Q.If three lines whose equations are $y = m_1 x + c_1$, $y = m_2 x + c_2$ and $y = m_3 x + c_3$ are concurrent, then show that $m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$.

Showing the 12 most recent of 28 on this concept.

1. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The straight line which is parallel to X-axis and passing through the intersection of the lines $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$, $(a, b) \neq (0, 0)$ is (A) above the X-axis at a distance of $\dfrac{3}{2}$ units from it (B) above the X-axis at a distance of $\dfrac{2}{3}$ units from it (C) below the X-axis at a distance of $\dfrac{3}{2}$ units from it (D) below the X-axis at a distance of $\dfrac{2}{3}$ units from it

   ›Reveal solutionSolution

   This tests solving a family of lines' intersection point independent of the parameters, then reading off a horizontal line through it. The intersection point turns out to be $(0,-3/2)$ for every valid $a,b$, so the required line is $y=-3/2$: below the X-axis, distance $3/2$.

   Concept and Intuition

   The two given lines are members of a family parametrised by $a,b$ (with $(a,b)\neq(0,0)$), but the problem asks about a fixed geometric object (a horizontal line through their intersection) — which strongly suggests their point of intersection is actually independent of $a$ and $b$. Recognising and exploiting that structure (rather than solving symbolically for a general intersection formula) is the key idea. A line parallel to the X-axis just has the form $y=k$, so once we know the intersection point's $y$-coordinate, we're done.

   Step-by-Step Solution

   1. The lines are

   $ax+2by+3b=0 \qquad (1)$

   $bx-2ay-3a=0 \qquad (2)$

   2. Multiply (1) by $a$ and (2) by $b$:

   $a^2x+2aby+3ab=0,\qquad b^2x-2aby-3ab=0.$

   3. Add these two equations — the $y$ terms cancel:

   $(a^2+b^2)x=0.$

   Since $(a,b)\neq(0,0)$, $a^2+b^2\neq0$, so $x=0$.

   4. Substitute $x=0$ into (1): $2by+3b=0\Rightarrow b(2y+3)=0$. If $b\neq 0$, then $y=-\dfrac32$. (If $b=0$, then $a\neq0$; equation (2) becomes $-2ay-3a=0\Rightarrow a(-2y-3)=0\Rightarrow y=-\dfrac32$ again — same point either way.)

   5. So the two lines always meet at $\left(0,-\dfrac32\right)$, no matter what $a,b$ are. …

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2. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.A line $L_1$ passing through the point of intersection of the lines $x - 2y + 3 = 0$ and $2x - y = 0$ is parallel to the Line $L_2$. If $L_2$ passes through origin and also through the point of intersection of the lines $3x - y + 2 = 0$ and $x - 3y - 2 = 0$, then the distance between the lines $L_1$ and $L_2$ is (A) $\dfrac{1}{\sqrt{2}}$ (B) $\sqrt{2}$ (C) $\sqrt{5}$ (D) $\dfrac{1}{\sqrt{5}}$

   ›Reveal solutionSolution

   Find both intersection points, pin down $L_2$ through the origin, build the parallel line $L_1$ through the first intersection point, then apply the parallel-line distance formula to get $\dfrac{1}{\sqrt2}$.

   Concept and Intuition

   Two intersecting lines meet at a unique point found by solving them simultaneously. Once a line's direction (slope) is known, a parallel line through any other point is easy to write. The perpendicular distance between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is $\dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}$.

   Step-by-Step Solution

   1. Solve $x-2y+3=0$ and $2x-y=0$: from the second, $y=2x$; substitute: $x-4x+3=0\Rightarrow x=1,\ y=2$. So the point is $(1,2)$.
   2. Solve $3x-y+2=0$ and $x-3y-2=0$: from the first, $y=3x+2$; substitute: $x-3(3x+2)-2=0\Rightarrow x-9x-6-2=0\Rightarrow x=-1,\ y=-1$. So the point is $(-1,-1)$.
   3. $L_2$ passes through the origin $(0,0)$ and $(-1,-1)$, so its direction is $(1,1)$ and its equation is $y=x$, i.e. $x-y=0$. …

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3. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the lines $x + y - 2 = 0$, $3x - 4y + 1 = 0$ and $5x + ky - 7 = 0$ are concurrent at $(\alpha, \beta)$, then equation of the line concurrent with the given lines and perpendicular to $kx + y - k = 0$ is (A) $x - 3y = -2$ (B) $x + 4y = 5$ (C) $x + 6y = 7$ (D) $x - 2y = -1$

   ›Reveal solutionSolution

   Find the common point of the first two lines, use concurrency to solve for $k$, then build the line through that point perpendicular to $kx+y-k=0$, giving $x-2y=-1$.

   Concept and Intuition

   Three lines are concurrent when they share one common point. Once two of them fix that point, substituting it into the third line's equation solves for any unknown parameter. Perpendicularity between two lines of slopes $m_1,m_2$ requires $m_1m_2=-1$.

   Step-by-Step Solution

   1. Solve $x+y-2=0$ and $3x-4y+1=0$: $y=2-x$, substitute: $3x-4(2-x)+1=0\Rightarrow 3x-8+4x+1=0\Rightarrow 7x=7\Rightarrow x=1,\ y=1$. So $(\alpha,\beta)=(1,1)$.
   2. Since all three lines are concurrent, $(1,1)$ also lies on $5x+ky-7=0$: $5+k-7=0\Rightarrow k=2$.
   3. The reference line becomes $2x+y-2=0$, slope $=-2$. A line perpendicular to it has slope $\dfrac12$. …

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4. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If the lines $x + 2ay + a = 0$, $x + 3by + b = 0$, $x + 4cy + c = 0$ are concurrent, then a, b, c are in (A) Arithmetic Progression (B) Geometric Progression (C) Harmonic Progression (D) Arithmetico-geometric Progression

   ›Reveal solutionSolution

   Apply the concurrency (determinant = 0) condition to the three lines; simplifying yields the AP condition on the reciprocals, i.e. $a,b,c$ are in HP.

   Concept and Intuition

   Three lines $a_ix+b_iy+c_i=0$ ($i=1,2,3$) are concurrent exactly when the $3\times3$ determinant of their coefficients is zero. Here each line is written with the same $x$-coefficient (1), which makes the determinant expansion clean and lets the progression fall out directly.

   Step-by-Step Solution

   1. The three lines are $x+2ay+a=0$, $x+3by+b=0$, $x+4cy+c=0$. Concurrency requires:

   $\begin{vmatrix}1 & 2a & a\\ 1 & 3b & b\\ 1 & 4c & c\end{vmatrix}=0$

   2. Expand along the first column:

   $1\cdot(3bc-4bc) - 2a\cdot(c-b) + a\cdot(4c-3b) = 0$

   $-bc -2ac+2ab+4ac-3ab = 0$

   $-bc+2ac-ab=0$

   3. Rearrange:

   $2ac = ab+bc$

   4. Divide through by $abc$ (all nonzero, as they're line coefficients): $\frac{2}{b} = \frac1c+\frac1a$ …

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5. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$L_1$ and $L_2$ are two lines having slopes 2 and $-\dfrac{1}{2}$ respectively. If both $L_1$ and $L_2$ are concurrent with the lines $x - y + 2 = 0$ and $2x + y + 3 = 0$, then sum of the absolute values of the intercepts made by the lines $L_1$ and $L_2$ on the coordinate axes is (A) $2$ (B) $7$ (C) $12$ (D) $9$

   ›Reveal solutionSolution

   Both lines are concurrent with the given pair, so both pass through their common intersection point; building each line from that point and its slope gives intercepts summing (in absolute value) to $7$.

   Concept and Intuition

   "$L_1$ and $L_2$ are concurrent with the lines $\dots$" means all lines meet at one common point — the intersection of the two fixed lines. Once that point and each slope are known, each line's equation, and hence its axis intercepts, follow directly.

   Step-by-Step Solution

   1. Solve $x-y+2=0$ and $2x+y+3=0$ together: adding gives $3x+5=0\Rightarrow x=-\dfrac53$; then $y=x+2=\dfrac13$. Common point $P=\left(-\dfrac53,\dfrac13\right)$.
   2. $L_1$ through $P$ with slope $2$: $y-\dfrac13=2\left(x+\dfrac53\right)\Rightarrow y=2x+\dfrac{11}{3}$. $x$-intercept: $0=2x+\dfrac{11}3\Rightarrow x=-\dfrac{11}{6}$. $y$-intercept: $\dfrac{11}{3}$. …

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6. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If A (1, 0, 2), B (2, 1, 0), C (2, -5, 3), D (0, 3, 2) are four points and the point of intersection of the lines AB and CD is P (a, b, c), then $a+b+c=$ (A) $3$ (B) $-5$ (C) $5$ (D) $-3$

   ›Reveal solutionSolution

   Write both lines in parametric form and solve for the parameters where they meet; a consistent third equation confirms the lines genuinely intersect.

   Concept and Intuition

   Two lines in 3D generally don't intersect (they're skew) unless a special consistency condition holds. Parametrising each line and equating coordinates gives an over-determined system; if it's consistent, the point found is the actual intersection.

   Step-by-Step Solution

   1. $A(1,0,2)$, $B(2,1,0)$: direction $B-A=(1,1,-2)$. Line $AB$: $(1+t,\,t,\,2-2t)$.
   2. $C(2,-5,3)$, $D(0,3,2)$: direction $D-C=(-2,8,-1)$. Line $CD$: $(2-2s,\,-5+8s,\,3-s)$.
   3. Equate: $1+t=2-2s$ … (i); $t=-5+8s$ … (ii); $2-2t=3-s$ … (iii).
   4. From (ii) into (i): $1+(-5+8s)=2-2s\Rightarrow-4+8s=2-2s\Rightarrow10s=6\Rightarrow s=3/5$.
   5. Then $t=-5+8(3/5)=-5+24/5=-1/5$. …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.The radical centre of the circles $x^2+y^2+2x+3y+1=0$, $x^2+y^2+x-y+3=0$, $x^2+y^2-3x+2y+5=0$ is (A) $\left(-\dfrac{7}{38},\dfrac{6}{19}\right)$ (B) $\left(\dfrac{6}{19},\dfrac{14}{19}\right)$ (C) $\left(\dfrac{14}{19},\dfrac{6}{19}\right)$ (D) $\left(\dfrac{2}{19},\dfrac{3}{19}\right)$

   ›Reveal solutionSolution

   The radical centre is the common intersection point of the three pairwise radical axes; find any two radical axes and solve them simultaneously.

   Concept and Intuition

   The radical axis of two circles $C_i=0,C_j=0$ (both written with unit coefficient of $x^2,y^2$) is simply $C_i-C_j=0$ — the quadratic terms cancel, leaving a straight line. The radical centre is where all three pairwise radical axes meet (any two suffice to find it).

   Step-by-Step Solution

   1. $C_1: x^2+y^2+2x+3y+1=0$; $C_2: x^2+y^2+x-y+3=0$; $C_3: x^2+y^2-3x+2y+5=0$.
   2. Radical axis of $C_1,C_2$: $C_1-C_2 = (2x+3y+1)-(x-y+3) = x+4y-2=0$.
   3. Radical axis of $C_2,C_3$: $C_2-C_3 = (x-y+3)-(-3x+2y+5) = x-y+3+3x-2y-5 = 4x-3y-2=0$.
   4. Solve $x+4y=2$ and $4x-3y=2$ simultaneously. From the first: $x=2-4y$.
   5. Substitute: $4(2-4y)-3y=2 \Rightarrow 8-16y-3y=2 \Rightarrow 8-19y=2 \Rightarrow 19y=6 \Rightarrow y=\dfrac{6}{19}$. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.For $\lambda, \mu \in \mathbb{R}$, $(x - 2y - 1) + \lambda(3x + 2y - 11) = 0$ and $(3x + 4y - 11) + \mu(-x + 2y - 3) = 0$ represent two families of lines. If the equation of the line common to both the families is $ax + by - 5 = 0$, then $2a + b =$ (A) $0$ (B) $1$ (C) $4$ (D) $3$

   ›Reveal solutionSolution

   Each family is a pencil of lines through a fixed point; a line common to both pencils must pass through both fixed points, which pins it down uniquely.

   Concept and Intuition

   An equation of the form $L_1+\lambda L_2=0$ represents every line through the intersection of $L_1=0$ and $L_2=0$ (except $L_2=0$ itself), as $\lambda$ varies. So Family 1 is the pencil of lines through one fixed point, and Family 2 is the pencil through another fixed point. A line that belongs to both families (for some $\lambda$ and some $\mu$) must pass through both fixed points — i.e. it's the unique line joining them.

   Step-by-Step Solution

   1. Base point of Family 1: solve $x-2y-1=0$ and $3x+2y-11=0$. Adding: $4x-12=0\Rightarrow x=3$; then $3-2y-1=0\Rightarrow y=1$. Point $(3,1)$.
   2. Base point of Family 2: solve $3x+4y-11=0$ and $-x+2y-3=0\Rightarrow x=2y-3$. Substituting: $3(2y-3)+4y-11=0 \Rightarrow 10y-20=0\Rightarrow y=2,\ x=1$. Point $(1,2)$.
   3. The common line passes through $(3,1)$ and $(1,2)$: slope $=\dfrac{2-1}{1-3}=-\dfrac12$.
   4. Equation: $y-1=-\dfrac12(x-3) \Rightarrow x+2y-5=0$. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If the reflection of a point $A(2,3)$ in X-axis is B; reflection of B in the line $x+y=0$ is C and the reflection of C in $x-y=0$ is D then the point of intersection of the lines CD, AB is (A) $(3,-2)$ (B) $(0,1)$ (C) $(4,-3)$ (D) $(2,-1)$

   ›Reveal solutionSolution

   Chasing the three reflections gives $B(2,-3)$, $C(3,-2)$, $D(-2,3)$; lines $AB$ and $CD$ intersect at $(2,-1)$.

   Concept and Intuition

   Reflection in the $x$-axis: $(x,y)\to(x,-y)$. Reflection in $x+y=0$: $(x,y)\to(-y,-x)$. Reflection in $x-y=0$ (i.e. $y=x$): $(x,y)\to(y,x)$. Chain these to find $B,C,D$, then intersect the two resulting lines.

   Step-by-Step Solution

   1. $A=(2,3)$. Reflect in $x$-axis: $B=(2,-3)$.
   2. Reflect $B$ in $x+y=0$: $(x,y)\to(-y,-x)$ gives $C=(-(-3),-(2))=(3,-2)$.
   3. Reflect $C$ in $x-y=0$: $(x,y)\to(y,x)$ gives $D=(-2,3)$.
   4. Line $AB$: both points have $x=2$, so it's the vertical line $x=2$.
   5. Line $CD$: slope $=\dfrac{3-(-2)}{-2-3}=\dfrac{5}{-5}=-1$; through $C(3,-2)$: $y+2=-(x-3)\Rightarrow x+y=1$. …

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10. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.The orthocentre of the triangle formed by lines $x + y + 1 = 0$; $x - y - 1 = 0$ and $3x + 4y + 5 = 0$ is (A) $(0, -1)$ (B) $(0, 0)$ (C) $(1, 1)$ (D) $(-1, 0)$

    ›Reveal solutionSolution

    This tests recognizing a right angle from perpendicular side-slopes, which immediately places the orthocentre at that right-angle vertex. Answer: $(0,-1)$.

    Concept and Intuition

    The orthocentre is the intersection of the triangle's three altitudes. In a right triangle, the two legs at the right-angle vertex are already mutually perpendicular, so each leg is itself the altitude from the opposite vertex — meaning both altitudes through the right angle already meet exactly at that vertex, so the orthocentre is that vertex.

    Step-by-Step Solution

    1. Find vertex $A$ = Line1 ($x+y+1=0$) $\cap$ Line2 ($x-y-1=0$): adding, $2x=0\Rightarrow x=0$, then $y=-1$. So $A=(0,-1)$.
    2. Find vertex $B$ = Line1 $\cap$ Line3 ($3x+4y+5=0$): from Line1, $y=-x-1$; substitute: $3x+4(-x-1)+5=0 \Rightarrow -x+1=0 \Rightarrow x=1, y=-2$. So $B=(1,-2)$.
    3. Find vertex $C$ = Line2 $\cap$ Line3: from Line2, $y=x-1$; substitute: $3x+4(x-1)+5=0 \Rightarrow 7x+1=0 \Rightarrow x=-1/7, y=-8/7$.
    4. Side $AB$ lies on Line1 (slope $-1$); side $AC$ lies on Line2 (slope $1$). Product of slopes $=-1\times1=-1$, so $AB\perp AC$, i.e. angle $A=90°$. …

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11. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.If all the normals drawn to the curve $y=\frac{1+3x^2}{3+x^2}$ at the points of intersection of $y=\frac{1+3x^2}{3+x^2}$ and $y=1$ pass through the point $(\alpha,\beta)$, then $3\alpha+2\beta=$ (A) $4$ (B) $2$ (C) $-2$ (D) $-4$

    ›Reveal solutionSolution

    Both normal lines at the two intersection points pass through the same fixed point $(0,2)$, so $3\alpha+2\beta=4$.

    Concept and Intuition

    Although the problem says "all normals," the curve $y=\frac{1+3x^2}{3+x^2}$ meets $y=1$ at exactly two points, so we just need the normals at both and find their common intersection.

    Step-by-Step Solution

    1. Set $\dfrac{1+3x^2}{3+x^2}=1\Rightarrow 1+3x^2=3+x^2\Rightarrow 2x^2=2\Rightarrow x=\pm1$, both giving $y=1$.
    2. Differentiate (quotient rule): $y'=\dfrac{6x(3+x^2)-(1+3x^2)(2x)}{(3+x^2)^2}=\dfrac{16x}{(3+x^2)^2}$.
    3. At $x=1$: $y'=\dfrac{16}{16}=1$; normal slope $=-1$. Normal at $(1,1)$: $y-1=-(x-1)\Rightarrow x+y=2$.
    4. At $x=-1$: $y'=\dfrac{-16}{16}=-1$; normal slope $=1$. Normal at $(-1,1)$: $y-1=1\cdot(x+1)\Rightarrow y-x=2$. …

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12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.The radical centre of the three circles $x^2 + y^2 - 1 = 0$, $x^2 + y^2 - 8x + 15 = 0$ and $x^2 + y^2 + 10y + 24 = 0$ is (A) $\left(2, \dfrac{-5}{2}\right)$ (B) $\left(2, \dfrac{5}{2}\right)$ (C) $\left(-2, \dfrac{5}{2}\right)$ (D) $\left(-2, \dfrac{-5}{2}\right)$

    ›Reveal solutionSolution

    The radical centre is found by intersecting any two of the three pairwise radical axes; it is $(2,-\frac52)$.

    Concept and Intuition

    For circles $S_1=0$, $S_2=0$ (each with unit coefficient of $x^2,y^2$), the radical axis is simply $S_1-S_2=0$, a linear equation. The radical centre is the common intersection point of all three pairwise radical axes.

    Step-by-Step Solution

    1. $S_1: x^2+y^2-1=0$, $S_2: x^2+y^2-8x+15=0$, $S_3: x^2+y^2+10y+24=0$.
    2. $S_1-S_2$: $(-1)-(-8x+15)=0\Rightarrow 8x-16=0\Rightarrow x=2$.
    3. $S_1-S_3$: $(-1)-(10y+24)=0\Rightarrow -10y-25=0\Rightarrow y=-\dfrac{5}{2}$.
    4. Radical centre: $(2,-\frac52)$. …

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