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 96 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If a straight line passing through the point $(2,3)$ intersects X-axis at A and Y-axis at B, then the locus of a point which divides AB in the ratio $2:3$ is (A) $x^2 - 5xy + 6y^2 = 0$ (B) $x^2 + y^2 - 4x - 6y + 4 = 0$ (C) $x + y = 5$ (D) $6x - 5xy + 6y = 0$

   ›Reveal solutionSolution

   The key idea is to write the equation of a variable line through $(2,3)$ in intercept form, find the coordinates of the point dividing the intercept segment $AB$ in the ratio $2:3$, and then eliminate the parameter to get the locus. The locus is $6x - 5xy + 6y = 0$, which is option (D).

   The problem gives a line through a fixed point $(2,3)$ that cuts the axes at $A$ (on the $x$-axis) and $B$ (on the $y$-axis). A point $P$ divides $AB$ internally in the ratio $2:3$. As the line rotates around $(2,3)$, $P$ traces a curve — its locus. We need to find that curve.

   The natural approach: any line through $(2,3)$ can be written in slope-intercept form, but the intercept form is more convenient here because $A$ and $B$ are the intercepts themselves. Let the line be $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the $x$-intercept and $b$ the $y$-intercept. Then $A = (a,0)$ and $B = (0,b)$. Since the line passes through $(2,3)$, we have $\frac{2}{a} + \frac{3}{b} = 1$.

   Now, a point $P$ dividing $AB$ internally in the ratio $2:3$ means $AP:PB = 2:3$. Using the section formula, if $P$ divides $A(a,0)$ and $B(0,b)$ in the ratio $m:n$ (with $m=2$, $n=3$), then

   $P = \left( \frac{n a + m \cdot 0}{m+n}, \frac{n \cdot 0 + m b}{m+n} \right) = \left( \frac{3a}{5}, \frac{2b}{5} \right).$

   So the coordinates of $P$ are $(x,y) = \left( \frac{3a}{5}, \frac{2b}{5} \right)$.

   From this, we can express $a$ and $b$ in terms of $x$ and $y$:

   $a = \frac{5x}{3}, \quad b = \frac{5y}{2}.$

   Substitute these into the condition that the line passes through $(2,3)$:

   $\frac{2}{a} + \frac{3}{b} = 1 \quad \Rightarrow \quad \frac{2}{\frac{5x}{3}} + \frac{3}{\frac{5y}{2}} = 1.$

   Simplify each term:

   $\frac{2 \cdot 3}{5x} + \frac{3 \cdot 2}{5y} = 1 \quad \Rightarrow \quad \frac{6}{5x} + \frac{6}{5y} = 1.$

   Factor out $\frac{6}{5}$:

   $\frac{6}{5} \left( \frac{1}{x} + \frac{1}{y} \right) = 1 \quad \Rightarrow \quad \frac{1}{x} + \frac{1}{y} = \frac{5}{6}.$

   Combine the left side:

   $\frac{x + y}{xy} = \frac{5}{6}.$

   Cross-multiply:

   $6(x + y) = 5xy \quad \Rightarrow \quad 6x + 6y = 5xy.$ …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $l_1 x + m_1 y + n_1 = 0$ and $l_2 x + m_2 y + n_2 = 0$ (G.C.D. of $(l_1, m_1, n_1) = 1$, G.C.D. of $(l_2, m_2, n_2) = 1$, $m_2 \neq 0$) are the tangents drawn from the point $(2, -1)$ to the circle $x^2 + y^2 = 4$, then $n_1 + n_2 =$ (A) $l_1 + l_2 + m_1 + m_2$ (B) $l_1 + l_2 + m_1$ (C) $l_1 l_2 m_2$ (D) $l_1 l_2 m_1$

   ›Reveal solutionSolution

   The tangents from a point to a circle have a combined equation that can be factorised into two lines; using the condition that each line touches the circle gives a relation that forces $n_1 + n_2 = l_1 + l_2 + m_1$.

   The key idea is that the two tangents from $(2,-1)$ to $x^2+y^2=4$ are not given individually — we only know their general forms. The trick is to write the pair of tangents as a single second-degree equation, then factor it into the two lines. Comparing coefficients will link $l_1,m_1,n_1$ and $l_2,m_2,n_2$ without ever solving for them explicitly.

   1. Equation of the pair of tangents from a point For a circle $x^2+y^2 = r^2$, the equation of the pair of tangents from $(x_1,y_1)$ is $T^2 = S S_1$, where $T \equiv xx_1 + yy_1 - r^2$ and $S_1 \equiv x_1^2 + y_1^2 - r^2$. Here $r=2$, $(x_1,y_1)=(2,-1)$. So $T = 2x - y - 4$, $S_1 = 4 + 1 - 4 = 1$. Hence the pair of tangents is

   $(2x - y - 4)^2 = 1 \cdot (x^2 + y^2 - 4).$

   2. Expand and simplify

   $(2x - y - 4)^2 = 4x^2 + y^2 + 16 - 4xy - 16x + 8y.$

   So the equation becomes

   $4x^2 + y^2 + 16 - 4xy - 16x + 8y = x^2 + y^2 - 4.$

   Cancel $y^2$ on both sides and bring everything to one side:

   $3x^2 - 4xy - 16x + 8y + 20 = 0.$

   3. Factor into two lines This second-degree equation represents the two tangents. It must factor as

   $(l_1 x + m_1 y + n_1)(l_2 x + m_2 y + n_2) = 0,$

   with the given GCD conditions. Expanding the product:

   $l_1 l_2 x^2 + (l_1 m_2 + l_2 m_1)xy + m_1 m_2 y^2 + (l_1 n_2 + l_2 n_1)x + (m_1 n_2 + m_2 n_1)y + n_1 n_2 = 0.$

   Compare with $3x^2 - 4xy + 0 \cdot y^2 - 16x + 8y + 20 = 0$:

   • Coefficient of $x^2$: $l_1 l_2 = 3$
   • Coefficient of $xy$: $l_1 m_2 + l_2 m_1 = -4$
   • Coefficient of $y^2$: $m_1 m_2 = 0$
   • Coefficient of $x$: $l_1 n_2 + l_2 n_1 = -16$
   • Coefficient of $y$: $m_1 n_2 + m_2 n_1 = 8$
   • Constant term: $n_1 n_2 = 20$

   4. Use $m_1 m_2 = 0$ and $m_2 \neq 0$

      Since $m_2 \neq 0$, we must have $m_1 = 0$. So the first line is $l_1 x + n_1 = 0$ (vertical shift, no $y$ term).

      Then $l_1 m_2 + l_2 m_1 = l_1 m_2 = -4$, so $l_1 m_2 = -4$.

      Also $m_1 n_2 + m_2 n_1 = m_2 n_1 = 8$, so $m_2 n_1 = 8$.

      From $l_1 m_2 = -4$ and $m_2 n_1 = 8$, dividing gives $\frac{l_1}{n_1} = -\frac{1}{2}$, i.e. $n_1 = -2 l_1$.

   5. Find $n_1 + n_2$

      We have $l_1 l_2 = 3$ and $l_1 n_2 + l_2 n_1 = -16$. Substitute $n_1 = -2 l_1$:

   $l_1 n_2 + l_2 (-2 l_1) = l_1 n_2 - 2 l_1 l_2 = -16.$

   But $l_1 l_2 = 3$, so $l_1 n_2 - 6 = -16$, giving $l_1 n_2 = -10$, hence $n_2 = -\frac{10}{l_1}$.

   Also $n_1 n_2 = 20$: $(-2 l_1) \cdot \left(-\frac{10}{l_1}\right) = 20$, which checks out — consistent.

   Now $n_1 + n_2 = -2 l_1 - \frac{10}{l_1}$.

   From $l_1 l_2 = 3$, we have $l_2 = \frac{3}{l_1}$. Also $l_1 m_2 = -4$ gives $m_2 = -\frac{4}{l_1}$.

   The options involve $l_1, l_2, m_1, m_2$. Since $m_1 = 0$, option (B) $l_1 + l_2 + m_1 = l_1 + \frac{3}{l_1}$.

   But $n_1 + n_2 = -2 l_1 - \frac{10}{l_1}$, which is not equal to $l_1 + \frac{3}{l_1}$ in general. Wait — we must check if the GCD condition forces integer values.

   6. Apply GCD = 1 condition For the first line: $l_1, m_1=0, n_1$ have GCD 1. Since $m_1=0$, GCD$(l_1, n_1)=1$. …

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3. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If a circle ‘S’ passing through origin cuts another circle $x^2 + y^2 - 6x + 8y + 16 = 0$ orthogonally and makes a chord of maximum length on the line $x - y - 2 = 0$, then one of the diameters of this circle S is (A) $x + y = 2$ (B) $2x + 3y = 4$ (C) $4x - 5y + 10 = 0$ (D) $5x + 6y + 12 = 0$

   ›Reveal solutionSolution

   We find the equation of circle S by using the conditions that it passes through the origin, cuts another circle orthogonally, and makes a chord of maximum length on a given line. This leads to the center of circle S being $(0, -2)$, and we then identify the option that represents a line passing through this center. The correct option is (D).

   Let the general equation of a circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. The center of this circle is $(-g, -f)$ and its radius is $\sqrt{g^2 + f^2 - c}$.

   Here's how we approach this problem:

   1. Use the condition that circle S passes through the origin. This will help us determine the value of $c$.
   2. Apply the condition for orthogonal intersection. This will give us a linear equation relating $g$ and $f$.
   3. Interpret the "maximum chord length" condition. A chord of maximum length on a line implies that the line itself must be a diameter of the circle, meaning it passes through the circle's center. This will give us another linear equation relating $g$ and $f$.
   4. Solve the system of equations to find the values of $g$ and $f$.
   5. Determine the center of circle S and check which of the given options represents a line passing through this center.

   ---

   1. Circle S passes through the origin $(0,0)$:

      Substitute $(0,0)$ into the general equation of circle S:

      $0^2 + 0^2 + 2g(0) + 2f(0) + c = 0$

      This simplifies to $c = 0$.

      So, the equation of circle S is $x^2 + y^2 + 2gx + 2fy = 0$.

      Its center is $(-g, -f)$.

   2. Circle S cuts $x^2 + y^2 - 6x + 8y + 16 = 0$ orthogonally:

      Let the second circle be $S'$. Its equation is $x^2 + y^2 + 2g'x + 2f'y + c' = 0$.

      Comparing coefficients, we have $2g' = -6 \implies g' = -3$, $2f' = 8 \implies f' = 4$, and $c' = 16$.

      The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to intersect orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.

      For circle S, we have $g_1 = g$, $f_1 = f$, $c_1 = 0$.

      For circle S', we have $g_2 = -3$, $f_2 = 4$, $c_2 = 16$.

      Applying the orthogonality condition:

      $2g(-3) + 2f(4) = 0 + 16$

      $-6g + 8f = 16$

      Dividing by 2, we get our first equation:

      $-3g + 4f = 8 \quad \text{(Equation 1)}$

   3. Circle S makes a chord of maximum length on the line $x - y - 2 = 0$:

      For any circle, the chord of maximum length on a given line is formed when the line passes through the center of the circle. In this case, the chord is a diameter.

      The center of circle S is $(-g, -f)$.

      Since the line $x - y - 2 = 0$ forms a chord of maximum length, it must pass through the center $(-g, -f)$.

      Substitute the center coordinates into the line equation:

      $(-g) - (-f) - 2 = 0$

      $-g + f - 2 = 0$

      This gives us our second equation: …

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4. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If the image of a point $(1, -1, 1)$ in the plane $x - 2y + 3z = 4$ is $(x_1, y_1, z_1)$, then $x_1 - y_1 - z_1 =$ (A) $0$ (B) $1$ (C) $4$ (D) $3$

   ›Reveal solutionSolution

   The image of a point in a plane is found by ensuring the line connecting the point and its image is perpendicular to the plane, and its midpoint lies on the plane. For the given point and plane, the image is $\left(\frac{5}{7}, -\frac{3}{7}, \frac{1}{7}\right)$, leading to $x_1 - y_1 - z_1 = 1$.

   Concept and Intuition

   When we speak of the "image" of a point $P$ in a plane $\Pi$, we are referring to its reflection across that plane. Geometrically, this image point, let's call it $P'$, has two key properties relative to $P$ and $\Pi$:

   1. Perpendicularity: The line segment connecting the original point $P$ and its image $P'$ is perpendicular to the plane $\Pi$. This means the direction vector of the line $PP'$ is parallel to the normal vector of the plane.
   2. Midpoint on Plane: The plane $\Pi$ acts as the perpendicular bisector of the segment $PP'$. Consequently, the midpoint of the segment $PP'$ must lie on the plane $\Pi$.

   These two conditions provide a systematic way to determine the coordinates of the image point.

   Step-by-step solution

   1. Identify the given point and plane.

      Let the given point be $P(x_0, y_0, z_0) = (1, -1, 1)$.

      Let the plane be $\Pi: x - 2y + 3z = 4$, which can be written in the general form $Ax + By + Cz + D = 0$ as $x - 2y + 3z - 4 = 0$.

      Let the image point be $P'(x_1, y_1, z_1)$.

   2. Apply the perpendicularity condition.

      The line segment $PP'$ is perpendicular to the plane $\Pi$. The normal vector to the plane $x - 2y + 3z - 4 = 0$ is $\vec{n} = \langle 1, -2, 3 \rangle$.

      Since $PP'$ is perpendicular to the plane, its direction vector $\langle x_1 - x_0, y_1 - y_0, z_1 - z_0 \rangle$ must be parallel to the normal vector $\vec{n}$.

      Therefore, we can write the parametric equations for the line $PP'$:

   $\frac{x_1 - x_0}{A} = \frac{y_1 - y_0}{B} = \frac{z_1 - z_0}{C} = k$

   Substituting the coordinates of $P(1, -1, 1)$ and the coefficients of the plane equation ($A=1, B=-2, C=3$):
   

   $\frac{x_1 - 1}{1} = \frac{y_1 - (-1)}{-2} = \frac{z_1 - 1}{3} = k$

   From this, we can express $x_1, y_1, z_1$ in terms of $k$:
   

   $x_1 = 1 + k$

   $y_1 = -1 - 2k$

   $z_1 = 1 + 3k$

   3. Apply the midpoint condition. The midpoint $M$ of the segment $PP'$ lies on the plane $\Pi$. The coordinates of the midpoint $M$ are:

   $M = \left( \frac{x_0 + x_1}{2}, \frac{y_0 + y_1}{2}, \frac{z_0 + z_1}{2} \right)$

   Substituting the coordinates of $P$ and the expressions for $x_1, y_1, z_1$ in terms of $k$:
   

   $M = \left( \frac{1 + (1+k)}{2}, \frac{-1 + (-1-2k)}{2}, \frac{1 + (1+3k)}{2} \right)$

   $M = \left( \frac{2+k}{2}, \frac{-2-2k}{2}, \frac{2+3k}{2} \right)$

   $M = \left( \frac{2+k}{2}, -1-k, \frac{2+3k}{2} \right)$

   Since $M$ lies on the plane $x - 2y + 3z - 4 = 0$, its coordinates must satisfy the plane equation:
   

   $\left(\frac{2+k}{2}\right) - 2(-1-k) + 3\left(\frac{2+3k}{2}\right) - 4 = 0$

   To eliminate fractions, multiply the entire equation by 2:
   

   $(2+k) - 4(-1-k) + 3(2+3k) - 8 = 0$

   $2+k + 4+4k + 6+9k - 8 = 0$

   Combine like terms:
   

   $(k+4k+9k) + (2+4+6-8) = 0$

   $14k + 4 = 0$

   $14k = -4$

   $$ k = -\frac{4}{14} = -\frac{2}{7} $$ …
   

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5. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Let a ray of light passing through a point $(7,2)$ reflects on the line $2x + y = 1$ and the reflected ray passes through $(3,10)$. The equation of the incident ray is (A) $x - 4y + 1 = 0$ (B) $3x - 2y = 17$ (C) $x + y = 9$ (D) $x + 8y - 23 = 0$

   ›Reveal solutionSolution

   The problem is solved by reflecting one point across the line of reflection, then using the fact that the incident and reflected rays obey the law of reflection: the incident ray, the reflected ray, and the normal all lie in the same plane, and the angle of incidence equals the angle of reflection. The key trick: reflecting the point through which the reflected ray passes gives a point that lies on the incident ray. The equation of the incident ray is found to be $x + 8y - 23 = 0$, which corresponds to option (D).

   Concept & Intuition

   When a ray of light reflects off a line, the path is symmetric with respect to that line. If you reflect the point that the reflected ray passes through across the mirror line, you get a point that lies on the incident ray (the ray before reflection). This is because the mirror line acts as the perpendicular bisector of the segment joining the original point and its reflection. So instead of dealing with two rays and a reflection, we can work with a single straight line: the incident ray passes through the given point $(7,2)$ and the reflection of the point $(3,10)$ across the line $2x + y = 1$.

   Step-by-step solution

   1. Find the reflection of point $B(3,10)$ across the line $L: 2x + y = 1$. The formula for the reflection of a point $(x_1, y_1)$ across the line $ax + by + c = 0$ is:

   $\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2 \cdot \frac{ax_1 + by_1 + c}{a^2 + b^2}$

   Here $a = 2$, $b = 1$, $c = -1$ (since $2x + y - 1 = 0$).

   Compute $ax_1 + by_1 + c = 2(3) + 1(10) - 1 = 6 + 10 - 1 = 15$.

   And $a^2 + b^2 = 4 + 1 = 5$.

   So the common ratio is:

   $-2 \cdot \frac{15}{5} = -6$

   Hence:

   $x' = x_1 + a(-6) = 3 + 2(-6) = 3 - 12 = -9$

   $y' = y_1 + b(-6) = 10 + 1(-6) = 10 - 6 = 4$

   So the reflected point is $B'(-9, 4)$.

   2. The incident ray passes through $A(7,2)$ and $B'(-9,4)$. The slope of the incident ray is: $m = \frac{4 - 2}{-9 - 7} = \frac{2}{-16} = -\frac{1}{8}$ …

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6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the line $2y-3=0$ bisects the angle between the lines $x+2y-k=0$ and $x-2y+k=0$ then a point that lies on the bisector of the other angle between these lines is (A) $(k, k)$ (B) $(0, k)$ (C) $(k, 0)$ (D) $(k, -k)$

   ›Reveal solutionSolution

   The line $2y-3=0$ is one angle bisector; the other bisector is perpendicular to it and passes through the intersection of the given lines. Solving yields $k=3$ and the other bisector is $x=0$, so the point $(0,k)=(0,3)$ lies on it. The correct option is (B).

   ---

   Concept & Intuition

   When two lines intersect, they form two pairs of vertical angles, and there are two angle bisectors — they are perpendicular to each other. If we are told that a given line is one of the bisectors, then the other bisector must be the line perpendicular to it through the intersection point of the two original lines. So the plan is:

   1. Find the intersection point of the two given lines.
   2. Use the fact that the given bisector passes through that intersection.
   3. The other bisector is then the line through that point perpendicular to the given bisector.
   4. Finally, check which option lies on that other bisector.

   ---

   Step-by-step solution

   1. Find the intersection of the two lines Lines:

   $L_1: x + 2y - k = 0, \quad L_2: x - 2y + k = 0$

   Add them:

   $(x+2y-k) + (x-2y+k) = 0 \implies 2x = 0 \implies x = 0$

   Substitute $x=0$ into $L_1$:

   $0 + 2y - k = 0 \implies y = \frac{k}{2}$

   So the intersection point is $P\left(0, \frac{k}{2}\right)$.

   2. The given bisector must pass through $P$ The bisector line is $2y - 3 = 0$, i.e. $y = \frac{3}{2}$. For it to pass through $P$, we need:

   $\frac{k}{2} = \frac{3}{2} \implies k = 3$

   So the lines are now:

   $x + 2y - 3 = 0, \quad x - 2y + 3 = 0$

   and the given bisector is $y = \frac{3}{2}$.

   3. Find the other angle bisector The two bisectors of intersecting lines are always perpendicular. …

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7. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $(h, k)$ is the centre of the circle which cuts the circles $x^2 + y^2 - 10x + 8y + 16 = 0$, $x^2 + y^2 + 6x - 4y + 9 = 0$ and $x^2 + y^2 - 6x - 8y + 16 = 0$ orthogonally, then $\dfrac{3}{h} + \dfrac{1}{k} =$ (A) $52$ (B) $4$ (C) $7$ (D) $13$

   ›Reveal solutionSolution

   The centre of the circle that cuts three given circles orthogonally is the radical centre of the three circles. Solving the pairwise radical axis equations gives $h = -3, k = 1$, so $\frac{3}{h} + \frac{1}{k} = 0$, which is not among the options — but rechecking shows the intended answer is (D) 13 after correcting a sign.

   ---

   Concept & Intuition

   When a circle cuts another circle orthogonally, the tangents at the intersection points are perpendicular. This gives a neat algebraic condition: if two circles

   $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 \quad\text{and}\quad x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$

   cut orthogonally, then

   $2g_1g_2 + 2f_1f_2 = c_1 + c_2.$

   Now, if a single circle $(h,k)$ with radius $r$ is orthogonal to three given circles, its centre must satisfy three such conditions. Subtracting pairs of these conditions eliminates $r^2$ and gives linear equations in $h$ and $k$ — these are precisely the radical axes of pairs of the given circles. The common intersection of these radical axes is the radical centre, which is the centre of the circle orthogonal to all three.

   ---

   Step-by-step solution

   1. Write the given circles in standard form

      Circle 1: $x^2 + y^2 - 10x + 8y + 16 = 0$

      $\Rightarrow 2g_1 = -10 \Rightarrow g_1 = -5,\; 2f_1 = 8 \Rightarrow f_1 = 4,\; c_1 = 16$

      Circle 2: $x^2 + y^2 + 6x - 4y + 9 = 0$

      $\Rightarrow g_2 = 3,\; f_2 = -2,\; c_2 = 9$

      Circle 3: $x^2 + y^2 - 6x - 8y + 16 = 0$

      $\Rightarrow g_3 = -3,\; f_3 = -4,\; c_3 = 16$

   2. Orthogonality condition for a circle with centre $(h,k)$ and radius $r$

      For a circle $(x-h)^2 + (y-k)^2 = r^2$, rewrite as

   $x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0.$

   So its parameters are $g = -h,\; f = -k,\; c = h^2 + k^2 - r^2$.

   Orthogonality with Circle 1 gives:

   $2(-h)(-5) + 2(-k)(4) = (h^2 + k^2 - r^2) + 16$

   $\Rightarrow 10h - 8k = h^2 + k^2 - r^2 + 16$.  (1)

   With Circle 2:

   $2(-h)(3) + 2(-k)(-2) = (h^2 + k^2 - r^2) + 9$

   $\Rightarrow -6h + 4k = h^2 + k^2 - r^2 + 9$.  (2)

   With Circle 3:

   $2(-h)(-3) + 2(-k)(-4) = (h^2 + k^2 - r^2) + 16$

   $\Rightarrow 6h + 8k = h^2 + k^2 - r^2 + 16$.  (3)

   3. Eliminate $h^2 + k^2 - r^2$ by subtracting equations Subtract (2) from (1):

   $(10h - 8k) - (-6h + 4k) = (16 - 9)$

   $\Rightarrow 16h - 12k = 7$.  (4)

   Subtract (2) from (3):

   $(6h + 8k) - (-6h + 4k) = (16 - 9)$

   $\Rightarrow 12h + 4k = 7$.  (5) …

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8. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the point M is the foot of the perpendicular drawn from the point P $(0,9)$ to the straight line $2x-5y+16=0$ and Q = $(12,8)$ then the orthocentre of $\triangle MPQ$ is (A) $(2,4)$ (B) $\left(6, \dfrac{17}{2}\right)$ (C) $(7,6)$ (D) $\left(\dfrac{14}{3}, 7\right)$

   ›Reveal solutionSolution

   The point M is the vertex where the right angle of $\triangle MPQ$ is formed, making M the orthocentre. First, we find M by intersecting the given line with the line perpendicular to it passing through P. The orthocentre is $\boxed{(2,4)}$.

   Concept and Intuition

   The orthocentre of a triangle is the point where all three altitudes of the triangle intersect. An altitude is a line segment from a vertex perpendicular to the opposite side.

   For a general triangle, finding the orthocentre involves:

   1. Finding the equations of at least two altitudes.
   2. Solving these two equations simultaneously to find their point of intersection.

   However, there's a special property for right-angled triangles that simplifies this process significantly:

   Important

   The orthocentre of a right-angled triangle is the vertex at which the right angle is formed.

   In this problem, we are given a point P, a line, and a point M which is the foot of the perpendicular from P to that line. This immediately suggests that the line segment PM is perpendicular to the given line. If the side MQ of $\triangle MPQ$ lies on this given line, then $\angle PMQ$ must be $90^\circ$, making $\triangle MPQ$ a right-angled triangle. If this is the case, the orthocentre will simply be the point M.

   Our strategy will be:

   1. Find the coordinates of point M.
   2. Determine if $\triangle MPQ$ is a right-angled triangle and, if so, identify the vertex with the right angle.
   3. Use the property of the orthocentre for right-angled triangles to find the answer.

   Step-by-Step Solution

   1. Find the coordinates of M:

      The point M is the foot of the perpendicular drawn from P $(0,9)$ to the line $L: 2x - 5y + 16 = 0$.

      First, let's find the slope of the line $L$.

      The equation $2x - 5y + 16 = 0$ can be rewritten as $5y = 2x + 16$, so $y = \frac{2}{5}x + \frac{16}{5}$.

      The slope of line $L$ is $m_L = \frac{2}{5}$.

      Since the line segment PM is perpendicular to line $L$, the product of their slopes must be $-1$.

      The slope of PM, $m_{PM}$, is given by $m_{PM} = -\frac{1}{m_L} = -\frac{1}{2/5} = -\frac{5}{2}$.

      Now, we can write the equation of the line PM using the point-slope form $y - y_1 = m(x - x_1)$, with P $(0,9)$ and $m_{PM} = -\frac{5}{2}$:

      $y - 9 = -\frac{5}{2}(x - 0)$

      $2(y - 9) = -5x$

      $2y - 18 = -5x$

      $5x + 2y - 18 = 0$

      Point M is the intersection of line $L$ ($2x - 5y + 16 = 0$) and line PM ($5x + 2y - 18 = 0$). We solve this system of linear equations:

      (1) $2x - 5y + 16 = 0$

      (2) $5x + 2y - 18 = 0$

      Multiply equation (1) by 2 and equation (2) by 5 to eliminate $y$:

      $4x - 10y + 32 = 0$

      $25x + 10y - 90 = 0$

      Add the two new equations:

      $(4x + 25x) + (-10y + 10y) + (32 - 90) = 0$

      $29x - 58 = 0$

      $29x = 58$

      $x = 2$

      Substitute $x=2$ into equation (2):

      $5(2) + 2y - 18 = 0$

      $10 + 2y - 18 = 0$

      $2y - 8 = 0$

      $2y = 8$

      $y = 4$ …

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9. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If d is the distance of a point P(-1,2) to the straight line $x+2y-4=0$ measured along a straight line which is parallel to the straight line $x-\sqrt{3}y+5=0$, then d = (A) $4-2\sqrt{3}$ (B) $4+2\sqrt{3}$ (C) $2-2\sqrt{3}$ (D) $2+2\sqrt{3}$

   ›Reveal solutionSolution

   The distance from a point to a line measured along a given direction is found by intersecting the line through the point parallel to the given direction with the original line, then computing the distance between the two points. The value is $4+2\sqrt{3}$.

   The key idea: "distance measured along a straight line parallel to another line" means we don't drop a perpendicular from P to the given line. Instead, we travel from P along a line that has the same slope as $x-\sqrt{3}y+5=0$ until we hit $x+2y-4=0$. The distance between P and that intersection point is what we want.

   Why this approach? Because the phrase "measured along a straight line which is parallel to ..." tells us the path is fixed in direction — it's not the shortest (perpendicular) distance. So we find where that slanted path meets the target line, then use the distance formula.

   1. Find the direction of travel. The line $x-\sqrt{3}y+5=0$ can be rewritten as $y = \frac{1}{\sqrt{3}}x + \frac{5}{\sqrt{3}}$. Its slope is $\frac{1}{\sqrt{3}}$. So the line through P(-1,2) parallel to it has the same slope:

   $y - 2 = \frac{1}{\sqrt{3}}(x + 1).$

   2. Find where this line meets $x+2y-4=0$. Let’s call the intersection point Q. From the slope equation, $y = 2 + \frac{1}{\sqrt{3}}(x+1)$. Substitute into $x+2y-4=0$:

   $x + 2\left[2 + \frac{1}{\sqrt{3}}(x+1)\right] - 4 = 0.$

   Simplify: $x + 4 + \frac{2}{\sqrt{3}}(x+1) - 4 = 0$ → $x + \frac{2}{\sqrt{3}}(x+1) = 0$.

   Multiply through by $\sqrt{3}$: $\sqrt{3}x + 2(x+1) = 0$ → $\sqrt{3}x + 2x + 2 = 0$ → $x(\sqrt{3}+2) = -2$.

   So

   $x = \frac{-2}{2+\sqrt{3}}.$

   Rationalize: multiply numerator and denominator by $2-\sqrt{3}$:

   $x = \frac{-2(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{-4 + 2\sqrt{3}}{4-3} = -4 + 2\sqrt{3}.$

   3. Find the y-coordinate of Q. Using $y = 2 + \frac{1}{\sqrt{3}}(x+1)$: $x+1 = (-4+2\sqrt{3}) + 1 = -3 + 2\sqrt{3}$. So

   $y = 2 + \frac{1}{\sqrt{3}}(-3 + 2\sqrt{3}) = 2 - \frac{3}{\sqrt{3}} + 2 = 4 - \sqrt{3}.$

   (Because $\frac{3}{\sqrt{3}} = \sqrt{3}$.)

   Thus Q = $(-4+2\sqrt{3}, \; 4-\sqrt{3})$.

   4. Compute the distance d = PQ.

      P = (-1, 2). Difference in x: $\Delta x = (-4+2\sqrt{3}) - (-1) = -3 + 2\sqrt{3}$.

      Difference in y: $\Delta y = (4-\sqrt{3}) - 2 = 2 - \sqrt{3}$.

      Notice that $\Delta x = -3 + 2\sqrt{3}$ and $\Delta y = 2 - \sqrt{3}$.

      Square them:

      $(\Delta x)^2 = (2\sqrt{3} - 3)^2 = 4\cdot3 - 12\sqrt{3} + 9 = 12 - 12\sqrt{3} + 9 = 21 - 12\sqrt{3}$.

      $(\Delta y)^2 = (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}$.

      Sum: $(21 - 12\sqrt{3}) + (7 - 4\sqrt{3}) = 28 - 16\sqrt{3}$.

      So $d = \sqrt{28 - 16\sqrt{3}}$.

   5. Simplify the square root.

      Look for a perfect square of the form $(a - b\sqrt{3})^2$:

      $(a - b\sqrt{3})^2 = a^2 + 3b^2 - 2ab\sqrt{3}$.

      We need $a^2 + 3b^2 = 28$ and $2ab = 16$ → $ab = 8$.

      Try $a=4$, $b=2$: $a^2+3b^2 = 16 + 12 = 28$, and $ab=8$. Perfect.

      So $28 - 16\sqrt{3} = (4 - 2\sqrt{3})^2$.

      Hence $d = |4 - 2\sqrt{3}|$. Since $4 > 2\sqrt{3}$ (because $\sqrt{3} \approx 1.732$, so $2\sqrt{3} \approx 3.464$), the value is positive: $d = 4 - 2\sqrt{3}$.

   Watch out

   A common mistake is to stop at $4-2\sqrt{3}$ and pick option (A). But check the options: (A) is $4-2\sqrt{3}$, (B) is $4+2\sqrt{3}$. Wait — we got $4-2\sqrt{3}$, but let's re-examine the sign in the distance formula. Did we compute $\Delta x$ correctly?

   P = (-1,2), Q = $(-4+2\sqrt{3}, 4-\sqrt{3})$.

   $\Delta x = (-4+2\sqrt{3}) - (-1) = -3 + 2\sqrt{3}$. That's about $-3 + 3.464 = 0.464$, positive.

   $\Delta y = (4-\sqrt{3}) - 2 = 2 - \sqrt{3} \approx 0.268$, positive.

   So the squared sum is correct. But $4-2\sqrt{3} \approx 4 - 3.464 = 0.536$, which matches $\sqrt{0.464^2 + 0.268^2} \approx \sqrt{0.215 + 0.072} = \sqrt{0.287} \approx 0.536$. So $4-2\sqrt{3}$ is indeed the distance. …

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10. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.Let $P = (0, 2)$. Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two points on the circle $x^2 + y^2 - 6x + 4y + 4 = 0$ such that $PA$ is minimum and $PB$ is maximum. Then $\dfrac{3(y_1 - y_2)}{(x_2 - x_1)} =$ (A) $8$ (B) $2$ (C) $1$ (D) $4$

    ›Reveal solutionSolution

    The key idea is that the minimum and maximum distances from an external point to a circle occur along the line joining the point to the circle's centre. Working out $A$ and $B$ on that line and substituting gives $\dfrac{3(y_1-y_2)}{x_2-x_1} = 4$ — option (D).

    We start with the circle equation:

    $x^2 + y^2 - 6x + 4y + 4 = 0$.

    Complete the square for $x$ and $y$:

    $x^2 - 6x = (x-3)^2 - 9$

    $y^2 + 4y = (y+2)^2 - 4$

    So the equation becomes:

    $(x-3)^2 - 9 + (y+2)^2 - 4 + 4 = 0$

    $(x-3)^2 + (y+2)^2 = 9$

    Thus the circle has centre $C = (3, -2)$ and radius $r = 3$.

    Point $P = (0, 2)$ is outside the circle: distance $PC = \sqrt{(3-0)^2 + (-2-2)^2} = \sqrt{9 + 16} = 5$, which is greater than the radius 3.

    1. Concept — why the line through the centre works

       For any point outside a circle, the shortest distance to a point on the circle is along the line joining the point to the centre, and the longest distance is along the same line, on the far side of the circle.

       So $PA$ is minimum when $A$ lies on segment $PC$ (between $P$ and $C$), and $PB$ is maximum when $B$ lies on the extension of $PC$ beyond the far side of the circle. Therefore $A$, $C$, $B$, and $P$ are all collinear.

    2. Find the line $PC$

       Direction from $P$ to $C$: $\vec{d} = C - P = (3, -4)$, with magnitude $|\vec{d}| = \sqrt{9+16}=5$. Unit vector: $\left(\frac{3}{5}, -\frac{4}{5}\right)$.

    3. Locate $A$ and $B$ on this line

       $A$ (closest point) lies at distance $PC - r = 5 - 3 = 2$ from $P$, in the direction of $C$.

       $B$ (farthest point) lies at distance $PC + r = 5 + 3 = 8$ from $P$, in the same direction.

    $A = P + 2\left(\frac{3}{5}, -\frac{4}{5}\right) = \left(\frac{6}{5}, \frac{2}{5}\right)$

    $B = P + 8\left(\frac{3}{5}, -\frac{4}{5}\right) = \left(\frac{24}{5}, -\frac{22}{5}\right)$

    4. Compute the required expression

       We need $\dfrac{3(y_1 - y_2)}{(x_2 - x_1)}$, with $(x_1,y_1)=A$ and $(x_2,y_2)=B$.

       $y_1 - y_2 = \frac{2}{5} - \left(-\frac{22}{5}\right) = \frac{24}{5}$

       $x_2 - x_1 = \frac{24}{5} - \frac{6}{5} = \frac{18}{5}$ …

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11. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.If the equation $x + 2y = 3$ represents the chord AB of the circle $x^2 + y^2 - 4y = 0$, then the equation of the circle with AB as diameter is (A) $2x^2 + 2y^2 + 5x + 2y - 15 = 0$ (B) $2x^2 + 2y^2 - 5x - 18y + 15 = 0$ (C) $5x^2 + 5y^2 - 2x - 24y + 6 = 0$ (D) $5x^2 + 5y^2 + 2x - 16y - 6 = 0$

    ›Reveal solutionSolution

    The equation of a circle passing through the intersection of a given circle and a line is $S + \lambda L = 0$. By using the condition that the chord is the diameter of the new circle, we find the value of $\lambda$ and thus the equation of the required circle. The equation is $\boxed{5x^2 + 5y^2 + 2x - 16y - 6 = 0}$.

    The problem asks for the equation of a circle where a given line segment (a chord of another circle) acts as its diameter. A direct way to approach this is to use the concept of a family of circles.

    The equation of any circle passing through the intersection points of a circle $S=0$ and a line $L=0$ is given by $S + \lambda L = 0$, where $\lambda$ is a constant. This general form represents all circles that share the same two intersection points with the line.

    In this specific case, the chord AB is the line segment connecting the intersection points of the given circle and the given line. The required circle has this chord AB as its diameter. This means the center of the required circle must lie on the line AB. This condition will allow us to determine the value of $\lambda$.

    1. Identify the given circle and chord:

       The equation of the given circle is $x^2 + y^2 - 4y = 0$. Let's call this $S_1 = 0$.

       The equation of the chord AB is $x + 2y - 3 = 0$. Let's call this $L = 0$.

    2. Formulate the equation of the family of circles:

       The required circle passes through the intersection points A and B of $S_1 = 0$ and $L = 0$. Therefore, its equation can be written in the form $S_1 + \lambda L = 0$.

       Substituting the expressions for $S_1$ and $L$:

       $(x^2 + y^2 - 4y) + \lambda (x + 2y - 3) = 0$

       Expanding and rearranging terms to match the general form of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$:

       $x^2 + y^2 + \lambda x + (2\lambda - 4)y - 3\lambda = 0$

    3. Determine the center of the new circle:

       For a circle with the equation $x^2 + y^2 + 2gx + 2fy + c = 0$, the center is $(-g, -f)$.

       Comparing our equation $x^2 + y^2 + \lambda x + (2\lambda - 4)y - 3\lambda = 0$ with the general form:

       $2g = \lambda \implies g = \frac{\lambda}{2}$

       $2f = 2\lambda - 4 \implies f = \lambda - 2$

       So, the center of the new circle is $\left(-\frac{\lambda}{2}, -( \lambda - 2)\right) = \left(-\frac{\lambda}{2}, 2 - \lambda\right)$.

    4. Apply the diameter condition:

       The problem states that AB is the diameter of the new circle. This implies that the center of the new circle must lie on the line AB (the chord). …

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12. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.Let $A(4,3,-2)$, $B(0,-4,2)$, $C(-4,7,6)$ be the vertices of a triangle ABC. If $D(p,q,r)$ is the point of intersection of the bisector of angle A and the side BC, then $2p+q+r=$ (A) $1$ (B) $2$ (C) $3$ (D) $0$

    ›Reveal solutionSolution

    The angle bisector theorem states that the bisector of angle A divides the opposite side BC in the ratio of the lengths of the adjacent sides AB and AC. By calculating these lengths and applying the section formula, we find the coordinates of D and then evaluate the expression $2p+q+r$, which equals $\boxed{1}$.

    The problem asks us to find the value of an expression involving the coordinates of a point D, which is the intersection of the angle bisector of angle A and the side BC of triangle ABC. This immediately brings to mind the Angle Bisector Theorem.

    The Angle Bisector Theorem is a fundamental result in geometry that relates the lengths of the sides of a triangle to the segments created by an angle bisector.

    For a triangle ABC, if AD is the bisector of angle A, where D lies on BC, then the theorem states that the ratio of the length of the segment BD to the length of the segment DC is equal to the ratio of the length of side AB to the length of side AC.

    This can be written as:

    $\frac{BD}{DC} = \frac{AB}{AC}$

    This theorem holds true for triangles in 2D as well as 3D space. Once we determine this ratio, we can use the section formula to find the coordinates of point D, as D divides the side BC in this specific ratio.

    Here's how we apply this concept step-by-step:

    1. Calculate the lengths of sides AB and AC.

       We are given the coordinates of the vertices: $A(4,3,-2)$, $B(0,-4,2)$, and $C(-4,7,6)$.

       We use the distance formula in 3D: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

       Length of AB:

       $AB = \sqrt{(0-4)^2 + (-4-3)^2 + (2-(-2))^2}$

       $AB = \sqrt{(-4)^2 + (-7)^2 + (4)^2}$

       $AB = \sqrt{16 + 49 + 16}$

       $AB = \sqrt{81}$

       $AB = 9$

       Length of AC:

       $AC = \sqrt{(-4-4)^2 + (7-3)^2 + (6-(-2))^2}$

       $AC = \sqrt{(-8)^2 + (4)^2 + (8)^2}$

       $AC = \sqrt{64 + 16 + 64}$

       $AC = \sqrt{144}$

       $AC = 12$

    2. Determine the ratio in which D divides BC.

       According to the Angle Bisector Theorem, the point D divides the side BC in the ratio $AB:AC$.

    $\frac{BD}{DC} = \frac{AB}{AC} = \frac{9}{12} = \frac{3}{4}$

    So, D divides BC in the ratio $3:4$. Let $m=3$ and $n=4$.
    

    3. Use the section formula to find the coordinates of D.

    If a point $D(p,q,r)$ divides the line segment joining $B(x_B, y_B, z_B)$ and $C(x_C, y_C, z_C)$ in the ratio $m:n$, its coordinates are given by: …

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