Q.In what ratio, the line joining $(-1, 1)$ and $(5, 7)$ is divided by the line $x + y = 4$?

Showing the 12 most recent of 62 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If a variable line passing through the point of intersection Q of the lines $2x + 3y = 6$ and $3x + 2y = 6$ also intersect the X and Y axes at A and B respectively, then the locus of a point which divides AB in the ratio $1 : 2$ is (A) $2x + 4y = 5xy$ (B) $4x + 2y = 5xy$ (C) $4x + 3y = 5xy$ (D) $3x + 4y = 5xy$

   ›Reveal solutionSolution

   Find the fixed point Q, write the intercept-form condition for lines through Q, express the dividing point's coordinates in terms of the intercepts, then eliminate $a,b$ to get the locus.

   Concept and Intuition

   Any line through a fixed point $Q$ imposes one linear constraint on its $X$- and $Y$-intercepts $a,b$ (from the intercept form $\frac xa+\frac yb=1$ passing through $Q$). As the line varies (varying $a,b$ subject to that constraint), the point dividing the intercept-segment in a fixed ratio traces a curve — found by expressing $a,b$ in terms of the moving point's coordinates and substituting into the constraint.

   Step-by-Step Solution

   1. Solve $2x+3y=6$ and $3x+2y=6$: subtracting gives $-x+y=0 \Rightarrow y=x$; substituting, $5x=6 \Rightarrow x=y=\frac65$. So $Q=\left(\frac65,\frac65\right)$.
   2. A variable line meeting the axes at $A=(a,0)$, $B=(0,b)$ has equation $\frac xa+\frac yb=1$. Since it passes through $Q$: $\dfrac{6/5}{a}+\dfrac{6/5}{b}=1 \Rightarrow \dfrac65\left(\dfrac1a+\dfrac1b\right)=1 \Rightarrow \dfrac1a+\dfrac1b=\dfrac56$.
   3. The point $P(x,y)$ dividing $AB$ in the ratio $1:2$ (from $A$) is $P = \left(\dfrac{1\cdot 0+2\cdot a}{3}, \dfrac{1\cdot b+2\cdot 0}{3}\right) = \left(\dfrac{2a}{3}, \dfrac b3\right)$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If the point $P$ which divides the line segment joining $A(1,1,1)$ and $B(2,2,2)$ in the ratio $1:m$ lies on the plane $x + 2y + 3z - 1 = 0$, then $m =$ (A) $-\dfrac{3}{2}$ (B) $\dfrac{4}{3}$ (C) $-\dfrac{11}{5}$ (D) $-\dfrac{1}{2}$

   ›Reveal solutionSolution

   Since $A$ and $B$ have identical $x=y=z$ patterns, $P$ also has equal coordinates $p$; substituting into the plane equation and solving for $m$ gives $-11/5$.

   Concept and Intuition

   The section formula for a point dividing $AB$ in ratio $l:m$ (from $A$) is $P=\dfrac{l\,B+m\,A}{l+m}$. Because $A=(1,1,1)$ and $B=(2,2,2)$ both have all three coordinates equal, $P$'s three coordinates are automatically equal too — this collapses the 3D problem to a single unknown $p$, which we then plug into the plane equation.

   Step-by-Step Solution

   1. With ratio $1:m$, $P=\dfrac{1\cdot(2,2,2)+m\cdot(1,1,1)}{1+m}=\left(\dfrac{2+m}{1+m},\dfrac{2+m}{1+m},\dfrac{2+m}{1+m}\right)$.
   2. Let $p=\dfrac{2+m}{1+m}$ (all coordinates of $P$).
   3. Plane: $x+2y+3z-1=0 \Rightarrow p+2p+3p-1=0 \Rightarrow 6p=1 \Rightarrow p=\dfrac16$. …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $\bar{i} + 2\bar{j} + \bar{k}$, $\alpha\bar{i} + 3\bar{j} + 2\bar{k}$, $-\bar{i} + 4\bar{j} + \beta\bar{k}$ are the position vectors of three points A, B, C, then the position vector of a point which divides BC in the ratio $\alpha + 1 : \beta$ is (A) $\left(\dfrac{-1}{4}, \dfrac{13}{4}, \dfrac{9}{4}\right)$ (B) $\left(\dfrac{-1}{3}, \dfrac{13}{3}, \dfrac{9}{3}\right)$ (C) $\left(\dfrac{5}{2}, \dfrac{7}{2}, \dfrac{6}{2}\right)$ (D) $\left(\dfrac{7}{3}, \dfrac{2}{3}, \dfrac{1}{3}\right)$

   ›Reveal solutionSolution

   With A, B, C collinear, matching direction vectors pins down $\alpha$ and $\beta$, after which the section-formula point on BC is computed directly. The answer is (A).

   Concept and Intuition

   For a division ratio expressed using unknown parameters $\alpha,\beta$ to yield one specific numeric point (as the answer choices demand), those parameters must be fixed by a geometric condition on A, B, C — here, that they are collinear (a standard setup for this style of vector problem). Once $\alpha,\beta$ are pinned down, the section formula $\dfrac{n\vec B+m\vec C}{m+n}$ for the point dividing $BC$ in ratio $m:n$ finishes the problem.

   Step-by-Step Solution

   1. $\vec{AB} = \vec B-\vec A = (\alpha-1)\bar i + (3-2)\bar j+(2-1)\bar k = (\alpha-1)\bar i+\bar j+\bar k$.
   2. $\vec{AC} = \vec C-\vec A = (-1-1)\bar i+(4-2)\bar j+(\beta-1)\bar k = -2\bar i+2\bar j+(\beta-1)\bar k$.
   3. Collinearity requires $\vec{AB}=t\,\vec{AC}$ for some scalar $t$. Matching the $\bar j$ components: $1 = 2t \Rightarrow t=\dfrac12$.
   4. Matching $\bar i$: $\alpha-1 = -2t = -1 \Rightarrow \alpha=0$.
   5. Matching $\bar k$: $1 = (\beta-1)t = \dfrac{\beta-1}{2} \Rightarrow \beta-1=2\Rightarrow \beta=3$.
   6. So the required ratio is $\alpha+1:\beta = 1:3$.
   7. With $\alpha=0$: $\vec B = (0,3,2)$; with $\beta=3$: $\vec C=(-1,4,3)$. …

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4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Let $\overline{OA} = \bar{i} + 2\bar{j} - 4\bar{k}$ and $\overline{OB} = 3\bar{i} - 4\bar{j} - 2\bar{k}$ be the position vectors of two points $A$ and $B$. If a point $C$ divides the line segment $AB$ in the ratio 1:3 externally, then the position vector of a point which divides $OC$ in the ratio 4:1 internally is (A) $5(\bar{i} - \bar{j})$ (B) $\bar{i} - 4\bar{j} + 2\bar{k}$ (C) $4\bar{i} - 2\bar{j} + \bar{k}$ (D) $4(\bar{j} - \bar{k})$

   ›Reveal solutionSolution

   Apply the external section formula to locate $C$ on line $AB$, then apply the internal section formula on segment $OC$. Answer: $4(\bar j-\bar k)$.

   Concept and Intuition

   For points with position vectors $A,B$, the point dividing $AB$ internally in ratio $m:n$ is $\dfrac{mB+nA}{m+n}$, while the point dividing it externally in ratio $m:n$ is $\dfrac{mB-nA}{m-n}$ — the external version effectively places the dividing point beyond one of the endpoints. Once $C$ is found this way, dividing $OC$ internally is just the ordinary internal-section formula applied to the segment from the origin to $C$.

   Step-by-Step Solution

   1. $A=\overline{OA}=(1,2,-4)$, $B=\overline{OB}=(3,-4,-2)$.
   2. $C$ divides $AB$ externally in ratio $1:3$ ($m=1,n=3$): $C=\dfrac{mB-nA}{m-n}=\dfrac{1\cdot B-3\cdot A}{1-3}=\dfrac{B-3A}{-2}=\dfrac{3A-B}{2}$.
   3. Compute $3A=(3,6,-12)$, then $3A-B=(3-3,\,6-(-4),\,-12-(-2))=(0,10,-10)$.
   4. So $C=\dfrac{(0,10,-10)}{2}=(0,5,-5)$. …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The external centre of similitude of the two circles $x^2+y^2-4x+6y+4=0$ and $x^2+y^2-2x+2y-2=0$ is (A) $(1,-3)$ (B) $(-1,3)$ (C) $(-1,-3)$ (D) $(1,3)$

   ›Reveal solutionSolution

   Find both circles' centres/radii, then divide the segment joining them externally in the ratio of the radii to get $(-1,3)$.

   Concept and Intuition

   The external centre of similitude of two circles is the point on the line joining their centres that divides that segment externally in the ratio of their radii, $r_1:r_2$ — the point from which the two circles subtend equal angles and appear as scaled copies of each other (external tangents from it touch both circles).

   Step-by-Step Solution

   1. Circle 1: $x^2+y^2-4x+6y+4=0$ has centre $C_1=(2,-3)$, radius $r_1=\sqrt{4+9-4}=3$.
   2. Circle 2: $x^2+y^2-2x+2y-2=0$ has centre $C_2=(1,-1)$, radius $r_2=\sqrt{1+1+2}=2$.
   3. External division of $C_1C_2$ in ratio $r_1:r_2=3:2$: $E=\dfrac{r_2C_1-r_1C_2}{r_2-r_1}$. …

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6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Let $A(1,2,1)$, $B\left(\dfrac{9}{5},\dfrac{8}{5},\dfrac{9}{5}\right)$, $C(\alpha,\beta,\gamma)$ and $D(-3,4,-3)$ be four collinear points. If B divides AC in the ratio of $m:n$, D divides AC in the ratio $m:-n$, then $\alpha+\beta+\gamma=$ (A) $7$ (B) $-2$ (C) $4$ (D) $\dfrac{26}{5}$

   ›Reveal solutionSolution

   Using the internal/external section-ratio formulas for $B$ and $D$ on segment $AC$, we solve for the ratio and recover $C=(3,1,3)$, giving $\alpha+\beta+\gamma=7$.

   Concept and Intuition

   "$B$ divides $AC$ in ratio $m:n$" means $B=\dfrac{nA+mC}{m+n}$ (internal division). "$D$ divides $AC$ in ratio $m:-n$" is the same formula with $n$ replaced by $-n$, i.e. $D=\dfrac{-nA+mC}{m-n}$ — this is external division. Adding/subtracting these two vector equations isolates $C$ cleanly in terms of $A$, $D$, and the ratio $k=n/m$, without ever needing $m,n$ individually.

   Step-by-Step Solution

   1. Internal division: $B(m+n)=nA+mC$.
   2. "External" (ratio $m:-n$) division: $D(m-n)=-nA+mC$.
   3. Add the two equations: $B(m+n)+D(m-n)=2mC \Rightarrow C=\dfrac{B(m+n)+D(m-n)}{2m}$.
   4. Equivalently, from equation (2) alone: $mC = D(m-n)+nA \Rightarrow C = D(1-k)+kA$, where $k=n/m$.
   5. Use the $x$-coordinates to pin down $k$: $A_x=1,\ D_x=-3$, and from the $B$-equation, $B_x(m+n)=nA_x+mC_x \Rightarrow C_x=\dfrac{9}{5}+\dfrac{4}{5}k$ (dividing the $B$-relation by $m$). Also from step 4, $C_x=-3(1-k)+k=-3+4k$.
   6. Equate: $-3+4k=\dfrac{9}{5}+\dfrac{4}{5}k\Rightarrow \dfrac{16}{5}k=\dfrac{24}{5}\Rightarrow k=\dfrac32$.
   7. So $C=D(1-\tfrac32)+\tfrac32 A=-\tfrac12 D+\tfrac32 A$. …

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7. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The locus of all the points which divide the line segment joining A (1, 2) and B (5, 7) in the ratio $m : 1-m$ $(m>1)$ is (A) $\left\{(x,y)/5x-4y+3=0, (x,y)\neq\left(\dfrac{5+k}{k+1},\dfrac{7+2k}{k+1}\right), k>0\right\}$ (B) $\{(x,y)/5x-4y+3=0\}$ (C) $\{(x,y)/5x-4y+3=0, x>5, y>7\}$ (D) $\{(x,y)/5x-4y+3=0, x<1, y<2\}$

   ›Reveal solutionSolution

   The affine combination $A+m(B-A)$ always lies on line $AB$; restricting $m>1$ restricts the locus to the ray beyond $B$, giving $5x-4y+3=0$ together with $x>5,y>7$.

   Concept and Intuition

   A point dividing $AB$ "in ratio $m:(1-m)$" (note the two parts sum to $1$) is simply the affine parametrisation $P(m)=(1-m)A+mB$: at $m=0$ we get $A$, at $m=1$ we get $B$, for $0<m<1$ we get interior points, and for $m>1$ we get points on the ray from $A$ through $B$ and beyond $B$.

   Step-by-Step Solution

   1. $P(m)=(1-m)A+mB=A+m(B-A)$. With $A=(1,2)$, $B-A=(4,5)$: $P=(1+4m,\,2+5m)$.
   2. Eliminate $m$: from $x=1+4m$, $m=\dfrac{x-1}{4}$. Substitute into $y=2+5m$: $y=2+\dfrac{5(x-1)}{4}\Rightarrow4y=8+5x-5\Rightarrow5x-4y+3=0$. This is the full line through $A$ and $B$.
   3. At $m=1$: $P=(5,7)=B$. For $m>1$, $x=1+4m>5$ and $y=2+5m>7$ (both coordinates are strictly increasing functions of $m$).
   4. So the locus is exactly the part of the line $5x-4y+3=0$ with $x>5$ and $y>7$ — the open ray beyond $B$, matching option (C). …

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8. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If the orthocentre and circumcentre of a triangle are $(-3,5,2), (6,2,5)$ respectively, then its centroid is (A) $(3,3,4)$ (B) $(3,4,3)$ (C) $(4,3,3)$ (D) $(0,0,3)$

   ›Reveal solutionSolution

   Uses the Euler-line ratio between orthocentre, centroid and circumcentre to get the centroid $(3,3,4)$.

   Concept and Intuition

   In any triangle the orthocentre $O$, centroid $G$, and circumcentre $C$ are collinear (the Euler line), and $G$ always divides $OC$ internally in the ratio $2:1$ (closer to $C$). This holds in 3D exactly as in the plane since the whole triangle (and hence its Euler line) lies in one plane.

   Step-by-Step Solution

   1. Euler line relation: $\vec{OG}=2\,\vec{GC}$, i.e. $G-O=2(C-G)$.
   2. Solve: $3G=O+2C\Rightarrow G=\dfrac{O+2C}{3}$.
   3. Substitute $O=(-3,5,2)$, $C=(6,2,5)$:
      • $x$: $\dfrac{-3+2(6)}{3}=\dfrac{-3+12}{3}=\dfrac{9}{3}=3$
      • $y$: $\dfrac{5+2(2)}{3}=\dfrac{5+4}{3}=\dfrac{9}{3}=3$ …

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9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $\bar i + \bar j - \bar k$, $7\bar i - 2\bar j - 3\bar k$ and $-5\bar i - 2\bar j + 5\bar k$ are the position vectors of the points $A, B, C$ respectively, then the position vector of the point of intersection of the bisector of $\angle BAC$ and side $BC$ is (A) $\dfrac{1}{16}(27\bar i - 32\bar j + 2\bar k)$ (B) $\dfrac{1}{4}(7\bar i - 8\bar j + 2\bar k)$ (C) $\dfrac{1}{4}(7\bar i + 8\bar j + 2\bar k)$ (D) $\dfrac{1}{16}(28\bar i - 32\bar j + 2\bar k)$

   ›Reveal solutionSolution

   The angle bisector from a vertex divides the opposite side in the ratio of the two adjacent sides ($BD:DC=AB:AC$); applying the section formula gives $D=\dfrac14(7\bar i-8\bar j+2\bar k)$.

   Concept and Intuition

   The internal angle-bisector theorem says the bisector of $\angle A$ meets side $BC$ at a point $D$ such that $BD:DC=AB:AC$. Once that ratio is known, the section formula (weighted average of the endpoints, weighted inversely to the adjacent segment) locates $D$ exactly.

   Step-by-Step Solution

   1. $A=\bar i+\bar j-\bar k,\ B=7\bar i-2\bar j-3\bar k,\ C=-5\bar i-2\bar j+5\bar k$.
   2. $\overrightarrow{AB}=B-A=(6,-3,-2)$, so $AB=\sqrt{36+9+4}=\sqrt{49}=7$.
   3. $\overrightarrow{AC}=C-A=(-6,-3,6)$, so $AC=\sqrt{36+9+36}=\sqrt{81}=9$.
   4. By the angle-bisector theorem, $BD:DC=AB:AC=7:9$.
   5. By the section formula for internal division in ratio $m:n=7:9$: $D=\dfrac{n\cdot B+m\cdot C}{m+n}=\dfrac{9B+7C}{16}$. …

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10. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.If N is the foot of the perpendicular drawn from the point P(5, −1, 3) to the line passing through the points A(1, 3, −5) and B(3, −1, 5), then the ratio in which N divides AB is (A) $-26 : 21$ (B) $13 : 2$ (C) $11 : 19$ (D) $-11 : 19$

    ›Reveal solutionSolution

    This tests finding the foot of a perpendicular onto a 3D line using a parameter $t$ along the line and the perpendicularity condition; the answer is that $N$ divides $AB$ in ratio $13:2$.

    Concept and Intuition

    Any point on line $AB$ can be written as $A+t(B-A)$, where $t$ is exactly the ratio $AN:AB$ (so $N$ divides $AB$ in the ratio $t:(1-t)$). The foot of the perpendicular from an external point $P$ is the value of $t$ for which $\vec{PN}$ is perpendicular to the direction of the line.

    Step-by-Step Solution

    1. Direction of $AB$: $B-A=(2,-4,10)$.
    2. $N = A+t(2,-4,10) = (1+2t,\,3-4t,\,-5+10t)$.
    3. $\vec{PN} = N-P = (2t-4,\,4-4t,\,10t-8)$.
    4. Perpendicularity: $\vec{PN}\cdot(2,-4,10)=0$: $2(2t-4) -4(4-4t) + 10(10t-8) = 0$ …

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11. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If the intercept of a straight line L made between the straight lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$, then the equation of L is (A) $35x - 83y + 92 = 0$ (B) $83x + 35y - 72 = 0$ (C) $63x - 35y + 82 = 0$ (D) $83x - 35y + 92 = 0$

    ›Reveal solutionSolution

    Find the two intercept points on the given lines whose midpoint is $(1,5)$, then write the line joining them.

    Concept and Intuition

    L meets $5x-y-4=0$ at some point $P$ and $3x+4y-4=0$ at some point $Q$; we're told $(1,5)$ bisects $PQ$. Parametrizing $P$ and $Q$ by their $x$-coordinates, the midpoint conditions give two linear equations that pin down $P$ and $Q$ exactly, after which L is just the line through them.

    Step-by-Step Solution

    1. Let $P=(p,\,5p-4)$ (on $5x-y-4=0$) and $Q=\left(q,\,\dfrac{4-3q}{4}\right)$ (on $3x+4y-4=0$).
    2. Midpoint $=(1,5)$: $\dfrac{p+q}{2}=1\Rightarrow p+q=2$, and $\dfrac{(5p-4)+\frac{4-3q}{4}}{2}=5$.
    3. Multiplying the second condition by 4: $20p-16+4-3q=40\Rightarrow 20p-3q=52$.
    4. Using $q=2-p$: $20p-3(2-p)=52\Rightarrow 23p=58\Rightarrow p=\dfrac{58}{23},\ q=-\dfrac{12}{23}$. …

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12. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If Q $(\alpha, \beta, \gamma)$ is the harmonic conjugate of the point $P(0,-7,1)$ with respect to the line segment joining the points $(2, -5, 3)$ and $(-1, -8, 0)$, then $\alpha - \beta + \gamma =$ (A) 4 (B) 3 (C) 2 (D) 1

    ›Reveal solutionSolution

    The harmonic conjugate of a point dividing a segment internally in ratio $m:n$ is the point dividing the same segment externally in the same ratio $m:n$. Working this out gives $\alpha-\beta+\gamma=4$.

    Concept and Intuition

    Four collinear points $A,B,P,Q$ are said to form a harmonic range (with $Q$ the harmonic conjugate of $P$ w.r.t. $A,B$) when the cross ratio $(A,B;P,Q)=-1$. Concretely, if $P$ divides $AB$ internally in ratio $m:n$, then $Q$ divides $AB$ externally in the same ratio $m:n$. So the recipe is: (1) find the internal ratio in which $P$ divides $AB$; (2) apply the external section formula with that same ratio to get $Q$.

    Step-by-Step Solution

    1. Let $A=(2,-5,3)$, $B=(-1,-8,0)$, so $B-A=(-3,-3,-3)$.
    2. Write $P = A + t(B-A)$ for the point $P(0,-7,1)$:

    $2-3t=0,\quad -5-3t=-7,\quad 3-3t=1$

    All three give $t=\dfrac{2}{3}$, confirming $P$ lies on line $AB$ and divides it internally with $AP:PB = t:(1-t) = \dfrac23:\dfrac13 = 2:1$.

    3. The harmonic conjugate $Q$ divides $AB$ externally in ratio $2:1$. Using the external section formula $Q=\dfrac{mB-nA}{m-n}$ with $m:n=2:1$: $Q = \frac{2B-A}{2-1} = 2B-A$ …

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