Q.Show that the area of the triangle formed by the lines $y = m_1 x + c_1$, $y = m_2 x + c_2$ and $x = 0$ is $\dfrac{(c_1 - c_2)^2}{2\,|m_1 - m_2|}$.

Showing the 12 most recent of 39 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Suppose $A$ is the point of intersection of the pair of lines $L \equiv 2x^2 - 3xy - 2y^2 + 18x - y + 28 = 0$. If $N, M$ are the feet of the perpendiculars from a point $P(3,4)$ on to the pair of lines $L = 0$, then area (in sq units) of the quadrilateral $APNM$ is (A) $\dfrac{23}{7}$ (B) $\dfrac{28}{5}$ (C) $15$ (D) $19$

   ›Reveal solutionSolution

   When the pair of lines through $A$ are perpendicular, the quadrilateral formed by $A$, $P$, and the two feet of perpendiculars is a rectangle whose area is simply the product of $P$'s distances to the two lines; the answer is $28/5$.

   Concept and Intuition

   Factoring the homogeneous part $2x^2-3xy-2y^2=(2x+y)(x-2y)$ suggests the full expression splits into two linear factors. Once we know the two lines and that they are perpendicular, dropping perpendiculars from any point $P$ onto them creates a rectangle with the vertex $A$ — and a rectangle's area is just the product of its two adjacent side lengths, which are exactly the two perpendicular distances from $P$.

   Step-by-Step Solution

   1. Try $(2x+y+p)(x-2y+q)=2x^2-3xy-2y^2+(2q+p)x+(q-2p)y+pq$. Matching with $18x-y+28$: $2q+p=18$, $q-2p=-1$, $pq=28$. Solving: $p=4,q=7$ (checks: $pq=28$ ✓).
   2. So $L_1:2x+y+4=0$ and $L_2:x-2y+7=0$. Their slopes are $-2$ and $1/2$ — product $=-1$, so $L_1\perp L_2$.
   3. Solving $L_1,L_2$ simultaneously: $y=-2x-4$; substitute into $x-2y+7=0$: $x+4x+8+7=0\Rightarrow x=-3,y=2$. So $A=(-3,2)$. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If 'O' is the origin and 'H' is the orthocenter of a triangle formed by the lines $x+y=1$, $6x^2-13xy+6y^2=0$, then $OH=$ (A) $12\sqrt{2}$ (B) $\dfrac{12\sqrt{2}}{25}$ (C) $\dfrac{24\sqrt{2}}{25}$ (D) $24\sqrt{2}$

   ›Reveal solutionSolution

   Factor the homogeneous pair into two lines through the origin, find the triangle's third side from $x+y=1$, and intersect two altitudes to get $H=(\frac{12}{25},\frac{12}{25})$, so $OH=\frac{12\sqrt2}{25}$.

   Concept and Intuition

   A homogeneous second-degree equation $ax^2+2hxy+by^2=0$ represents a pair of straight lines through the origin. Combined with a third line, it forms a triangle with one vertex at $O$. The orthocenter is the intersection of any two altitudes — using $O$ itself as a vertex often simplifies one altitude to a very clean line.

   Step-by-Step Solution

   1. Factor $6x^2-13xy+6y^2=(3x-2y)(2x-3y)$, giving lines $L_1: 3x-2y=0$ and $L_2: 2x-3y=0$, both through $O(0,0)$.
   2. Intersect $L_1$ with $x+y=1$: substitute $y=\frac32x$: $x+\frac32x=1\Rightarrow x=\frac25,\ y=\frac35$. So $A=\left(\frac25,\frac35\right)$.
   3. Intersect $L_2$ with $x+y=1$: substitute $y=\frac23x$: $x+\frac23x=1\Rightarrow x=\frac35,\ y=\frac25$. So $B=\left(\frac35,\frac25\right)$.
   4. Side $AB$ is exactly the line $x+y=1$ (slope $-1$), so the altitude from $O$ perpendicular to $AB$ has slope $1$: $y=x$.
   5. Side $OB$ lies along $L_2$ (slope $2/3$), so the altitude from $A$ perpendicular to $OB$ has slope $-3/2$: $y-\frac35=-\frac32\left(x-\frac25\right)$. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $A,B$ are the feet of the perpendiculars drawn from $P(3,1)$ on to the pair of lines $2x^2+3xy-2y^2+3x+y+1=0$ and $Q$ is the point of intersection of the pair of lines, then the area of the quadrilateral $PAQB$ is (A) $25$ (B) $\dfrac{36}{5}$ (C) $\dfrac{25}{6}$ (D) $\dfrac{49}{8}$

   ›Reveal solutionSolution

   The pair of lines splits into two perpendicular lines through $Q$; PAQB is then two right triangles sharing hypotenuse $PQ$, giving area $\frac{36}{5}$.

   Concept and Intuition

   A homogeneous-plus-linear second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines. Once factored into two linear factors, $Q$ (their intersection) and $P$ together with the feet of perpendiculars $A,B$ from $P$ onto each line form a quadrilateral with right angles at $A$ and $B$ — and if the two lines happen to be perpendicular, also at $Q$, making $PAQB$ a "kite-like" figure built from two right triangles on the common hypotenuse $PQ$.

   Step-by-Step Solution

   1. Factor the quadratic part: $2x^2+3xy-2y^2=(2x-y)(x+2y)$.
   2. Write the full pair as $(2x-y+a)(x+2y+b)=0$ and match coefficients with $2x^2+3xy-2y^2+3x+y+1$: solving $2b+a=3,\ 2a-b=1,\ ab=1$ gives $a=1,b=1$. So the lines are $L_1: 2x-y+1=0$ and $L_2: x+2y+1=0$.
   3. Slopes are $2$ and $-\frac12$ — their product is $-1$, so $L_1\perp L_2$, meaning $\angle AQB=90^\circ$.
   4. Solve $L_1,L_2$ simultaneously: $Q=\left(-\frac35,-\frac15\right)$.
   5. $PA$ = distance from $P(3,1)$ to $L_1$: $\dfrac{|2(3)-1+1|}{\sqrt5}=\dfrac{6}{\sqrt5}$. $PB$ = distance from $P$ to $L_2$: $\dfrac{|3+2+1|}{\sqrt5}=\dfrac{6}{\sqrt5}$ (equal, by the near-symmetry of the coefficients).
   6. $PQ^2 = \left(3+\frac35\right)^2+\left(1+\frac15\right)^2=\left(\frac{18}5\right)^2+\left(\frac65\right)^2=\frac{324+36}{25}=\frac{360}{25}$. …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Area of triangle formed by the lines $2x^2 + 5xy + 3y^2 = 0$ and $3x + 4y - 1 = 0$ is (A) $1/2$ (B) $5/2$ (C) $12$ (D) $16$

   ›Reveal solutionSolution

   This tests factoring a homogeneous pair of lines through the origin, then computing the area of the triangle they form with a third line. Answer: $1/2$.

   Concept and Intuition

   $2x^2+5xy+3y^2=0$ is a homogeneous second-degree equation, so it always represents a pair of straight lines through the origin. Factoring it gives the two individual lines explicitly; then, together with the given third line, we have three lines whose pairwise intersections are the triangle's three vertices, and the standard vertex-coordinate area formula finishes it.

   Step-by-Step Solution

   1. Factor: $2x^2+5xy+3y^2=(2x+3y)(x+y)$ — check: $(2x+3y)(x+y)=2x^2+2xy+3xy+3y^2=2x^2+5xy+3y^2$ ✓.
   2. The two lines are $2x+3y=0$ and $x+y=0$, both through the origin, so their intersection is $(0,0)$.
   3. Intersect $2x+3y=0$ with $3x+4y-1=0$: from the first, $x=-\tfrac32y$; substitute: $3(-\tfrac32y)+4y-1=0 \Rightarrow -\tfrac12y=1 \Rightarrow y=-2, x=3$. Vertex $(3,-2)$. …

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5. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.In $\triangle ABC$, coordinates of $A$ are $(1, 2)$. If the equations of the medians through $B$ and $C$ are $x + y = 5$ and $x = 4$ respectively, then the area of $\triangle ABC$ (in sq. units) is (A) $12$ (B) $9$ (C) $6$ (D) $4$

   ›Reveal solutionSolution

   This tests reconstructing a triangle's vertices from its centroid and two given median lines, then finding the area. Vertex $A$ is given; $B$ and $C$ are pinned down using the median-line equations plus the centroid formula, and the shoelace formula gives area $=9$.

   Concept and Intuition

   Every median passes through the centroid $G$, which divides each median in ratio $2:1$ and is also the average of all three vertices: $G=\dfrac{A+B+C}{3}$. Here we're told the lines containing two medians (through $B$ and through $C$). Since $B$ lies on its own median, and $C$ lies on its own median, we get one linear constraint on each. Combined with the vector centroid identity (two scalar equations, for $x$ and $y$), we have exactly enough equations to pin down $B$ and $C$ uniquely.

   Step-by-Step Solution

   1. The centroid $G$ is the common point of the two given median lines:

   $x+y=5,\qquad x=4 \;\Rightarrow\; y=1.$

   So $G=(4,1)$.

   2. Let $B=(b_1,b_2)$. Since $B$ lies on the median through $B$ (line $x+y=5$):

   $b_1+b_2=5.$

   3. Let $C=(c_1,c_2)$. Since $C$ lies on the median through $C$ (line $x=4$):

   $c_1=4.$

   4. Centroid formula with $A=(1,2)$:

   $\frac{1+b_1+c_1}{3}=4 \;\Rightarrow\; b_1+c_1=11 \;\Rightarrow\; b_1=11-4=7.$

   $\frac{2+b_2+c_2}{3}=1 \;\Rightarrow\; b_2+c_2=1.$

   5. From step 2: $b_2=5-b_1=5-7=-2$. So $B=(7,-2)$.
   6. Then $c_2=1-b_2=1-(-2)=3$. So $C=(4,3)$. …

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6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The area (in square units) of the triangle formed by the X-axis, the tangent and the normal drawn at (1,1) to the curve $x^3+y^3=2xy$ is (A) $\dfrac{1}{2}$ (B) $1$ (C) $2$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   Implicit differentiation gives the tangent/normal slopes at $(1,1)$ on the folium-like curve $x^3+y^3=2xy$; both lines' X-intercepts, together with $(1,1)$, form a triangle of area $1$.

   Concept and Intuition

   The tangent and normal at a point are perpendicular lines through that point. Their X-intercepts, plus the point of tangency itself, form a right triangle whose legs are easy to read off once both lines are known — no need for the general area-of-triangle-from-3-points formula.

   Step-by-Step Solution

   1. Differentiate $x^3+y^3=2xy$ implicitly w.r.t. $x$:

   $3x^2+3y^2\frac{dy}{dx}=2y+2x\frac{dy}{dx} \implies \frac{dy}{dx}\left(3y^2-2x\right)=2y-3x^2 \implies \frac{dy}{dx}=\frac{2y-3x^2}{3y^2-2x}.$

   2. At $(1,1)$: $\dfrac{dy}{dx}=\dfrac{2(1)-3(1)}{3(1)-2(1)}=\dfrac{-1}{1}=-1$.
   3. Tangent (slope $-1$) through $(1,1)$: $y-1=-1(x-1)\Rightarrow y=-x+2$, i.e. $x+y=2$. Setting $y=0$: X-intercept at $(2,0)$.
   4. Normal (slope $=-1/(-1)=1$, since normal $\perp$ tangent) through $(1,1)$: $y-1=1\cdot(x-1)\Rightarrow y=x$. Setting $y=0$: X-intercept at $(0,0)$.
   5. The three vertices of the triangle are $(1,1)$ (the point of tangency), $(2,0)$ (tangent's foot), and $(0,0)$ (normal's foot) — the latter two both lie on the X-axis. …

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7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If $\left(\dfrac{2}{3},0\right)$ is the centroid of the triangle formed by the lines $4x^2-y^2=0$ and $lx+my+n=0$, then $l+m+n=$ (A) 1 (B) 0 (C) -1 (D) 2

   ›Reveal solutionSolution

   The pair of lines through the origin plus a third line form a triangle; matching the given centroid forces the third line to be vertical, $x=1$. Answer: $l+m+n=0$.

   Concept and Intuition

   $4x^2-y^2=0$ factors into two lines through the origin: $y=2x$ and $y=-2x$. Together with $lx+my+n=0$, these three lines form a triangle with one vertex at the origin. The centroid's coordinates constrain $l,m,n$ (up to an overall scalar, since a line's equation is defined only up to a nonzero multiple).

   Step-by-Step Solution

   1. Factor $4x^2-y^2=(2x-y)(2x+y)$, giving lines $y=2x$ and $y=-2x$, meeting at the origin $O=(0,0)$.
   2. Intersection with $y=2x$: substituting into $lx+my+n=0$ gives $x(l+2m)=-n\Rightarrow x=\dfrac{-n}{l+2m}$, $y=2x=\dfrac{-2n}{l+2m}$.
   3. Intersection with $y=-2x$: $x(l-2m)=-n\Rightarrow x=\dfrac{-n}{l-2m}$, $y=-2x=\dfrac{2n}{l-2m}$.
   4. Centroid's $y$-coordinate $=\dfrac{0+y_1+y_2}{3}=0$, so $y_1+y_2=0$: $\dfrac{-2n}{l+2m}+\dfrac{2n}{l-2m}=0\Rightarrow 2n\left[\dfrac{1}{l-2m}-\dfrac{1}{l+2m}\right]=0\Rightarrow \dfrac{8mn}{l^2-4m^2}=0$. Since $n\ne0$ (else the line passes through the origin, degenerating the triangle), we need $m=0$.
   5. With $m=0$, the third line is $lx+n=0$, i.e. $x=-n/l=k$ (a vertical line). It meets $y=2x$ at $(k,2k)$ and $y=-2x$ at $(k,-2k)$. …

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8. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If the locus of a point which is equidistant from the coordinate axes forms a triangle with the line y = 3, then the area of the triangle is (A) 18 (B) 9 (C) 6 (D) 3

   ›Reveal solutionSolution

   The locus $|x|=|y|$ is the pair of lines $y=\pm x$; together with $y=3$ they bound a triangle of area 9.

   Concept and Intuition

   Equidistance from the two coordinate axes means the perpendicular distances $|x|$ and $|y|$ are equal, which is satisfied by the two diagonal lines $y=x$ and $y=-x$ — together they look like an "X" through the origin, and cutting across with a horizontal line produces an isosceles triangle.

   Step-by-Step Solution

   1. Locus equidistant from the axes: $|x|=|y| \Rightarrow y=x$ or $y=-x$.
   2. Intersection of $y=x$ with $y=3$: point $(3,3)$.
   3. Intersection of $y=-x$ with $y=3$: point $(-3,3)$.
   4. Intersection of $y=x$ and $y=-x$: the origin $(0,0)$.
   5. These three points $(0,0), (3,3), (-3,3)$ form the triangle. Base $=$ distance between $(3,3)$ and $(-3,3)=6$; height $=$ perpendicular distance from $(0,0)$ to the line $y=3$, which is $3$. …

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9. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If the tangents of the parabola $y^2 = 8x$ passing through the point P(1, 3) touches the parabola at A and B, then the area (in sq. units) of $\triangle ABC$ is (A) 1 (B) $\dfrac{3}{4}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Parametrize the two points of tangency using the tangent-parameter equation from the external point, then compute the triangle area directly from the coordinates. Answer: $\dfrac14$ sq. units.

   Concept and Intuition

   From an external point, two tangents touch a parabola $y^2=4ax$ at points that can be parametrized as $(at^2,2at)$, where the parameters $t_1,t_2$ satisfy a quadratic obtained by requiring the tangent line $ty=x+at^2$ to pass through the external point. Once we know $t_1+t_2$ and $t_1t_2$, we can get the exact area of the triangle formed by the external point and the two tangency points using coordinate geometry — without needing $t_1,t_2$ individually.

   Step-by-Step Solution

   1. Parabola $y^2=8x=4(2)x \Rightarrow a=2$.
   2. Tangent at parameter $t$: $ty=x+2t^2$. For it to pass through $P(1,3)$: $3t=1+2t^2 \Rightarrow 2t^2-3t+1=0$.
   3. So $t_1+t_2=\dfrac32$, $t_1t_2=\dfrac12$.
   4. Tangency points: $A=(2t_1^2,4t_1)$, $B=(2t_2^2,4t_2)$; $P=(1,3)$.
   5. Area of $\triangle PAB = \dfrac12\big|(x_A-x_P)(y_B-y_P)-(x_B-x_P)(y_A-y_P)\big|$.
   6. Expanding: $(2t_1^2-1)(4t_2-3)-(2t_2^2-1)(4t_1-3) = (t_1-t_2)\big[8t_1t_2-6(t_1+t_2)+4\big]$. …

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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.Tangents are drawn at three points P($t_1$), Q($t_2$), R($t_3$) on the parabola $y^2 = x$. Let these tangents intersect each other at the points L, M, N. If $t_1 = 2, t_2 = -4, t_3 = 6$, then the area of the triangle LMN is (A) $24$ (B) $18.5$ (C) $7.5$ (D) $12$

    ›Reveal solutionSolution

    This tests the known result that tangents at three points of a parabola meet pairwise at points given by a simple parameter formula, then computing a triangle's area. Answer: area $=7.5$ sq. units.

    Concept and Intuition

    For the parabola $y^2=4ax$, parametrize points as $(at^2,2at)$. The tangent at parameter $t$ is $ty=x+at^2$. Solving the tangents at $t_i$ and $t_j$ simultaneously shows they always intersect at $\big(a\,t_it_j,\ a(t_i+t_j)\big)$ — a clean, memorable formula that avoids re-deriving each tangent pair from scratch.

    Step-by-Step Solution

    1. $y^2=x=4ax$ gives $a=1/4$.
    2. Intersection of tangents at $t_i,t_j$: $\left(\dfrac{t_it_j}{4},\dfrac{t_i+t_j}{4}\right)$.
    3. For $(t_1,t_2)=(2,-4)$: $t_1t_2=-8,\ t_1+t_2=-2$ → point $L=(-2,-0.5)$.
    4. For $(t_2,t_3)=(-4,6)$: $t_2t_3=-24,\ t_2+t_3=2$ → point $M=(-6,0.5)$.
    5. For $(t_1,t_3)=(2,6)$: $t_1t_3=12,\ t_1+t_3=8$ → point $N=(3,2)$. …

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11. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.The area (in sq. units) of the triangle formed by the tangent and normal to the ellipse $9x^2 + 4y^2 = 72$ at the point (2, 3) with the X-axis is (A) $\dfrac{25}{2}$ (B) $\dfrac{39}{4}$ (C) $\dfrac{35}{4}$ (D) $\dfrac{45}{4}$

    ›Reveal solutionSolution

    This tests writing the tangent and normal to an ellipse at a point, finding where each meets the x-axis, and computing the enclosed triangle's area. Answer: $\dfrac{39}{4}$.

    Concept and Intuition

    At any point on an ellipse $\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$, the tangent and normal are perpendicular lines through that point with standard forms. Since both lines pass through the same point on the curve, the triangle formed with the x-axis has that point as its apex and the two x-axis intercepts as its base — so the height is simply the point's $y$-coordinate.

    Step-by-Step Solution

    1. Divide by 72: $\dfrac{x^2}{8}+\dfrac{y^2}{18}=1$, so $A^2=8,\,B^2=18$. Check $(2,3)$: $\tfrac{4}{8}+\tfrac{9}{18}=0.5+0.5=1$ ✓.
    2. Tangent at $(x_1,y_1)$: $\dfrac{xx_1}{A^2}+\dfrac{yy_1}{B^2}=1$. At $(2,3)$: $\dfrac{2x}{8}+\dfrac{3y}{18}=1\Rightarrow \dfrac{x}{4}+\dfrac{y}{6}=1$. Set $y=0$: $x=4$, giving tangent's x-intercept $T=(4,0)$.
    3. Normal at $(x_1,y_1)$: $\dfrac{A^2x}{x_1}-\dfrac{B^2y}{y_1}=A^2-B^2$. At $(2,3)$: $\dfrac{8x}{2}-\dfrac{18y}{3}=8-18\Rightarrow 4x-6y=-10$. Set $y=0$: $4x=-10\Rightarrow x=-2.5$, giving normal's x-intercept $N=(-2.5,0)$. …

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12. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If the tangent of the curve $4y^3 = 3ax^2 + x^3$ drawn at the point $(a, a)$ forms a triangle of area $\dfrac{25}{24}$ sq.units with the coordinate axes then $a =$ (A) $\pm 10$ (B) $\pm 5$ (C) $\pm 6$ (D) $\pm 3$

    ›Reveal solutionSolution

    Finding the tangent to $4y^3=3ax^2+x^3$ at $(a,a)$ and using the area of the triangle it cuts with the axes gives $a=\pm5$.

    Concept and Intuition

    The slope of the tangent at a point on an implicit curve comes from implicit differentiation. Once we know the tangent line, its intercepts with the coordinate axes form a right triangle whose area we can compute directly, and set equal to the given value to solve for the unknown parameter.

    Step-by-Step Solution

    1. Verify $(a,a)$ lies on the curve: $4a^3 \stackrel{?}{=} 3a\cdot a^2+a^3=3a^3+a^3=4a^3$. ✓ (true for any $a$).
    2. Differentiate $4y^3=3ax^2+x^3$ implicitly: $12y^2\,y' = 6ax+3x^2$.
    3. At $(a,a)$: $12a^2\,y' = 6a\cdot a+3a^2 = 9a^2 \Rightarrow y' = \dfrac{9a^2}{12a^2}=\dfrac34$ (for $a\neq0$).
    4. Tangent line: $y-a=\dfrac34(x-a) \Rightarrow y=\dfrac34x - \dfrac34a+a = \dfrac34x+\dfrac{a}{4}$.
    5. $x$-intercept (set $y=0$): $0=\dfrac34x+\dfrac a4 \Rightarrow x=-\dfrac a3$.
    6. $y$-intercept (set $x=0$): $y=\dfrac a4$. …

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