Q.Find the angle between the x-axis and the line joining the points $(3, -1)$ and $(4, -2)$.

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Let $L_1 \equiv x + 2y + 1 = 0, L_2 \equiv 2x + y - 3 = 0, L_3 \equiv ax + by + 1 = 0, a,b \in \mathbb{Z}$ represent the sides of an isosceles triangle. If $L_1 = 0$ is the base and $(5,1)$ is a point on $L_3 = 0$, then $a-b=$ (A) 13 (B) -9 (C) 12 (D) -7

   ›Reveal solutionSolution

   Isosceles-with-base-$L_1$ forces $L_3$ to make the same angle with $L_1$ as $L_2$ does; that plus "$(5,1)$ on $L_3$" pins down $a=2,b=-11$, giving $a-b=13$.

   Concept and Intuition

   In an isosceles triangle the two base angles (between the base and each of the equal sides) are equal. So if $L_1$ is the base and $L_2,L_3$ are the two slant sides, $L_1$ must be equally inclined to $L_2$ and to $L_3$ — this is the key geometric fact that lets us find the slope of $L_3$ without knowing the triangle's vertices.

   Step-by-Step Solution

   1. Slopes: $L_1:x+2y+1=0\Rightarrow m_1=-\tfrac12$. $L_2:2x+y-3=0\Rightarrow m_2=-2$.
   2. Angle between $L_1,L_2$: $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|=\left|\dfrac{-\tfrac12+2}{1+1}\right|=\dfrac{3/2}{2}=\dfrac34$.
   3. Let $m_3$ be the slope of $L_3$. Require $\left|\dfrac{m_1-m_3}{1+m_1m_3}\right|=\dfrac34$, i.e. $\dfrac{-\tfrac12-m_3}{1-\tfrac12 m_3}=\pm\dfrac34$.
   4. The "+" sign reproduces $m_3=-2=m_2$ (the trivial/parallel case, not a genuine third side). The "$-$" sign: $4(-\tfrac12-m_3)=-3(1-\tfrac12m_3)\Rightarrow-2-4m_3=-3+1.5m_3\Rightarrow1=5.5m_3\Rightarrow m_3=\dfrac{2}{11}$.
   5. $L_3:ax+by+1=0$ has slope $-a/b=\dfrac2{11}$, so $b=-\dfrac{11a}{2}$. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If the sum of the slopes of the lines given by $x^2 - 2cxy - 7y^2 = 0$ is four times their product, then the value of 'c' is (A) 2 (B) -2 (C) 1 (D) -1

   ›Reveal solutionSolution

   Using the standard slope-sum and slope-product formulas for a homogeneous pair of lines through the origin, the given condition reduces to a single linear equation in $c$, giving $c=2$.

   Concept and Intuition

   A homogeneous second-degree equation $ax^2+2hxy+by^2=0$ always represents a pair of straight lines through the origin. Dividing by $x^2$ and setting $m=y/x$ gives $bm^2+2hm+a=0$, whose roots $m_1,m_2$ are the slopes of the two lines. By Vieta's formulas, $m_1+m_2 = -2h/b$ and $m_1 m_2 = a/b$ — this lets us translate a condition on slopes directly into a condition on the coefficients.

   Step-by-Step Solution

   1. Given: $x^2 - 2cxy - 7y^2 = 0$. Here $a=1,\ 2h=-2c \Rightarrow h=-c,\ b=-7$.
   2. Sum of slopes: $m_1+m_2 = -\dfrac{2h}{b} = -\dfrac{2(-c)}{-7} = -\dfrac{2c}{7}$.
   3. Product of slopes: $m_1 m_2 = \dfrac{a}{b} = \dfrac{1}{-7} = -\dfrac17$. …

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3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If the slope of one of the lines is twice the slope of the other in the pair of straight lines $6x^2 + 2hxy + y^2 = 0$ then $|h| =$ (A) $\dfrac{-3\sqrt{3}}{2}$ (B) $\dfrac{3\sqrt{2}}{3}$ (C) $\dfrac{3\sqrt{3}}{2}$ (D) $\dfrac{3\sqrt{5}}{2}$

   ›Reveal solutionSolution

   Pair-of-lines slope-sum/product relations with the constraint "one slope is twice the other" gives $|h|=\frac{3\sqrt3}{2}$.

   Concept and Intuition

   A homogeneous equation $ax^2+2hxy+by^2=0$ represents two lines through the origin with slopes $m_1,m_2$ satisfying $m_1+m_2=-\dfrac{2h}{b}$ and $m_1m_2=\dfrac{a}{b}$ (dividing through by $x^2$ and solving the resulting quadratic in $m=y/x$). Given a ratio between the slopes, both relations combine to pin down $h$.

   Step-by-Step Solution

   1. Equation: $6x^2+2hxy+y^2=0$, so $a=6,\,b=1$. Dividing by $x^2$: $y/x=m$ satisfies $m^2+2hm+6=0$, giving $m_1+m_2=-2h$ and $m_1m_2=6$.
   2. Let $m_2=2m_1$. Then $m_1+2m_1=3m_1=-2h$, and $m_1(2m_1)=2m_1^2=6\Rightarrow m_1^2=3\Rightarrow m_1=\pm\sqrt3$.
   3. From $3m_1=-2h$: $h=-\dfrac{3m_1}{2}$, so $h^2=\dfrac{9m_1^2}{4}=\dfrac{9\cdot3}{4}=\dfrac{27}{4}$.
   4. $|h|=\sqrt{27/4}=\dfrac{3\sqrt3}{2}$.

   Common Mistakes …

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4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The angle between the tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 4x$ is (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   This tests the tangent-pair formula for a parabola from an external point; the angle comes out to $\pi/3$.

   Concept and Intuition

   Any tangent to $y^2=4ax$ has the form $y=mx+\dfrac{a}{m}$ for some slope $m$. From an external point, exactly two values of $m$ satisfy the condition that the line passes through that point — these give the two tangent slopes, and the angle between them is found from the standard formula for the angle between two lines.

   Step-by-Step Solution

   1. Here $4a=4\Rightarrow a=1$, so a tangent is $y=mx+\dfrac1m$.
   2. This tangent passes through $(1,4)$: $4=m(1)+\dfrac1m\Rightarrow 4m=m^2+1\Rightarrow m^2-4m+1=0$.
   3. So $m_1+m_2=4$ and $m_1m_2=1$.
   4. Angle between the two tangent lines: $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$.
   5. $(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2=16-4=12\Rightarrow|m_1-m_2|=2\sqrt3$.
   6. $1+m_1m_2=1+1=2$. …

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5. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The equation of the base of an equilateral triangle is $x+y=2$ and its opposite vertex is $(2,1)$. If $m_1, m_2$ are the slopes of the other two sides and the length of its side is a, then $|m_1-m_2|+a\sqrt{2}=$ (A) $8\sqrt{3}$ (B) $\dfrac{8}{\sqrt{3}}$ (C) $4\sqrt{\dfrac{2}{3}}$ (D) $8\sqrt{\dfrac{2}{3}}$

   ›Reveal solutionSolution

   Use the perpendicular height to get the side length $a$, and the fixed $60^\circ$ base angles to get the two slopes; combine. Answer: $8/\sqrt3$.

   Concept and Intuition

   In an equilateral triangle, the altitude from any vertex to the opposite side relates to the side length by $h=\dfrac{\sqrt3}{2}a$, and each base angle is exactly $60^\circ$. So the two "other sides" are simply the two lines through the apex making a $60^\circ$ angle with the base line.

   Step-by-Step Solution

   1. Height from $(2,1)$ to $x+y-2=0$: $h=\dfrac{|2+1-2|}{\sqrt{1^2+1^2}}=\dfrac{1}{\sqrt2}$.
   2. Since $h=\dfrac{\sqrt3}{2}a$: $a=\dfrac{2h}{\sqrt3}=\dfrac{2}{\sqrt2\,\sqrt3}=\dfrac{\sqrt2}{\sqrt3}=\sqrt{\dfrac23}$.
   3. So $a\sqrt2 = \sqrt{\dfrac23}\cdot\sqrt2=\sqrt{\dfrac43}=\dfrac{2}{\sqrt3}$.
   4. Base line slope $m=-1$. The two other sides pass through $(2,1)$ making $60^\circ$ with this base, so their slopes $m'$ satisfy $\tan60^\circ=\left|\dfrac{m'-(-1)}{1+(-1)m'}\right|=\left|\dfrac{m'+1}{1-m'}\right|=\sqrt3$.
   5. Case $(+)$: $m'+1=\sqrt3(1-m')\Rightarrow m'(1+\sqrt3)=\sqrt3-1\Rightarrow m'=\dfrac{\sqrt3-1}{\sqrt3+1}=2-\sqrt3$ (after rationalizing). …

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6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The triangle formed by the lines $2x^2+xy-6y^2=0$ and $x+y-1=0$ is (A) equilateral (B) isosceles (C) right angled (D) scalene

   ›Reveal solutionSolution

   Factor the homogeneous pair of lines, find all three vertices with the transversal line, then compare the three side lengths. Answer: scalene.

   Concept and Intuition

   $2x^2+xy-6y^2=0$ is a homogeneous second-degree equation representing a pair of straight lines through the origin. Factoring it, along with the given line $x+y-1=0$, gives three lines whose pairwise intersections are the triangle's vertices — from there, side lengths settle the triangle's type.

   Step-by-Step Solution

   1. Factor: $2x^2+xy-6y^2=(2x-3y)(x+2y)$ — check: $(2x-3y)(x+2y)=2x^2+4xy-3xy-6y^2=2x^2+xy-6y^2$. ✓. So the two lines are $L_1: 2x-3y=0$ and $L_2: x+2y=0$; the third line is $L_3: x+y-1=0$.
   2. Vertex $O$ (intersection of $L_1,L_2$): both pass through the origin, so $O=(0,0)$.
   3. Vertex $P$ (intersection of $L_1,L_3$): from $2x=3y$, $x=\tfrac{3y}{2}$; substitute into $x+y=1$: $\tfrac{3y}{2}+y=1\Rightarrow\tfrac{5y}{2}=1\Rightarrow y=\tfrac25,\ x=\tfrac35$. So $P=(3/5,2/5)$.
   4. Vertex $Q$ (intersection of $L_2,L_3$): $x=-2y$; substitute into $x+y=1$: $-2y+y=1\Rightarrow y=-1,\ x=2$. So $Q=(2,-1)$.
   5. Side lengths squared: $OP^2=(3/5)^2+(2/5)^2=9/25+4/25=13/25$; $OQ^2=2^2+1^2=5$; $PQ^2=(2-3/5)^2+(-1-2/5)^2=(7/5)^2+(-7/5)^2=98/25$. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\alpha$ is the angle made by the perpendicular drawn from origin to the line $12x - 5y + 13 = 0$ with the positive X-axis in anti-clockwise direction, then $\alpha =$ (A) $\mathrm{Tan}^{-1}\dfrac{5}{12}$ (B) $2\pi - \mathrm{Tan}^{-1}\dfrac{5}{12}$ (C) $\pi - \mathrm{Tan}^{-1}\dfrac{5}{12}$ (D) $\pi + \mathrm{Tan}^{-1}\dfrac{5}{12}$

   ›Reveal solutionSolution

   Convert the line to normal form $x\cos\alpha+y\sin\alpha=p$ ($p\ge0$) to read off $\alpha$ directly; the negative-cosine, positive-sine signs place $\alpha$ in Q2. Answer: $\pi-\mathrm{Tan}^{-1}\frac{5}{12}$.

   Concept and Intuition

   The perpendicular dropped from the origin onto a line makes some angle $\alpha$ with the positive $x$-axis; $\alpha$ and the perpendicular distance $p$ together define the line's normal form $x\cos\alpha+y\sin\alpha=p$, always written with $p\ge0$. Reading $\cos\alpha,\sin\alpha$ off this form tells us both the magnitude and the quadrant of $\alpha$ — we can't just take $\mathrm{Tan}^{-1}$ of the raw slope ratio, because that only returns a principal value in $(-\pi/2,\pi/2)$.

   Step-by-Step Solution

   1. Line: $12x-5y+13=0 \Rightarrow 12x-5y=-13$.
   2. Since $\sqrt{12^2+5^2}=13$, dividing by $13$ gives RHS $=-1$ (negative) — not the required normal form. Divide by $-13$ instead so RHS becomes $+1$:

   $-\frac{12}{13}x+\frac{5}{13}y = 1$

   3. So $\cos\alpha=-\dfrac{12}{13}$, $\sin\alpha=\dfrac{5}{13}$, $p=1$.
   4. $\cos\alpha<0,\ \sin\alpha>0 \Rightarrow \alpha$ is in the second quadrant ($90^\circ<\alpha<180^\circ$). …

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8. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If one of the lines given by the pair of lines $3x^2 - 2y^2 + axy = 0$ is making an angle $60°$ with x-axis then $a =$ (A) $\sqrt{3}$ (B) $\dfrac{1}{\sqrt{3}}$ (C) $3$ (D) $\dfrac{1}{3}$

   ›Reveal solutionSolution

   Converting the homogeneous pair-of-lines equation into a quadratic in slope $m$ and plugging in $m=\tan60°$ gives $a=\sqrt3$.

   Concept and Intuition

   A homogeneous second-degree equation $px^2+qxy+ry^2=0$ represents a pair of straight lines through the origin. Dividing by $x^2$ and writing $m=y/x$ turns it into a quadratic in $m$, whose two roots are the slopes of the two lines. Knowing one line's slope lets us solve for the unknown coefficient.

   Step-by-Step Solution

   1. Given: $3x^2 - 2y^2 + axy = 0$, i.e. $3x^2+axy-2y^2=0$.
   2. Divide by $x^2$ (with $m=y/x$): $3+am-2m^2=0$, i.e. $2m^2-am-3=0$.
   3. One line makes $60°$ with the $x$-axis, so $m=\tan60°=\sqrt3$ is a root. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.A line L passing through the point $P(-5,-4)$ cuts the lines $x-y-5=0$ and $x+3y+2=0$ respectively at Q and R such that $\frac{18}{PQ}+\frac{15}{PR}=2$, then slope of the line L is (A) $\pm 1$ (B) $\pm\frac{1}{\sqrt{3}}$ (C) $\pm\sqrt{3}$ (D) $\pm\frac{2}{\sqrt{3}}$

   ›Reveal solutionSolution

   Using a directional parametrization of the line through $P$, the given condition reduces to $\cos\theta=\frac12$, giving slope $\pm\sqrt3$.

   Concept and Intuition

   Parametrize points on line $L$ through $P(-5,-4)$ as $(x,y)=(-5+t\cos\theta,\ -4+t\sin\theta)$, where $t$ is the signed distance from $P$ and $\theta$ is the inclination. Substituting into each cutting line's equation gives the parameter value $t$ at the intersection point directly.

   Step-by-Step Solution

   1. Intersection with $x-y-5=0$: $(-5+t\cos\theta)-(-4+t\sin\theta)-5=0 \Rightarrow t(\cos\theta-\sin\theta)=6 \Rightarrow t_Q=\dfrac{6}{\cos\theta-\sin\theta}$, so $PQ=|t_Q|$.
   2. Intersection with $x+3y+2=0$: $(-5+t\cos\theta)+3(-4+t\sin\theta)+2=0 \Rightarrow t(\cos\theta+3\sin\theta)=15 \Rightarrow t_R=\dfrac{15}{\cos\theta+3\sin\theta}$, so $PR=|t_R|$.
   3. Then $\dfrac{18}{PQ}=3|\cos\theta-\sin\theta|$ and $\dfrac{15}{PR}=|\cos\theta+3\sin\theta|$. …

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10. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.Let P = (-1,0), Q = (0,0) and R = (3, 3\sqrt{3}) be three points. Then the equation of the bisector of the $\angle PQR$ is (A) $\sqrt{3}x + y = 0$ (B) $x + \dfrac{\sqrt{3}}{2}y = 0$ (C) $\dfrac{-\sqrt{3}}{2}x + y = 0$ (D) $x + \sqrt{3}y = 0$

    ›Reveal solutionSolution

    The angle bisector at vertex $Q$ runs along the sum of the unit vectors toward $P$ and $R$; this gives slope $-\sqrt3$ through the origin, i.e. $\sqrt3 x+y=0$.

    Concept and Intuition

    Just as with the earlier vector-bisector question, the bisector of the angle at a vertex of a triangle (or here, angle $PQR$) points along the sum of the unit vectors from that vertex toward the two other points defining the angle.

    Step-by-Step Solution

    1. $P=(-1,0)$, $Q=(0,0)$, $R=(3,3\sqrt3)$.
    2. $\overline{QP}=P-Q=(-1,0)$, magnitude $1$, so unit vector is $(-1,0)$ itself.
    3. $\overline{QR}=R-Q=(3,3\sqrt3)$, magnitude $\sqrt{9+27}=\sqrt{36}=6$, unit vector $=\left(\dfrac36,\dfrac{3\sqrt3}{6}\right)=\left(\dfrac12,\dfrac{\sqrt3}{2}\right)$.
    4. Bisector direction $=(-1,0)+\left(\dfrac12,\dfrac{\sqrt3}{2}\right)=\left(-\dfrac12,\dfrac{\sqrt3}{2}\right)$.
    5. Slope of bisector $=\dfrac{\sqrt3/2}{-1/2}=-\sqrt3$. …

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11. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.If the angle between the curves $y = e^{2(1+x)-4}$ and $x^2y=1$ at the point (1,1) is $\theta$, then $|\sin\theta|+|\cos\theta| =$ (A) 7/5 (B) 3/5 (C) 8/7 (D) 6/5

    ›Reveal solutionSolution

    Find the tangent slopes of both curves at $(1,1)$, use the angle-between-two-lines formula, and convert the resulting $\tan\theta=4/3$ into $|\sin\theta|+|\cos\theta|$.

    Concept and Intuition

    The angle between two curves at a common point is defined as the angle between their tangent lines there. Once we have both slopes, this reduces to a standard angle-between-lines calculation, followed by translating $\tan\theta$ into $\sin\theta,\cos\theta$ via a right triangle.

    Step-by-Step Solution

    1. First curve: $y=e^{2(1+x)-4}=e^{2x-2}$. At $x=1$: $y=e^0=1$ ✓ point $(1,1)$. $\dfrac{dy}{dx}=2e^{2x-2}$, at $x=1$: $m_1=2$.
    2. Second curve: $x^2y=1 \Rightarrow y=1/x^2$. $\dfrac{dy}{dx}=-2/x^3$, at $x=1$: $m_2=-2$.
    3. $\tan\theta = \left|\dfrac{m_1-m_2}{1+m_1m_2}\right| = \left|\dfrac{2-(-2)}{1+2(-2)}\right| = \left|\dfrac{4}{-3}\right| = \dfrac43$. …

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12. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.The equal sides of an isosceles triangle are given by equations $7x - y + 3 = 0$ and $x + y - 3 = 0$. If the slope m of the third side is an integer, then m = (A) $-3$ (B) $3$ (C) $4$ (D) $-1$

    ›Reveal solutionSolution

    The base of an isosceles triangle makes equal angles with both equal sides; equating the two tangent expressions and solving gives $m=\frac13$ or $m=-3$ — the integer one is $-3$.

    Concept and Intuition

    Base angles of an isosceles triangle are equal. The base angle at a vertex is the angle between that equal side and the base (third side). So if the equal sides have slopes $m_1=7$, $m_2=-1$, and the base has slope $m$, the acute angle between (line$_1$, base) must equal the acute angle between (line$_2$, base). This is captured by setting the absolute tangent-of-angle-between-lines formula equal for both pairs.

    Step-by-Step Solution

    1. Slopes of equal sides: from $7x-y+3=0$, $m_1=7$; from $x+y-3=0$, $m_2=-1$.
    2. Equal-angle condition: $\left|\dfrac{m_1-m}{1+m_1m}\right|=\left|\dfrac{m_2-m}{1+m_2m}\right|$, i.e. $\left|\dfrac{7-m}{1+7m}\right|=\left|\dfrac{-1-m}{1-m}\right|$.
    3. One sign case gives $8m^2+8=0$ (no real root) — rejected.
    4. The other sign case: $(7-m)(1-m)=(1+m)(1+7m)$. LHS $=7-8m+m^2$; RHS $=1+8m+7m^2$. …

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