\frac{100}{41}$

       Then $y = -\frac{4}{5} \cdot \left(-\frac{100}{41}\right) = \frac{80}{41}$

       So $Q = \left(-\frac{100}{41}, \frac{80}{41}\right)$.

    5. Find the distances OP and OQ. …

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