Q.If $l_1, m_1, n_1$; $l_2, m_2, n_2$; $l_3, m_3, n_3$ are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to $l_1 + l_2 + l_3$, $m_1 + m_2 + m_3$, $n_1 + n_2 + n_3$ makes equal angles with them.

Showing the 12 most recent of 23 on this concept.

1. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If the vectors $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}, \vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$ and $\vec{c} = \lambda\hat{i} + \hat{j} + \mu\hat{k}$ are mutually orthogonal then $(\lambda, \mu)$ is equal to (A) $(-3, 2)$ (B) $(2, -3)$ (C) $(-2, 3)$ (D) $(3, -2)$

   ›Reveal solutionSolution

   Setting the two dot products with $\vec c$ to zero gives a linear system solved by $(\lambda,\mu)=(-3,2)$.

   Concept and Intuition

   Three vectors are mutually orthogonal when every pair's dot product is zero. Since $\vec a\cdot\vec b$ is already zero (given), we only need $\vec a\cdot\vec c=0$ and $\vec b\cdot\vec c=0$ to pin down $\lambda,\mu$.

   Step-by-Step Solution

   1. Check: $\vec a\cdot\vec b = (1)(2)+(-1)(4)+(2)(1) = 2-4+2=0$ ✓ (consistent with mutual orthogonality).
   2. $\vec a\cdot\vec c = \lambda(1)+1(-1)+\mu(2) = \lambda-1+2\mu=0 \Rightarrow \lambda+2\mu=1$.
   3. $\vec b\cdot\vec c = \lambda(2)+1(4)+\mu(1) = 2\lambda+4+\mu=0 \Rightarrow 2\lambda+\mu=-4$. …

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2. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If $ad \neq 0$ and two of the lines represented by $ax^3+3bx^2y+3cxy^2+dy^3=0$ are perpendicular, then (A) $a^2+ac+bd+d^2=0$ (B) $a^2+3ac+3bd+d^2=0$ (C) $a^2-3ac-3bd+d^2=0$ (D) $a^2+3ac-3bd+d^2=0$

   ›Reveal solutionSolution

   Converting the homogeneous cubic into a cubic in the slope $m=y/x$ and applying Vieta's formulas with the perpendicularity condition $m_1m_2=-1$ gives the identity $a^2+3ac+3bd+d^2=0$.

   Concept and Intuition

   A homogeneous cubic $ax^3+3bx^2y+3cxy^2+dy^3=0$ represents three concurrent (through-origin) lines. Dividing through by $x^3$ turns it into a cubic equation in $m=y/x$, whose three roots are exactly the three lines' slopes — so Vieta's relations connect the coefficients to the slopes, and any geometric condition on the slopes (like perpendicularity) becomes an algebraic condition on $a,b,c,d$.

   Step-by-Step Solution

   1. Divide by $x^3$: $d m^3+3cm^2+3bm+a=0$ where $m=y/x$.
   2. Vieta: $m_1+m_2+m_3=-\dfrac{3c}{d}$, $m_1m_2+m_2m_3+m_3m_1=\dfrac{3b}{d}$, $m_1m_2m_3=-\dfrac{a}{d}$.
   3. Let the perpendicular pair be $m_1,m_2$, so $m_1m_2=-1$. From the product relation: $-1\cdot m_3=-\dfrac ad\Rightarrow m_3=\dfrac ad$.
   4. From the sum: $m_1+m_2=-\dfrac{3c}d-m_3=-\dfrac{3c+a}d$.
   5. From the pairwise sum: $-1+m_3(m_1+m_2)=\dfrac{3b}d\Rightarrow m_3(m_1+m_2)=\dfrac{3b+d}d$. …

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3. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If one of the lines given by the equation $2x^2 + axy + 3y^2 = 0$ coincide with one of those given by the equation $2x^2 + bxy - 3y^2 = 0$, while the other two lines are perpendicular to each other, then the values of $a$ and $b$ are ________ (A) $a=-5$ & $b=1$ (B) $a=-4$ & $b=-1$ (C) $a=4$ & $b=1$ (D) $a=-5$ & $b=-1$

   ›Reveal solutionSolution

   Converting each homogeneous pair of lines into a quadratic in slope $t=y/x$, using a shared root for the coinciding line and the perpendicularity condition on the remaining two lines pins down $a$ and $b$.

   Concept and Intuition

   A homogeneous equation $Ax^2+Bxy+Cy^2=0$ represents two lines through the origin with slopes that are roots of $Ct^2+Bt+A=0$ (dividing by $x^2$ and setting $t=y/x$). Sharing one line between two such pairs means sharing one root; the "other two lines perpendicular" condition means the product of the two other slopes is $-1$.

   Step-by-Step Solution

   1. From $2x^2+axy+3y^2=0$: dividing by $x^2$, $3t^2+at+2=0$, with roots $t_1,t_2$ satisfying $t_1+t_2=-\dfrac a3,\ t_1t_2=\dfrac23$.
   2. From $2x^2+bxy-3y^2=0$: $-3t^2+bt+2=0 \Rightarrow 3t^2-bt-2=0$, roots $s_1,s_2$ with $s_1+s_2=\dfrac b3,\ s_1s_2=-\dfrac23$.
   3. Let the shared (coincident) slope be $t_1=s_1=t$. The other two lines have slopes $t_2$ and $s_2$, and perpendicularity gives $t_2 s_2=-1$.
   4. From step 1: $t_2=\dfrac{2}{3t}$. From step 2: $s_2=\dfrac{-2}{3t}$.
   5. Perpendicularity: $\dfrac{2}{3t}\cdot\dfrac{-2}{3t}=-1 \Rightarrow \dfrac{-4}{9t^2}=-1 \Rightarrow t^2=\dfrac49 \Rightarrow t=\pm\dfrac23$. …

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4. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.Suppose the pairs of straight lines $2x^2 + axy + 3y^2 = 0$ and $2x^2 + bxy - 3y^2 = 0$ are such that they have one common line with the other two remaining perpendicular. Then the values of a and b respectively are (A) $-5, 1$ (B) $5, -1$ (C) $5, 1$ (D) $5, \dfrac{1}{5}$

   ›Reveal solutionSolution

   Use the common root of the two pairs of lines through the origin, then the perpendicularity condition on the remaining two lines, to pin down $a$ and $b$.

   Concept and Intuition

   Each homogeneous equation $Ax^2+Bxy+Cy^2=0$ represents a pair of lines $y=m_1x$, $y=m_2x$ through the origin, where $m_1,m_2$ are roots of $Cm^2+Bm+A=0$ (dividing by $x^2$ and substituting $y=mx$). "One common line" means one root is shared between the two quadratics; "remaining two perpendicular" means the other two roots multiply to $-1$.

   Step-by-Step Solution

   1. For $2x^2+axy+3y^2=0$, put $y=mx$: $3m^2+am+2=0$, roots $m,m_2$ with $m+m_2=-a/3$, $mm_2=2/3$.
   2. For $2x^2+bxy-3y^2=0$, put $y=mx$: $-3m^2+bm+2=0$, roots $m,m_4$ with $m+m_4=b/3$, $mm_4=-2/3$.
   3. Common root is $m$ (same in both). From step 1: $m_2 = \dfrac{2}{3m}$. From step 2: $m_4=-\dfrac{2}{3m}$.
   4. Perpendicularity of the remaining lines: $m_2 m_4=-1 \Rightarrow \dfrac{2}{3m}\cdot\left(-\dfrac{2}{3m}\right)=-1 \Rightarrow \dfrac{4}{9m^2}=1 \Rightarrow m^2=\dfrac{4}{9}$, so $m=\pm\dfrac23$. …

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5. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.Two of the lines represented by the equation $ay^4 + bxy^3 + cx^2y^2 + dx^3y + ex^4 = 0$ will be perpendicular, then ________ (A) $(b+d)(ad+be) + (e-a)^2(a+c+e) = 0$ (B) $(b+d)(ad+be) + (e+a)^2(a+c+e) = 0$ (C) $(b-d)(ad-be) + (e-a)^2(a+c+e) = 0$ (D) $(b-d)(ad-be) + (e+a)^2(a+c+e) = 0$

   ›Reveal solutionSolution

   This is the standard textbook condition for two lines among the four represented by the quartic to be perpendicular: $(b+d)(ad+be)+(e-a)^2(a+c+e)=0$.

   Concept and Intuition

   Dividing the homogeneous quartic by $x^4$ turns it into a polynomial in $m=y/x$ whose four roots are the slopes of the four lines. If two of these slopes multiply to $-1$ (perpendicular condition), we can factor the quartic as a product of two quadratics — one having that perpendicular pair as roots (with product $-1$) — and match coefficients using Vieta's relations to derive the required condition purely in terms of $a,b,c,d,e$.

   Step-by-Step Solution

   1. Let $m=y/x$: $am^4+bm^3+cm^2+dm+e=0$, roots $m_1,m_2,m_3,m_4$.
   2. Suppose $m_1m_2=-1$ (perpendicular pair). Factor the monic quartic (dividing by $a$) as $(m^2-s_1m-1)(m^2-s_2m+p_2)$ where $s_1=m_1+m_2$, $s_2=m_3+m_4$, $p_2=m_3m_4$. …

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6. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.If the bisectors of the pair of lines $x^2 - 2mxy - y^2 = 0$ is represented by $x^2 - 2nxy - y^2 = 0$, then _______ (A) $mn + 1 = 0$ (B) $mn - 1 = 0$ (C) $m + n = 0$ (D) $m - n = 0$

   ›Reveal solutionSolution

   Compute the bisector of the given pair using the standard formula, then match coefficients with the stated bisector pair to relate $m$ and $n$.

   Concept and Intuition

   Since $x^2-2mxy-y^2=0$ already has coefficient of $x^2$ equal to $1$ and of $y^2$ equal to $-1$ (so $a+b=0$), the two lines it represents are perpendicular to each other. Its bisector pair is found with the same formula as before, and comparing that result to the given bisector pair $x^2-2nxy-y^2=0$ (same "shape") lets us equate coefficients of $xy$ directly.

   Step-by-Step Solution

   1. For $x^2-2mxy-y^2=0$: $a=1$, $b=-1$, $2h=-2m \Rightarrow h=-m$.
   2. Bisector formula: $\dfrac{x^2-y^2}{a-b} = \dfrac{xy}{h} \Rightarrow \dfrac{x^2-y^2}{1-(-1)} = \dfrac{xy}{-m} \Rightarrow \dfrac{x^2-y^2}{2}=\dfrac{xy}{-m}$.
   3. Cross-multiply: $-m(x^2-y^2) = 2xy \Rightarrow mx^2 - my^2 + 2xy = 0$ (multiplying by $-1$), i.e. $mx^2+2xy-my^2=0$. …

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7. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.Suppose the pairs of straight lines $x^2 - 2axy - y^2 = 0$ and $x^2 - 2bxy - y^2 = 0$ are such that each pair bisects the angles between the other two. Then $ab =$ (A) $1$ (B) $-1$ (C) $2$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Using the standard angle-bisector formula for a homogeneous pair of lines and matching the bisector of one pair to the other pair gives $ab=-1$.

   Concept and Intuition

   For a pair of straight lines through the origin, $Ax^2+2Hxy+By^2=0$, the combined equation of their two angle bisectors is $\dfrac{x^2-y^2}{A-B}=\dfrac{xy}{H}$. "Each pair bisects the angle between the other" means: (bisector of pair 1) $\equiv$ pair 2, and by symmetry (bisector of pair 2) $\equiv$ pair 1.

   Step-by-Step Solution

   1. Pair 1: $x^2-2axy-y^2=0\Rightarrow A=1,\ B=-1,\ 2H=-2a\Rightarrow H=-a$.
   2. Bisector of pair 1: $\dfrac{x^2-y^2}{1-(-1)}=\dfrac{xy}{-a}\Rightarrow\dfrac{x^2-y^2}{2}=-\dfrac{xy}{a}\Rightarrow a(x^2-y^2)=-2xy\Rightarrow ax^2+2xy-ay^2=0$.
   3. This must represent the same lines as pair 2: $x^2-2bxy-y^2=0$. Comparing ratios of coefficients of $x^2, xy, y^2$: $\dfrac{a}{1}=\dfrac{2}{-2b}=\dfrac{-a}{-1}$.
   4. From $\dfrac{a}{1}=\dfrac{2}{-2b}$: $a=-\dfrac1b\Rightarrow ab=-1$. …

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8. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.The equation of the circle, which cuts orthogonally each of the three circles $x^2+y^2-2x+3y-7=0$, $x^2+y^2+5x-5y+9=0$ and $x^2+y^2+7x-9y+29=0$ ______. (A) $x^2+y^2-16x-18y-4=0$ (B) $x^2+y^2=a^2$ (C) $x^2+y^2-16x=0$ (D) $y^2-x^2+2x=0$

   ›Reveal solutionSolution

   Use the orthogonality condition $2g_1g_2+2f_1f_2=c_1+c_2$ against all three given circles to pin down the unknown circle's coefficients. Answer: $x^2+y^2-16x-18y-4=0$.

   Concept and Intuition

   Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally iff $2g_1g_2+2f_1f_2=c_1+c_2$. Requiring orthogonality with three circles gives three linear equations in the unknown circle's $(G,F,C)$.

   Step-by-Step Solution

   Let the required circle be $x^2+y^2+2Gx+2Fy+C=0$.

   1. Against $x^2+y^2-2x+3y-7=0$ ($g_1=-1,f_1=1.5,c_1=-7$): $-2G+3F=C-7$.
   2. Against $x^2+y^2+5x-5y+9=0$ ($g_2=2.5,f_2=-2.5,c_2=9$): $5G-5F=C+9$.
   3. Against $x^2+y^2+7x-9y+29=0$ ($g_3=3.5,f_3=-4.5,c_3=29$): $7G-9F=C+29$.
   4. From (1): $C=-2G+3F+7$. Substitute into (2): $5G-5F=-2G+3F+16\Rightarrow 7G-8F=16$.
   5. Substitute into (3): $7G-9F=-2G+3F+36\Rightarrow 9G-12F=36\Rightarrow 3G-4F=12$. …

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9. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.If the lines joining the origin to the points of intersection of $2x + 3y = k$ and $3x^2 - xy + 3y^2 + 2x - 3y - 4 = 0$ are at right angles, then (A) $6k^2 + 5k + 52 = 0$ (B) $6k^2 + 5k - 52 = 0$ (C) $6k^2 - 5k + 52 = 0$ (D) $6k^2 - 5k - 52 = 0$

   ›Reveal solutionSolution

   Homogenize the conic with the line to get the pair of lines through the origin, then apply the perpendicularity condition "coefficient of $x^2$ + coefficient of $y^2=0$". Answer: (D).

   Concept and Intuition

   The lines joining the origin to the intersection points of a conic and a line form a homogeneous second-degree equation (a pair of straight lines through the origin), obtained by making every term of the conic degree-2 using the line equation (which equals $1$ when divided by its constant). For a pair of lines $ax^2+2hxy+by^2=0$ to be perpendicular, the well-known condition is $a+b=0$.

   Step-by-Step Solution

   1. Write $\dfrac{2x+3y}{k}=1$ and use it to homogenize $3x^2-xy+3y^2+2x-3y-4=0$: replace the linear part $(2x-3y)$ by $(2x-3y)\cdot\dfrac{2x+3y}{k}$ and the constant $-4$ by $-4\left(\dfrac{2x+3y}{k}\right)^2$.
   2. Multiply through by $k^2$: $3k^2x^2-k^2xy+3k^2y^2+k(2x-3y)(2x+3y)-4(2x+3y)^2=0$.
   3. Expand $(2x-3y)(2x+3y)=4x^2-9y^2$ and $(2x+3y)^2=4x^2+12xy+9y^2$. …

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10. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.The condition that the lines joining the origin to the points of intersection of the line $\frac{x}{a}+\frac{y}{b}=2$ and the circle $(x-a)^2+(y-b)^2=r^2$ are at right angles is (A) $a^2+b^2=r^2$ (B) $a^2-b^2=r^2$ (C) $a^2-b^2+r^2=0$ (D) $a^2+b^2+r^2=0$

    ›Reveal solutionSolution

    Homogenizing the circle with the chord's line equation and applying "coefficient of $x^2$ + coefficient of $y^2$ = 0" (perpendicular lines through the origin) gives $a^2+b^2=r^2$.

    Concept and Intuition

    The pair of lines joining the origin to the points where a line meets a curve is obtained by making the curve's equation homogeneous of degree 2 using the line equation (raised to the power needed to match each term's degree). Two lines $Ax^2+2Hxy+By^2=0$ through the origin are perpendicular exactly when $A+B=0$ — this is the standard test applied after homogenizing.

    Step-by-Step Solution

    1. Circle: $x^2+y^2-2ax-2by+(a^2+b^2-r^2)=0$.
    2. Line: $\dfrac xa+\dfrac yb=2\Rightarrow\dfrac{1}{2}\left(\dfrac xa+\dfrac yb\right)=1$.
    3. Homogenize the degree-1 and degree-0 terms of the circle using this: $x^2+y^2-2ax\cdot\dfrac12\left(\dfrac xa+\dfrac yb\right)-2by\cdot\dfrac12\left(\dfrac xa+\dfrac yb\right)+(a^2+b^2-r^2)\cdot\dfrac14\left(\dfrac xa+\dfrac yb\right)^2=0$.
    4. The linear-homogenizing terms simplify to $-x^2-y^2-\left(\dfrac ab+\dfrac ba\right)xy$, cancelling the leading $x^2+y^2$, leaving: $\dfrac{a^2+b^2-r^2}{4}\left(\dfrac{x^2}{a^2}+\dfrac{2xy}{ab}+\dfrac{y^2}{b^2}\right)-\left(\dfrac{a^2+b^2}{ab}\right)xy=0$. …

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11. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.If $ax^2 + 6xy + by^2 - 10x + 10y - 6 = 0$ represents a pair of perpendicular lines then the value of $|a|$ equals ______ (A) $6$ (B) $4$ (C) $2$ (D) $3$

    ›Reveal solutionSolution

    Tests the perpendicular-pair condition ($A+B=0$) together with the general second-degree determinant condition for a genuine pair of straight lines.

    Concept and Intuition

    A general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ represents a pair of straight lines only when the $3\times3$ determinant of its coefficients vanishes ($ABC+2FGH-AF^2-BG^2-CH^2=0$). Independently, the pair is perpendicular exactly when the coefficient of $x^2$ plus the coefficient of $y^2$ is zero ($A+B=0$), since perpendicular lines' slope product is $-1$ and for the homogeneous part this translates to $A+B=0$.

    Step-by-Step Solution

    1. Match $ax^2+6xy+by^2-10x+10y-6=0$ to $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$: $A=a$, $2H=6\Rightarrow H=3$, $B=b$, $2G=-10\Rightarrow G=-5$, $2F=10\Rightarrow F=5$, $C=-6$.
    2. Perpendicular lines: $A+B=0 \Rightarrow a+b=0 \Rightarrow b=-a$.
    3. Pair-of-lines condition: $ABC+2FGH-AF^2-BG^2-CH^2=0$. …

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12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.If the angle between the straight lines joining the foci and the ends of the minor axis of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is $90^0$, then its eccentricity is (A) $\dfrac{1}{2}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{\sqrt{2}}$

    ›Reveal solutionSolution

    The right-angle condition at the minor-axis end forces $b^2=a^2e^2$, and combined with the standard eccentricity relation gives $e=\frac{1}{\sqrt2}$.

    Concept and Intuition

    The angle referred to is the angle at an end of the minor axis, $B=(0,b)$, subtended by the two foci $F_1=(ae,0)$ and $F_2=(-ae,0)$. Setting the dot product of $\vec{BF_1}$ and $\vec{BF_2}$ to zero encodes the $90^\circ$ condition.

    Step-by-Step Solution

    1. $\vec{BF_1}=(ae,-b)$, $\vec{BF_2}=(-ae,-b)$.
    2. Perpendicularity: $\vec{BF_1}\cdot\vec{BF_2}=-a^2e^2+b^2=0\Rightarrow b^2=a^2e^2$.
    3. Standard relation for an ellipse: $b^2=a^2(1-e^2)$. …

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