Q.Find the equations of the two lines through the origin which intersect the line $\dfrac{x-3}{2} = \dfrac{y-3}{1} = \dfrac{z}{1}$ at angles of $\dfrac{\pi}{3}$ each.

Showing the 12 most recent of 44 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the direction cosines of two lines satisfy the equations $2l + m - n = 0$, $l^2 - 2m^2 + n^2 = 0$ and $\theta$ is the angle between the lines then $\cos \theta =$ (A) $\dfrac{1}{5}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   The direction cosines of each line satisfy two given equations; solving them yields two distinct direction vectors, and the cosine of the angle between them is found via dot product, giving $\cos\theta = \frac{1}{5}$.

   We are given that the direction cosines $(l, m, n)$ of two lines satisfy

   $2l + m - n = 0 \quad\text{and}\quad l^2 - 2m^2 + n^2 = 0.$

   The angle $\theta$ between the lines is the angle between their direction vectors. Since direction cosines satisfy $l^2 + m^2 + n^2 = 1$, each line’s $(l, m, n)$ is a unit vector. The two equations above must hold for both lines, but they define a set of possible unit vectors; the two distinct solutions give the two lines.

   Why this approach works:

   We treat the equations as a system in $l, m, n$ with the constraint $l^2+m^2+n^2=1$. Solving gives two unit vectors. Their dot product is $\cos\theta$.

   ---

   1. Express one variable in terms of another

      From $2l + m - n = 0$, we have $n = 2l + m$.

   2. Substitute into the second equation

   $l^2 - 2m^2 + (2l + m)^2 = 0.$

   Expand:

   $l^2 - 2m^2 + 4l^2 + 4lm + m^2 = 0 \quad\Rightarrow\quad 5l^2 + 4lm - m^2 = 0.$

   3. Solve the quadratic relation between $l$ and $m$ Treat $5l^2 + 4lm - m^2 = 0$ as quadratic in $l$:

   $5l^2 + 4m\,l - m^2 = 0.$

   Using the quadratic formula:

   $l = \frac{-4m \pm \sqrt{16m^2 + 20m^2}}{10} = \frac{-4m \pm 6m}{10}.$

   So the two possibilities are:

   $l = \frac{2m}{10} = \frac{m}{5} \quad\text{or}\quad l = \frac{-10m}{10} = -m.$

   4. Find corresponding $(l, m, n)$ for each case

      • Case 1: $l = \frac{m}{5}$ Then $n = 2l + m = \frac{2m}{5} + m = \frac{7m}{5}$. The unit vector condition $l^2 + m^2 + n^2 = 1$ gives:

   $\left(\frac{m}{5}\right)^2 + m^2 + \left(\frac{7m}{5}\right)^2 = 1 \quad\Rightarrow\quad \frac{m^2}{25} + m^2 + \frac{49m^2}{25} = 1.$

    Combine: $\frac{1+25+49}{25}m^2 = \frac{75}{25}m^2 = 3m^2 = 1$, so $m^2 = \frac{1}{3}$.  
    Choose $m = \frac{1}{\sqrt{3}}$ (sign doesn’t matter for direction). Then  
   

   $l = \frac{1}{5\sqrt{3}},\quad n = \frac{7}{5\sqrt{3}}.$

    So one direction vector is  
   

   $\mathbf{v}_1 = \left(\frac{1}{5\sqrt{3}},\; \frac{1}{\sqrt{3}},\; \frac{7}{5\sqrt{3}}\right).$

   • Case 2: $l = -m$ Then $n = 2(-m) + m = -m$. …

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2. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.A line makes angles $60^\circ$, $45^\circ$, $\theta$ with positive X, Y, Z-axes respectively. If $\theta$ is an acute angle, then $\tan \theta =$ (A) $\dfrac{1}{\sqrt{3}}$ (B) $2$ (C) $1$ (D) $\sqrt{3}$

   ›Reveal solutionSolution

   The direction cosines of a line satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. Using the given angles, we find $\cos\theta = \frac12$, so $\theta = 60^\circ$ and $\tan\theta = \sqrt{3}$. The correct option is (D).

   The key idea is that for any line in 3D space, the cosines of the angles it makes with the coordinate axes are called direction cosines, and they always satisfy the fundamental identity $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. This is because the direction vector’s components are proportional to these cosines, and the squared length of that vector is the sum of the squares of its components.

   Here we are given two angles directly and told the third angle $\theta$ is acute. We can use the identity to solve for $\cos\theta$, then find $\tan\theta$.

   1. Write the direction cosines. If a line makes angles $\alpha$, $\beta$, $\gamma$ with the positive X, Y, Z axes, then its direction cosines are $\cos\alpha$, $\cos\beta$, $\cos\gamma$. Here $\alpha = 60^\circ$, $\beta = 45^\circ$, $\gamma = \theta$ (acute). So:

   $\cos 60^\circ = \frac12,\quad \cos 45^\circ = \frac{\sqrt{2}}{2},\quad \cos\theta \text{ unknown}.$

   2. Apply the fundamental identity. For any line:

   $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.$

   Substitute the known values:

   $\left(\frac12\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + \cos^2\theta = 1.$

   3. Simplify.

   $\frac14 + \frac{2}{4} + \cos^2\theta = 1 \quad\Rightarrow\quad \frac34 + \cos^2\theta = 1.$

   So:

   $\cos^2\theta = 1 - \frac34 = \frac14.$

   4. Determine $\cos\theta$. …

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3. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If the direction cosines $(l, m, n)$ of two lines are connected by the relations $l + m + n = 0$ and $lm = 0$, then the angle between those lines is (A) $\dfrac{\pi}{3}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{2}$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The condition $l+m+n=0$ and $lm=0$ forces each line’s direction cosines to be a permutation of $(1,-1,0)/\sqrt{2}$, so the angle between them is $\pi/3$, making option (A) correct.

   ---

   We are given two lines whose direction cosines $(l, m, n)$ satisfy:

   $l + m + n = 0 \quad\text{and}\quad lm = 0.$

   We need the angle between these two lines.

   Concept & Intuition

   Direction cosines satisfy $l^2 + m^2 + n^2 = 1$. The conditions $l+m+n=0$ and $lm=0$ are symmetric but not fully symmetric — they force one of $l$ or $m$ to be zero. That gives us a family of possible triples, but the angle between two distinct lines from this family is fixed. The trick is to find two distinct triples that satisfy both conditions, then compute the dot product to get the cosine of the angle between them.

   ---

   Step-by-step reasoning

   1. Use the normalization condition Since $(l,m,n)$ are direction cosines, we have:

   $l^2 + m^2 + n^2 = 1.$

   Together with $l+m+n=0$, we can eliminate $n$: $n = -l - m$.

   2. Apply $lm = 0$

      This means either $l = 0$ or $m = 0$ (or both, but both zero would force $n=0$ from $l+m+n=0$, which is impossible because then $l^2+m^2+n^2=0 \neq 1$). So we have two cases:

      • Case 1: $l = 0$. Then $m + n = 0 \Rightarrow n = -m$. Normalization: $0^2 + m^2 + (-m)^2 = 2m^2 = 1 \Rightarrow m = \pm \frac{1}{\sqrt{2}}$. So one line has direction cosines $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ or $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. These are essentially the same line (opposite direction), so pick one representative:

   $\text{Line } A: \left(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right).$

   • Case 2: $m = 0$. Then $l + n = 0 \Rightarrow n = -l$. Normalization: $l^2 + 0^2 + (-l)^2 = 2l^2 = 1 \Rightarrow l = \pm \frac{1}{\sqrt{2}}$. Pick the representative:

   $\text{Line } B: \left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right).$

   These are two distinct lines satisfying the given relations.

   3. Compute the angle between them …

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4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.A line makes angles $60^\circ$, $45^\circ$, $\theta$ with positive X, Y, Z-axes respectively. If $\theta$ is an acute angle, then $\tan \theta =$ (A) $\sqrt{3}$ (B) $\dfrac{1}{\sqrt{3}}$ (C) $1$ (D) $2$

   ›Reveal solutionSolution

   The direction cosines of a line satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. Given $\alpha = 60^\circ$, $\beta = 45^\circ$, and $\gamma = \theta$ acute, we find $\cos\theta = \frac12$, so $\tan\theta = \sqrt{3}$. The correct option is (A).

   The key idea is that any line in 3D space has direction cosines — the cosines of the angles it makes with the positive coordinate axes. These three cosines are not independent; they satisfy a fundamental Pythagorean-like identity. Once we know two angles, we can solve for the third, and then compute its tangent.

   1. Recall the direction cosine identity If a line makes angles $\alpha$, $\beta$, $\gamma$ with the positive X, Y, Z axes, then

   $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.$

   This holds because the direction vector’s components are proportional to these cosines, and its squared length is the sum of squares of those components.

   2. Plug in the given angles We have $\alpha = 60^\circ$, $\beta = 45^\circ$, $\gamma = \theta$ (acute).

   $\cos 60^\circ = \frac12, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}.$

   So

   $\left(\frac12\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + \cos^2\theta = 1.$

   3. Simplify to find $\cos\theta$

   $\frac14 + \frac{2}{4} + \cos^2\theta = 1 \quad\Rightarrow\quad \frac34 + \cos^2\theta = 1.$

   Hence

   $$$$

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5. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The direction cosines of two lines are connected by the relations $l - m + n = 0$ and $2l - 3m + nl = 0$. If $\theta$ is the angle between these two lines, then $\cos \theta =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{\sqrt{19}}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{3\sqrt{2}}$

   ›Reveal solutionSolution

   Eliminating $m$ gives $2l^2=3n^2$, so the two lines have direction ratios $\left(\pm\sqrt3,\ \pm\sqrt3+\sqrt2,\ \sqrt2\right)$. Their dot product is $-2$ and the product of the magnitudes is $2\sqrt{19}$, giving $\cos\theta=\dfrac{1}{\sqrt{19}}$ — option (B).

   The concept first

   When two direction cosines relations are given — one linear and one homogeneous quadratic — the standard recipe is:

   1. use the linear relation to express one variable in terms of the other two;
   2. substitute into the quadratic, which becomes a homogeneous quadratic in the two remaining variables — hence an equation for a ratio;
   3. its two roots give the direction ratios of the two lines;
   4. finally

   $\cos\theta=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\ \sqrt{l_2^2+m_2^2+n_2^2}} ,$

   where we may use direction ratios (not necessarily normalised) provided we divide by the magnitudes.

   Step-by-step

   1. Eliminate $m$. From $l-m+n=0$,

   $m=l+n .$

   2. Substitute into $2lm-3mn+nl=0$:

   $2l(l+n)-3(l+n)n+nl=2l^2+2ln-3ln-3n^2+nl$

   $=2l^2+(2-3+1)log-3n^2=2l^2-3n^2=0 .$

   The $log$ terms cancel exactly — that is what makes this problem tractable.

   3. Solve for the ratio.

   $2l^2=3n^2\ \Longrightarrow\ \frac{l}{n}=\pm\sqrt{\frac32}.$

   Choose the convenient scaling $n=\sqrt2$, so $l=\pm\sqrt3$, and $m=l+n$:

   $\text{Line 1: } \left(\sqrt3,\ \sqrt3+\sqrt2,\ \sqrt2\right),\qquad \text{Line 2: } \left(-\sqrt3,\ \sqrt2-\sqrt3,\ \sqrt2\right).$

   4. Dot product. …

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6. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Let $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$ lying along X, Y, Z-axes respectively represent the coterminous edges of a rectangular parallelepiped. If $OA = 1$, $OB = 2$, $OC = 3$ then the angle between a pair of diagonals of the parallelepiped drawn through the vertices O and A is (A) $\dfrac{\pi}{3}$ (B) $\cos^{-1}\left(\dfrac{5}{7}\right)$ (C) $\cos^{-1}\left(\dfrac{6}{7}\right)$ (D) $\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   The angle between the space diagonals through O and A of a rectangular box is found using the dot product of their direction vectors. The correct answer is $\cos^{-1}\left(\frac{6}{7}\right)$, option (C).

   The problem gives us a rectangular parallelepiped (a box) with edges along the coordinate axes. The three coterminous edges from O are $\overrightarrow{OA}$ along X, $\overrightarrow{OB}$ along Y, and $\overrightarrow{OC}$ along Z, with lengths 1, 2, and 3 respectively.

   The key idea: a rectangular box has four space diagonals. Two of them pass through O and A (opposite vertices). The angle between any pair of space diagonals can be found by writing their direction vectors and using the dot product formula.

   1. Set up coordinates. Place O at the origin $(0,0,0)$. Then:

      • A is at $(1,0,0)$ (since OA = 1 along X)
      • B is at $(0,2,0)$
      • C is at $(0,0,3)$ The opposite vertex to O is the one with all three coordinates: $(1,2,3)$. Call it D.

   2. Identify the two diagonals through O and A. The diagonal through O goes from O to D: vector $\overrightarrow{OD} = (1,2,3)$. The diagonal through A goes from A to the vertex opposite A, which is the vertex with coordinates $(0,2,3)$ — call it E. So the diagonal through A is $\overrightarrow{AE} = (0-1, 2-0, 3-0) = (-1,2,3)$.

   3. Find the angle between these two diagonals. Use the dot product:

   $\overrightarrow{OD} \cdot \overrightarrow{AE} = (1)(-1) + (2)(2) + (3)(3) = -1 + 4 + 9 = 12$

   Magnitudes:

   $|\overrightarrow{OD}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$

   $|\overrightarrow{AE}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{14}$

   So:

   $$$$

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7. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the angle between the straight lines whose direction cosines satisfy the equations $l - 2m + n = 0$ and $2l^2 - 3m^2 + n^2 = 0$ is $\theta$, then $\cos \theta =$ (A) $\dfrac{9}{\sqrt{105}}$ (B) $\dfrac{3\sqrt{3}}{\sqrt{7}}$ (C) $\dfrac{\pi}{2}$ (D) $\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   The problem asks for the cosine of the angle between two lines whose direction cosines satisfy two given equations. By solving the system for possible direction ratios and using the dot product formula, we find $\cos\theta = \frac{9}{\sqrt{105}}$, which corresponds to option (A).

   We are given two conditions that the direction cosines $(l, m, n)$ of each line must satisfy:

   $l - 2m + n = 0 \quad \text{and} \quad 2l^2 - 3m^2 + n^2 = 0.$

   Since direction cosines also satisfy $l^2 + m^2 + n^2 = 1$, but we don’t need that directly — we only need the ratios of direction cosines to find the angle between the lines. The key idea: each line corresponds to a set $(l, m, n)$ (up to a common factor) that satisfies both equations. The angle between two such lines is found from the dot product of their direction vectors.

   ---

   1. Eliminate one variable using the linear equation

   From $l - 2m + n = 0$, we have

   $n = 2m - l.$

   2. Substitute into the quadratic equation

   Plug into $2l^2 - 3m^2 + n^2 = 0$:

   $2l^2 - 3m^2 + (2m - l)^2 = 0.$

   Expand $(2m - l)^2 = 4m^2 - 4lm + l^2$, so:

   $2l^2 - 3m^2 + 4m^2 - 4lm + l^2 = 0,$

   $3l^2 + m^2 - 4lm = 0.$

   3. Treat as a quadratic in $l/m$

   Divide through by $m^2$ (assuming $m \neq 0$; we’ll check later):

   $3\left(\frac{l}{m}\right)^2 - 4\left(\frac{l}{m}\right) + 1 = 0.$

   Let $t = l/m$. Then:

   $3t^2 - 4t + 1 = 0.$

   Solve:

   $t = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6}.$

   So $t = 1$ or $t = \frac{1}{3}$.

   4. Find direction ratios for each case

   Case 1: $l/m = 1 \Rightarrow l = m$.

   From $n = 2m - l = 2m - m = m$.

   So direction ratios are $(l, m, n) = (1, 1, 1)$.

   Case 2: $l/m = 1/3 \Rightarrow l = m/3$.

   Then $n = 2m - l = 2m - m/3 = \frac{5m}{3}$.

   So direction ratios are $(1/3, 1, 5/3)$, or multiply by 3: $(1, 3, 5)$.

   Thus the two lines have direction vectors $\mathbf{a} = (1, 1, 1)$ and $\mathbf{b} = (1, 3, 5)$. …

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8. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If the d.r.'s of two lines are connected by the relations $a - b + c = 0$, $a^2 - b^2 + 2c^2 = 0$ and $\theta$ is the angle between these lines then $\cos \theta =$ (A) $\dfrac{2}{\sqrt{7}}$ (B) $\dfrac{3}{2\sqrt{7}}$ (C) $\dfrac{3}{4\sqrt{2}}$ (D) $\dfrac{1}{3\sqrt{2}}$

   ›Reveal solutionSolution

   The relations give the two direction ratios $(1,1,0)$ and $(1,3,2)$; the angle between them has $\cos\theta=\dfrac{2}{\sqrt7}$.

   From $a-b+c=0$ we get $b=a+c$. Substitute into $a^2-b^2+2c^2=0$:

   $a^2-(a+c)^2+2c^2=0 \;\Rightarrow\; -2ac+c^2=0 \;\Rightarrow\; c(c-2a)=0.$

   So $c=0$ or $c=2a$, giving the two lines:

   • $c=0 \Rightarrow b=a$: direction ratios $(a,a,0)\propto(1,1,0)$.
   • $c=2a \Rightarrow b=3a$: direction ratios $(a,3a,2a)\propto(1,3,2)$.

   Angle between them: …

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9. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.If $\theta$ is the acute angle between the two lines whose direction cosines are connected by the relations $l + m + n = 0$ and $2lm + 2nl - mn = 0$, then $\cos \theta =$ (A) $\dfrac{1}{2}$ (B) $\dfrac{\sqrt{3}}{2}$ (C) $\dfrac{5}{6}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   The acute angle between the two lines is found by solving the given constraints for direction cosines, then using the dot product formula; the result is $\cos\theta = \frac{1}{2}$, so the correct option is (A).

   We are given two lines whose direction cosines $(l, m, n)$ satisfy two conditions:

   1. $l + m + n = 0$
   2. $2lm + 2nl - mn = 0$

   We need the cosine of the acute angle between these two lines.

   ---

   Concept and Intuition

   Direction cosines of a line satisfy $l^2 + m^2 + n^2 = 1$. For two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$, the cosine of the angle between them is

   $\cos\theta = l_1 l_2 + m_1 m_2 + n_1 n_2.$

   Here, both lines share the same pair of equations, so we must find two distinct sets $(l, m, n)$ that satisfy both constraints. The trick: treat the equations as a system that yields a relation between the ratios of $l, m, n$, then find two independent direction vectors.

   ---

   Step-by-step solution

   1. Eliminate one variable using $l + m + n = 0$

      From $l + m + n = 0$, we have $n = -l - m$.

   2. Substitute into the second equation

      The second condition is $2lm + 2nl - mn = 0$. Substitute $n$:

   $2lm + 2(-l - m)l - m(-l - m) = 0.$

   Simplify:

   $2lm - 2l^2 - 2lm + ml + m^2 = 0.$

   The $2lm$ and $-2lm$ cancel. We get:

   $-2l^2 + ml + m^2 = 0.$

   Multiply by $-1$:

   $2l^2 - ml - m^2 = 0.$

   3. Solve the quadratic in $l$ and $m$ Treat this as a quadratic in $l$:

   $2l^2 - m l - m^2 = 0.$

   Using the quadratic formula:

   $l = \frac{m \pm \sqrt{m^2 + 8m^2}}{4} = \frac{m \pm 3m}{4}.$

   So the two possibilities are:

   $l = \frac{m + 3m}{4} = m \quad \text{or} \quad l = \frac{m - 3m}{4} = -\frac{m}{2}.$

   4. Find the corresponding direction ratios

      • Case 1: $l = m$. Then $n = -l - m = -2l$. So direction ratios are $(l, l, -2l)$, i.e., proportional to $(1, 1, -2)$.
      • Case 2: $l = -\frac{m}{2}$. Then $m = -2l$, and $n = -l - (-2l) = l$. So direction ratios are $(l, -2l, l)$, i.e., proportional to $(1, -2, 1)$.

      These are two distinct lines.

   5. Compute $\cos\theta$ using the dot product …

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10. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
    Q.The slope of a line $L$ is $2$. If $m_1, m_2$ are slopes of two lines which are inclined at an angle of $\dfrac{\pi}{6}$ with $L$, then $m_1 + m_2 =$ (A) $-11$ (B) $16$ (C) $11$ (D) $-16$

    ›Reveal solutionSolution

    The key idea is to use the angle-between-lines formula $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$ with $\theta = \frac{\pi}{6}$ and the known slope $2$, then solve the resulting quadratic to find the two slopes and their sum. The sum is $-16$.

    When a line makes a fixed angle with a given line, there are always two such lines — one on each side of the given line. Their slopes are the two roots of a quadratic equation that comes from the angle formula. The sum of those slopes can be read directly from the quadratic without finding each slope individually.

    ---

    1. Set up the angle condition Let the slope of the given line $L$ be $m = 2$. Let the slope of a line inclined at $\frac{\pi}{6}$ to $L$ be $m_1$ (or $m_2$). The formula for the acute angle $\theta$ between two lines with slopes $m$ and $m_1$ is:

    $\tan\theta = \left| \frac{m_1 - m}{1 + m_1 m} \right|$

    Here $\theta = \frac{\pi}{6}$, so $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

    2. Write the equation without the absolute value The absolute value means there are two possibilities:

    $\frac{m_1 - 2}{1 + 2m_1} = \pm \frac{1}{\sqrt{3}}$

    These two equations give the two distinct slopes $m_1$ and $m_2$.

    3. Combine into a single quadratic Square both sides (or handle the two cases separately — both lead to the same quadratic). From $\frac{m_1 - 2}{1 + 2m_1} = \frac{1}{\sqrt{3}}$:

    $\sqrt{3}(m_1 - 2) = 1 + 2m_1 \quad\Rightarrow\quad \sqrt{3}\,m_1 - 2\sqrt{3} = 1 + 2m_1$

    $(\sqrt{3} - 2)m_1 = 1 + 2\sqrt{3}$

    This gives one slope. The other case with the negative sign gives the other slope.

    Instead of solving each, multiply the two equations:

    $\left(\frac{m_1 - 2}{1 + 2m_1}\right)\left(\frac{m_2 - 2}{1 + 2m_2}\right) = \left(\frac{1}{\sqrt{3}}\right)\left(-\frac{1}{\sqrt{3}}\right) = -\frac{1}{3}$

    But a cleaner method: treat $m_1$ and $m_2$ as the two roots of the quadratic obtained by removing the absolute value.

    4. Form the quadratic From $\frac{m - 2}{1 + 2m} = \pm \frac{1}{\sqrt{3}}$, cross-multiply and square:

    $3(m - 2)^2 = (1 + 2m)^2$

    Expand: …

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11. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.The slope of a line L passing through the point $(-2, -3)$ is not defined. If the angle between the lines L and $ax - 2y + 3 = 0$ $(a > 0)$ is $45^\circ$, then the angle made by the line $x + ay - 4 = 0$ with positive X-axis in the anticlockwise direction is (A) $\pi - \tan^{-1}\left(\dfrac{1}{2}\right)$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{2\pi}{3}$ (D) $\tan^{-1}\left(\dfrac{1}{2}\right)$

    ›Reveal solutionSolution

    The undefined slope means L is vertical. Using the angle condition with the given line, we find $a = 2$. Then the line $x + 2y - 4 = 0$ has slope $-\frac12$, so its angle with the positive X-axis is $\pi - \tan^{-1}(1/2)$, which is option (A).

    ---

    Concept & Intuition

    When a line’s slope is “not defined,” it is vertical (parallel to the Y-axis). The angle between two lines can be found using the formula $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$. Here, one slope is infinite, so we handle it by thinking geometrically: a vertical line makes a $90^\circ$ angle with the X-axis. The angle between L and the given line is $45^\circ$, so the given line must make either $45^\circ$ or $135^\circ$ with the vertical. That gives its slope, which lets us find $a$. Finally, we find the slope of the third line and its angle with the positive X-axis.

    ---

    Step-by-step solution

    1. Interpret “slope not defined”

       A line through $(-2,-3)$ with undefined slope is vertical: $x = -2$. Its direction is straight up/down, making an angle of $90^\circ$ with the positive X-axis.

    2. Angle condition with the second line

       The second line is $ax - 2y + 3 = 0$. Rewrite in slope-intercept form:

    $-2y = -ax - 3 \quad\Rightarrow\quad y = \frac{a}{2}x + \frac{3}{2}$

    So its slope is $m = \frac{a}{2}$.

    The angle between a vertical line (slope infinite) and a line of slope $m$ is $45^\circ$.

    Geometrically, a vertical line makes $90^\circ$ with the horizontal. If the other line makes an angle $\theta$ with the horizontal, then the angle between them is $|90^\circ - \theta|$.

    We require $|90^\circ - \theta| = 45^\circ$, so $\theta = 45^\circ$ or $\theta = 135^\circ$.

    3. Find $a$ from the slope
       • If $\theta = 45^\circ$, then $\tan 45^\circ = 1 = \frac{a}{2} \Rightarrow a = 2$.
       • If $\theta = 135^\circ$, then $\tan 135^\circ = -1 = \frac{a}{2} \Rightarrow a = -2$. …

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12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.If the slope of one of the lines represented by $5x^2 + \frac{40}{3}xy + ky^2 = 0$ is $3$, then the angle between the pair of lines is (A) $0^\circ$ (B) $\frac{\pi}{4}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{2}$

    ›Reveal solutionSolution

    Given one slope of a homogeneous pair of lines, we find the other slope using the relation between coefficients, then compute the angle between them. The angle is $\frac{\pi}{4}$.

    The equation $5x^2 + \frac{40}{3}xy + ky^2 = 0$ represents a pair of straight lines through the origin. For such a homogeneous second-degree equation $ax^2 + 2hxy + by^2 = 0$, the slopes $m_1$ and $m_2$ of the two lines satisfy:

    • $m_1 + m_2 = -\frac{2h}{b}$
    • $m_1 m_2 = \frac{a}{b}$

    Here, comparing, we have $a = 5$, $2h = \frac{40}{3}$ so $h = \frac{20}{3}$, and $b = k$.

    We are told one slope is $3$. Let $m_1 = 3$. Then $m_2$ is the other slope.

    1. Find the other slope using the sum of slopes.

       $m_1 + m_2 = -\frac{2h}{b} = -\frac{2 \cdot \frac{20}{3}}{k} = -\frac{40}{3k}$

       So $3 + m_2 = -\frac{40}{3k}$.

    2. Find the other slope using the product of slopes.

       $m_1 m_2 = \frac{a}{b} = \frac{5}{k}$

       So $3 m_2 = \frac{5}{k}$, giving $m_2 = \frac{5}{3k}$.

    3. Equate the two expressions for $m_2$.

       From step 1: $m_2 = -\frac{40}{3k} - 3$

       From step 2: $m_2 = \frac{5}{3k}$

       Set them equal:

       $\frac{5}{3k} = -\frac{40}{3k} - 3$

       Multiply through by $3k$ (assuming $k \neq 0$):

       $5 = -40 - 9k$

       $9k = -45$

       $k = -5$

    4. Now find $m_2$ using $k = -5$.

       $m_2 = \frac{5}{3(-5)} = -\frac{1}{3}$

    So the two slopes are $3$ and $-\frac{1}{3}$. …

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