Q.Find the vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines: $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$.

Showing the 12 most recent of 27 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $A(2,1)$, $B(4,k)$, $C(3,4)$ are the vertices of a triangle right angled at $B$ and $k$ is not an odd number, then the equation of the line joining the orthocentre and circumcentre of $\triangle ABC$ is (A) $x + y = 6$ (B) $x - y = 0$ (C) $3x + y = 14$ (D) $x + 3y = 10$

   ›Reveal solutionSolution

   For a right triangle, the orthocentre is at the vertex of the right angle and the circumcentre is at the midpoint of the hypotenuse. Using the right-angle condition at $B$ to find $k$, then locating both centres, the line through them gives equation $x + y = 6$.

   The problem gives three points $A(2,1)$, $B(4,k)$, $C(3,4)$ forming a triangle right-angled at $B$, with $k$ not odd. We need the line joining the orthocentre and circumcentre.

   The key insight: in any right triangle, the orthocentre is exactly the vertex where the right angle occurs. The circumcentre is the midpoint of the hypotenuse. So once we identify which side is the hypotenuse, both centres are trivial to locate — no heavy coordinate geometry needed.

   ---

   1. Use the right-angle condition at $B$ Since $\angle B = 90^\circ$, vectors $BA$ and $BC$ are perpendicular. $BA = A - B = (2-4, 1-k) = (-2, 1-k)$ $BC = C - B = (3-4, 4-k) = (-1, 4-k)$ Perpendicular means dot product = 0:

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

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

   Expand: $(1-k)(4-k) = 4 - k - 4k + k^2 = 4 - 5k + k^2$

   So $2 + 4 - 5k + k^2 = 0 \implies k^2 - 5k + 6 = 0$

   Factor: $(k-2)(k-3) = 0 \implies k = 2$ or $k = 3$.

   The problem says $k$ is not odd, so $k = 2$.

   2. Identify the hypotenuse

      With $k=2$, $B = (4,2)$. The right angle is at $B$, so the hypotenuse is $AC$.

      $A = (2,1)$, $C = (3,4)$.

   3. Locate the orthocentre

      In a right triangle, the orthocentre is the vertex of the right angle.

      So orthocentre $H = B = (4,2)$.

   4. Locate the circumcentre …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the pair of lines joining the origin to the points of intersection of the curve $2x^2 - 5xy - 3y^2 - 10x + 27y - 8 = 0$ and the straight line $ax + y = 1$ are perpendicular to each other and if ‘a’ is not an integer, then $a =$ (A) $\dfrac{3}{2}$ (B) $\dfrac{9}{4}$ (C) $-\dfrac{9}{4}$ (D) $-\dfrac{3}{2}$

   ›Reveal solutionSolution

   The condition that the pair of lines from the origin to the intersection points of a curve and a line are perpendicular leads to a relation between coefficients. Substituting the line into the curve and homogenizing gives a quadratic whose sum of coefficients of $x^2$ and $y^2$ is zero. Solving yields $a = -\frac{9}{4}$.

   The key idea here is that when we want the lines joining the origin to the intersection points of a curve and a line to be perpendicular, we first make the equation of the curve homogeneous by eliminating the constant terms using the line equation. Then the condition for perpendicular lines from the origin is that the sum of the coefficients of $x^2$ and $y^2$ in the resulting homogeneous quadratic is zero.

   Let’s work through it step by step.

   1. Understand the geometry.

      The curve is a conic: $2x^2 - 5xy - 3y^2 - 10x + 27y - 8 = 0$.

      The line is $ax + y = 1$.

      Their intersection gives two points (in general). The lines joining the origin to these two points form a pair of straight lines through the origin. We are told these two lines are perpendicular.

   2. Make the equation homogeneous.

      To get the pair of lines through the origin, we must eliminate the constant and linear terms. We do this by using the line equation to replace every constant term with a multiple of $1$, and then replace $1$ by $(ax + y)$ because on the line, $ax + y = 1$.

      So rewrite the curve as:

   $2x^2 - 5xy - 3y^2 - 10x(1) + 27y(1) - 8(1)^2 = 0$

   Now substitute $1 = ax + y$:

   $2x^2 - 5xy - 3y^2 - 10x(ax + y) + 27y(ax + y) - 8(ax + y)^2 = 0$

   3. Expand carefully.

      First term: $-10x(ax + y) = -10ax^2 - 10xy$

      Second: $27y(ax + y) = 27a xy + 27y^2$

      Third: $-8(ax + y)^2 = -8(a^2 x^2 + 2a xy + y^2) = -8a^2 x^2 - 16a xy - 8y^2$

      Now collect all terms:

      $x^2$ terms: $2x^2 - 10a x^2 - 8a^2 x^2 = (2 - 10a - 8a^2)x^2$

      $xy$ terms: $-5xy - 10xy + 27a xy - 16a xy = (-5 - 10 + 27a - 16a)xy = (-15 + 11a)xy$

      $y^2$ terms: $-3y^2 + 27y^2 - 8y^2 = (16)y^2$

      So the homogeneous equation is:

   $(2 - 10a - 8a^2)x^2 + (-15 + 11a)xy + 16y^2 = 0$

   4. Apply the perpendicular condition. …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The straight line joining the points $(2, -2, 1)$ and $(-2, 1, -1)$ is perpendicular to the plane $\pi_1$ which passes through the point $(1, 1, 1)$ and the equation of $\pi_1$ is $ax + by + cz + d = 0$ (G.C.D of $a, b, c, d = 1$). If $(b, c, d)$ are the direction ratios of a perpendicular drawn to the plane $\pi_2$ which passes through the point $(1, 2, 3)$, then the equation of $\pi_2$ is (A) $3x - 2y + 3z - 8 = 0$ (B) $3x - 2y + 3z + 8 = 0$ (C) $3x + 2y + 3z + 8 = 0$ (D) $3x + 2y + 3z - 8 = 0$

   ›Reveal solutionSolution

   We first find the direction ratios of the line perpendicular to plane $\pi_1$, which become the normal vector components $(a, b, c)$ for $\pi_1$. Using a point on $\pi_1$, we determine its full equation $ax+by+cz+d=0$. Then, we use $(b, c, d)$ from $\pi_1$ as the normal vector for plane $\pi_2$ and, with a point on $\pi_2$, derive its equation. The final equation for $\pi_2$ is $\boxed{3x - 2y + 3z - 8 = 0}$.

   Concept and Intuition

   The core idea here revolves around the relationship between a line and a plane when they are perpendicular, and how to form the equation of a plane.

   1. Normal Vector of a Plane: The equation of a plane is typically given as $Ax + By + Cz + D = 0$. The vector $\vec{n} = (A, B, C)$ is called the normal vector to the plane. This vector is perpendicular to every line lying in the plane.
   2. Line Perpendicular to a Plane: If a straight line is perpendicular to a plane, then the direction ratios of that line are identical to (or proportional to) the direction ratios of the normal vector of the plane. This is a crucial link: the line is the normal direction.
   3. Equation of a Plane: To find the equation of a plane, you need two pieces of information:
      • A point $(x_0, y_0, z_0)$ that lies on the plane.
      • The direction ratios $(A, B, C)$ of a vector normal to the plane. The equation is then $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.

   In this problem, we first use the given line to find the normal vector of $\pi_1$. Once we have the normal vector and a point on $\pi_1$, we can write its equation and identify the coefficients $a, b, c, d$. These coefficients then provide the normal vector for $\pi_2$, allowing us to find its equation using the point it passes through.

   Step-by-Step Solution

   1. Find the direction ratios of the line joining the given points.

      The line passes through $P_1(2, -2, 1)$ and $P_2(-2, 1, -1)$.

      The direction ratios of this line are given by the differences in the coordinates:

      $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (-2 - 2, 1 - (-2), -1 - 1) = (-4, 3, -2)$.

   2. Determine the normal vector for plane $\pi_1$.

      The problem states that this line is perpendicular to plane $\pi_1$. Therefore, the direction ratios of the line are the direction ratios of the normal vector to $\pi_1$.

      Let the normal vector to $\pi_1$ be $\vec{n_1} = (a, b, c)$.

      So, we can take $(a, b, c)$ as $(-4, 3, -2)$.

   3. Form the equation of plane $\pi_1$.

      Plane $\pi_1$ passes through the point $(1, 1, 1)$ and has a normal vector $\vec{n_1} = (-4, 3, -2)$.

      Using the formula $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$:

      $-4(x - 1) + 3(y - 1) - 2(z - 1) = 0$

      $-4x + 4 + 3y - 3 - 2z + 2 = 0$

      $-4x + 3y - 2z + 3 = 0$

   4. Identify the coefficients $a, b, c, d$ for $\pi_1$.

      The equation of $\pi_1$ is given as $ax + by + cz + d = 0$.

      Comparing this with $-4x + 3y - 2z + 3 = 0$, we get:

      $a = -4$, $b = 3$, $c = -2$, $d = 3$.

      The problem states that the G.C.D of $a, b, c, d = 1$. Here, G.C.D($-4, 3, -2, 3$) is $1$, so these values are valid. …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.A line L perpendicular to the line $5x - 12y + 6 = 0$ makes positive intercept on the Y-axis. If the distance from the origin to the line L is 2 units and the angle made by the perpendicular drawn from the origin to the line L with positive X-axis is $\theta$, then $\tan \theta + \cot \theta =$ (A) $\dfrac{25}{12}$ (B) $\dfrac{625}{168}$ (C) $\dfrac{169}{60}$ (D) $\dfrac{1681}{360}$

   ›Reveal solutionSolution

   The key idea is to use the condition of perpendicularity to get the slope of L, then use the distance-from-origin condition to fix its intercept, and finally relate the angle of the perpendicular from the origin to the line’s geometry. The required value is $\frac{169}{60}$.

   The line we start with is $5x - 12y + 6 = 0$. Its slope is $\frac{5}{12}$. A line perpendicular to it has slope $m = -\frac{12}{5}$, because the product of slopes of perpendicular lines is $-1$.

   Since L makes a positive intercept on the Y-axis, its equation can be written as

   $y = -\frac{12}{5}x + c$, with $c > 0$.

   In standard form: $12x + 5y - 5c = 0$ (multiplying through by 5 and rearranging).

   Now, the distance from the origin to L is given as 2 units. For a line $Ax + By + C = 0$, the distance from $(0,0)$ is $\frac{|C|}{\sqrt{A^2 + B^2}}$. Here $A = 12$, $B = 5$, $C = -5c$. So:

   $\frac{|-5c|}{\sqrt{12^2 + 5^2}} = \frac{5c}{13} = 2$

   Thus $c = \frac{26}{5}$. So L is $12x + 5y - 26 = 0$. …

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Every point on the curve $3x + 2y - 3xy = 0$ is the centroid of a triangle formed by the coordinate axes and a line (L) intersecting both the coordinate axes. Then all such lines (L) (A) are parallel (B) are concurrent (C) intersect each other at different points (D) are perpendicular to the tangents to the curve

   ›Reveal solutionSolution

   The key idea is that the centroid of the triangle formed by the axes and a line L is given by a point on the curve; by expressing the line’s intercepts in terms of the centroid, we find that all such lines pass through a fixed point, so they are concurrent. The correct option is (B).

   ---

   Concept and Intuition

   We are given a curve: $3x + 2y - 3xy = 0$. Every point $(h, k)$ on this curve is the centroid of a triangle formed by the coordinate axes and a line $L$ that cuts both axes. The triangle’s vertices are $(0,0)$, $(a,0)$ (x-intercept of L), and $(0,b)$ (y-intercept of L). The centroid of this triangle is $\left( \frac{a}{3}, \frac{b}{3} \right)$. So if $(h,k)$ is on the curve, then $h = a/3$ and $k = b/3$, meaning $a = 3h$ and $b = 3k$. The line L has equation $\frac{x}{a} + \frac{y}{b} = 1$, i.e., $\frac{x}{3h} + \frac{y}{3k} = 1$. Since $(h,k)$ lies on the given curve, we can substitute to find a relationship that all such lines satisfy — and that will tell us whether they are parallel, concurrent, etc.

   ---

   Step-by-step reasoning

   1. Set up the triangle and centroid Let line L have x-intercept $a$ and y-intercept $b$ (both nonzero). The triangle’s vertices are $(0,0)$, $(a,0)$, $(0,b)$. The centroid is the average of the vertices:

   $\left( \frac{0+a+0}{3}, \frac{0+0+b}{3} \right) = \left( \frac{a}{3}, \frac{b}{3} \right).$

   This point lies on the curve $3x + 2y - 3xy = 0$. So let $x = a/3$, $y = b/3$.

   2. Substitute into the curve equation

   $3\left(\frac{a}{3}\right) + 2\left(\frac{b}{3}\right) - 3\left(\frac{a}{3}\right)\left(\frac{b}{3}\right) = 0.$

   Simplify:

   $a + \frac{2b}{3} - \frac{ab}{3} = 0.$

   Multiply through by 3:

   $3a + 2b - ab = 0.$

   This is the relation between the intercepts $a$ and $b$ of any such line L.

   3. Write the equation of line L The line with intercepts $a$ and $b$ is:

   $\frac{x}{a} + \frac{y}{b} = 1.$

   Using the relation $3a + 2b - ab = 0$, we can solve for one intercept in terms of the other. For instance, solve for $b$:

   $2b - ab = -3a \quad\Rightarrow\quad b(2 - a) = -3a \quad\Rightarrow\quad b = \frac{3a}{a - 2} \quad (\text{provided } a \neq 2).$

   (If $a = 2$, then from $3a + 2b - ab = 0$ we get $6 + 2b - 2b = 6 = 0$, impossible, so $a \neq 2$.)

   4. Substitute $b$ into the line equation

   $\frac{x}{a} + \frac{y}{\frac{3a}{a-2}} = 1.$

   Simplify the second term:

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

   Multiply through by $3a$:

   $3x + (a-2)y = 3a.$

   Rearranging:

   $3x - 2y + a y - 3a = 0 \quad\Rightarrow\quad 3x - 2y + a(y - 3) = 0.$

   5. Interpret the family of lines …

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6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If the foot of the perpendicular drawn from the point $(2,0,-3)$ to the plane $\pi$ is $(1,-2,0)$ and the equation of the plane $\pi$ is $ax+by-3z+d=0$ then $a+b+d=$ (A) $0$ (B) $1$ (C) $6$ (D) $2$

   ›Reveal solutionSolution

   The foot of the perpendicular gives both a point on the plane and the direction of the normal; using these we find the plane’s equation and then compute $a+b+d = 2$.

   We are given a point $P(2,0,-3)$ and the foot $F(1,-2,0)$ of the perpendicular from $P$ to the plane $\pi: ax+by-3z+d=0$.

   The key idea: the vector $\overrightarrow{PF}$ is parallel to the normal vector of the plane, and $F$ lies on the plane. This gives two equations to determine the unknown coefficients.

   1. Find the normal direction from the perpendicular The vector from $P$ to $F$ is

   $\overrightarrow{PF} = (1-2,\,-2-0,\,0-(-3)) = (-1,\,-2,\,3).$

   Since $F$ is the foot of the perpendicular, $\overrightarrow{PF}$ is parallel to the plane’s normal vector $\mathbf{n} = (a,b,-3)$.

   Therefore the components must be proportional:

   $\frac{a}{-1} = \frac{b}{-2} = \frac{-3}{3}.$

   The last ratio gives $\frac{-3}{3} = -1$. Hence

   $a = (-1)(-1) = 1,\qquad b = (-2)(-1) = 2.$

   2. Use the foot point to find $d$ The point $F(1,-2,0)$ lies on the plane, so it satisfies

   $a(1) + b(-2) - 3(0) + d = 0.$

   Substituting $a=1,\; b=2$:

   $1(1) + 2(-2) + d = 0 \quad\Rightarrow\quad 1 - 4 + d = 0 \quad\Rightarrow\quad d = 3.$ …

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7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If the foot of the perpendicular drawn from the point $(2, 0, -3)$ to the plane $\pi$ is $(1, -2, 0)$ and the equation of the plane $\pi$ is $ax + by - 3z + d = 0$ then $a + b + d =$ (A) $2$ (B) $1$ (C) $6$ (D) $0$

   ›Reveal solutionSolution

   The foot of the perpendicular gives both a point on the plane and the direction of the normal; using these we find the plane’s equation and then compute $a+b+d = 1$.

   We are given a point $P(2,0,-3)$ and the foot $F(1,-2,0)$ of the perpendicular from $P$ to the plane $\pi: ax+by-3z+d=0$.

   The key idea: The vector from $P$ to $F$ is parallel to the plane’s normal vector $(a,b,-3)$. Also, $F$ lies on the plane, so its coordinates satisfy the plane equation. This gives us two equations to find $a,b,d$.

   1. Find the direction of the normal The foot $F$ is the projection of $P$ onto the plane, so $\overrightarrow{PF}$ is perpendicular to the plane.

   $\overrightarrow{PF} = (1-2,\,-2-0,\,0-(-3)) = (-1,\,-2,\,3)$

   This vector is parallel to the normal $\vec{n} = (a,b,-3)$. Therefore, the components are proportional:

   $\frac{a}{-1} = \frac{b}{-2} = \frac{-3}{3}$

   From the last ratio: $\frac{-3}{3} = -1$. So:

   $a = (-1)(-1) = 1, \quad b = (-2)(-1) = 2$

   2. Use that $F$ lies on the plane Substitute $F(1,-2,0)$ into $ax+by-3z+d=0$: $1(1) + 2(-2) - 3(0) + d = 0 \quad\Rightarrow\quad 1 - 4 + d = 0 \quad\Rightarrow\quad d = 3$ …

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8. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The number of values of ‘k’ for which the points $(-4, 9, k)$, $(-1, 6, k)$, $(0, 7, 10)$ form a right-angled isosceles triangle is (A) 0 (B) 4 (C) 1 (D) 2

   ›Reveal solutionSolution

   We use the distance formula and the conditions for a right‑angled isosceles triangle (two equal sides and a right angle) to set up equations in $k$. Solving yields exactly two values of $k$, so the answer is 2.

   We are given three points:

   $A(-4,9,k)$, $B(-1,6,k)$, $C(0,7,10)$.

   For them to form a right‑angled isosceles triangle, two sides must be equal in length and the angle between them must be $90^\circ$. The most direct way is to compute all three squared distances (to avoid square roots) and then impose the two conditions.

   ---

   1. Compute the squared distances

   $AB^2 = (-1+4)^2 + (6-9)^2 + (k-k)^2 = 3^2 + (-3)^2 + 0 = 9+9 = 18.$

   $AC^2 = (0+4)^2 + (7-9)^2 + (10-k)^2 = 4^2 + (-2)^2 + (10-k)^2 = 16+4+(10-k)^2 = 20 + (10-k)^2.$

   $BC^2 = (0+1)^2 + (7-6)^2 + (10-k)^2 = 1^2 + 1^2 + (10-k)^2 = 2 + (10-k)^2.$

   Notice that $AB^2$ is constant (18), while $AC^2$ and $BC^2$ depend on $k$.

   2. Identify which sides could be equal

      Since $AB^2$ is fixed, the equal sides must involve $AB$ or not. There are three possibilities for the two equal sides:

      • $AB = AC$
      • $AB = BC$
      • $AC = BC$

      We also need a right angle. The right angle is opposite the hypotenuse, so the two equal sides must be the legs (the sides forming the right angle). That means the equal sides are the ones meeting at the right‑angle vertex.

   3. Case 1: Right angle at $A$

      Then legs are $AB$ and $AC$, so $AB = AC$ and $BC$ is the hypotenuse.

   $AB^2 = AC^2 \quad\Rightarrow\quad 18 = 20 + (10-k)^2 \quad\Rightarrow\quad (10-k)^2 = -2,$

   impossible. No solution.

   4. Case 2: Right angle at $B$ Then legs are $AB$ and $BC$, so $AB = BC$ and $AC$ is the hypotenuse.

   $18 = 2 + (10-k)^2 \quad\Rightarrow\quad (10-k)^2 = 16 \quad\Rightarrow\quad 10-k = \pm 4.$

   So $k = 6$ or $k = 14$.

   Check the right angle: For a right angle at $B$, we need $AB^2 + BC^2 = AC^2$. …

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9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $(1,1),(-2,2),(2,-2)$ are $3$ points on a circle $S$, then the perpendicular distance from the centre of the circle $S$ to the line $3x-4y+1=0$ is (A) $\dfrac{1}{2}$ (B) $1$ (C) $\dfrac{23}{10}$ (D) $2$

   ›Reveal solutionSolution

   The three given points determine a unique circle; we find its centre by solving the perpendicular bisector intersection, then compute the distance from that centre to the given line. The result is 1, so the correct option is (B).

   Concept & Intuition

   Any three non-collinear points define a unique circle. The centre is the intersection of the perpendicular bisectors of any two chords. Once we have the centre, the distance to a line is a straightforward formula application. The trick is to avoid messy algebra by noticing symmetries in the given points.

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   1. Find the circle’s centre using perpendicular bisectors Let the points be $A(1,1)$, $B(-2,2)$, $C(2,-2)$. Compute the midpoint of $AB$:

   $M_{AB} = \left( \frac{1 + (-2)}{2}, \frac{1+2}{2} \right) = \left( -\frac12, \frac32 \right)$

   Slope of $AB$:

   $m_{AB} = \frac{2-1}{-2-1} = \frac{1}{-3} = -\frac13$

   So the perpendicular bisector of $AB$ has slope $3$ (negative reciprocal). Its equation through $M_{AB}$:

   $y - \frac32 = 3\left(x + \frac12\right) \quad\Rightarrow\quad y = 3x + 3$

   2. Now do the same for chord $AC$ Midpoint of $AC$:

   $M_{AC} = \left( \frac{1+2}{2}, \frac{1+(-2)}{2} \right) = \left( \frac32, -\frac12 \right)$

   Slope of $AC$:

   $m_{AC} = \frac{-2-1}{2-1} = -3$

   Perpendicular slope = $\frac13$. Equation through $M_{AC}$:

   $y + \frac12 = \frac13\left(x - \frac32\right) \quad\Rightarrow\quad y = \frac13 x - 1$

   3. Intersect the two bisectors to get the centre Set $3x + 3 = \frac13 x - 1$: Multiply by 3: $9x + 9 = x - 3$ → $8x = -12$ → $x = -\frac32$. Then $y = 3\left(-\frac32\right) + 3 = -\frac92 + 3 = -\frac32$. So centre $O = \left(-\frac32, -\frac32\right)$. …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.A rectangle is formed by the lines $x = 4$, $x = -2$, $y = 5$, $y = -2$ and a circle is drawn through the vertices of this rectangle. The pole of the line $y + 2 = 0$ with respect to this circle is (A) $\left(1, \dfrac{-85}{14}\right)$ (B) $\left(1, \dfrac{-32}{7}\right)$ (C) $(-2, -2)$ (D) $(1, -4)$

    ›Reveal solutionSolution

    The rectangle’s vertices define a circle whose center is the rectangle’s center (1, 1.5) and radius is half the diagonal. The pole of the horizontal line y = –2 is found via the polar formula, giving (1, –85/14), which matches option (A).

    Concept & Intuition

    The problem gives a rectangle from x = –2 to x = 4 and y = –2 to y = 5. The circle through its four vertices is the circumcircle of the rectangle. For any rectangle, the circumcenter is the intersection of the diagonals — i.e., the rectangle’s center. Once we have the circle’s equation, the pole of a given line with respect to that circle is found using the standard polar relation: if the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$, then the pole of $lx + my + n = 0$ satisfies $x_1 x + y_1 y + g(x + x_1) + f(y + y_1) + c = 0$ matching the line. Here the line is $y + 2 = 0$, so we can solve directly.

    ---

    Step-by-step solution

    1. Find the center of the rectangle (and the circle)

       The rectangle’s x‑range: from –2 to 4 → center x‑coordinate = $\frac{-2 + 4}{2} = 1$.

       y‑range: from –2 to 5 → center y‑coordinate = $\frac{-2 + 5}{2} = 1.5 = \frac{3}{2}$.

       So the circle’s center is $C(1, \frac{3}{2})$.

    2. Find the radius

       The distance from the center to any vertex, say $(4,5)$:

    $R = \sqrt{(4-1)^2 + \left(5 - \frac{3}{2}\right)^2} = \sqrt{3^2 + \left(\frac{7}{2}\right)^2} = \sqrt{9 + \frac{49}{4}} = \sqrt{\frac{36}{4} + \frac{49}{4}} = \sqrt{\frac{85}{4}} = \frac{\sqrt{85}}{2}.$

    3. Write the circle equation in standard form Center $(1, \frac{3}{2})$, radius $\frac{\sqrt{85}}{2}$:

    $(x-1)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{85}{4}.$

    Expand:

    $x^2 - 2x + 1 + y^2 - 3y + \frac{9}{4} = \frac{85}{4}.$

    $x^2 + y^2 - 2x - 3y + \left(1 + \frac{9}{4}\right) = \frac{85}{4}.$

    $x^2 + y^2 - 2x - 3y + \frac{13}{4} = \frac{85}{4}.$

    Subtract $\frac{85}{4}$:

    $x^2 + y^2 - 2x - 3y + \frac{13}{4} - \frac{85}{4} = 0$

    $x^2 + y^2 - 2x - 3y - \frac{72}{4} = 0$

    $x^2 + y^2 - 2x - 3y - 18 = 0.$

    So $2g = -2 \Rightarrow g = -1$, $2f = -3 \Rightarrow f = -\frac{3}{2}$, $c = -18$.

    4. Pole of a line with respect to a circle For circle $x^2 + y^2 + 2gx + 2fy + c = 0$, the polar of point $(x_1, y_1)$ is

    $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0.$

    Conversely, the pole of line $lx + my + n = 0$ satisfies that this polar equation is proportional to the given line.

    Our line is $y + 2 = 0$, i.e., $0\cdot x + 1\cdot y + 2 = 0$. So $l=0, m=1, n=2$.

    5. Set up the pole condition Let the pole be $(x_1, y_1)$. The polar is:

    $x x_1 + y y_1 + (-1)(x + x_1) + \left(-\frac{3}{2}\right)(y + y_1) - 18 = 0.$

    Simplify:

    $x x_1 + y y_1 - x - x_1 - \frac{3}{2}y - \frac{3}{2}y_1 - 18 = 0.$

    Group x‑terms, y‑terms, constant:

    $(x_1 - 1)x + \left(y_1 - \frac{3}{2}\right)y + \left(-x_1 - \frac{3}{2}y_1 - 18\right) = 0.$

    This must be proportional to $0\cdot x + 1\cdot y + 2 = 0$.

    6. Match coefficients For the x‑coefficient: $x_1 - 1 = 0 \Rightarrow x_1 = 1$. For the y‑coefficient: $y_1 - \frac{3}{2}$ must be proportional to 1, and the constant term proportional to 2. Let the proportionality constant be $k$. Then:

    $y_1 - \frac{3}{2} = k \cdot 1, \quad -x_1 - \frac{3}{2}y_1 - 18 = k \cdot 2.$

    Substitute $x_1 = 1$:

    $-1 - \frac{3}{2}y_1 - 18 = 2k.$

    Also $k = y_1 - \frac{3}{2}$. Substitute:

    $-19 - \frac{3}{2}y_1 = 2\left(y_1 - \frac{3}{2}\right) = 2y_1 - 3.$

    Bring terms:

    $-19 + 3 = 2y_1 + \frac{3}{2}y_1 \quad\Rightarrow\quad -16 = \frac{7}{2}y_1.$

    So $y_1 = -\frac{32}{7}$.

    7. Check the result

       The pole is $(1, -\frac{32}{7})$. But $-\frac{32}{7} = -\frac{64}{14}$, and option (A) is $\left(1, -\frac{85}{14}\right)$. Wait — that’s different. Let’s re-check the constant term carefully.

       Re‑evaluate constant term in polar:

       Original polar:

    $x x_1 + y y_1 - x - x_1 - \frac{3}{2}y - \frac{3}{2}y_1 - 18 = 0.$

    Constant term (no x or y) is $-x_1 - \frac{3}{2}y_1 - 18$.

    With $x_1=1$: constant = $-1 - \frac{3}{2}y_1 - 18 = -19 - \frac{3}{2}y_1$.

    The line is $0\cdot x + 1\cdot y + 2 = 0$. So we require:

    $\frac{x_1 - 1}{0} = \frac{y_1 - \frac{3}{2}}{1} = \frac{-19 - \frac{3}{2}y_1}{2}.$ …

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11. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.$\beta$ is the angle made by the perpendicular drawn from origin to the line $L \equiv x + y - 2 = 0$ with the positive X-axis in the anticlockwise direction. If '$a$' is the X-intercept of the line $L = 0$ and $p$ is the perpendicular distance from the origin to the line $L = 0$, then $a \tan \beta + p^2 =$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

    ›Reveal solutionSolution

    The key is to interpret $\beta$ as the angle of the perpendicular from the origin to the line, then compute $a$, $\tan\beta$, and $p$ directly from the line equation $x+y-2=0$; the expression simplifies to $2$, so the correct option is (B).

    ---

    We start with the line $L: x + y - 2 = 0$.

    We are given three quantities:

    • $a$ = the X-intercept of the line.
    • $p$ = the perpendicular distance from the origin to the line.
    • $\beta$ = the angle that the perpendicular from the origin to the line makes with the positive X-axis (measured anticlockwise).

    We need $a \tan \beta + p^2$.

    ---

    Concept and intuition:

    For a line, the perpendicular from the origin has a clear geometric meaning: its direction is given by the normal vector of the line. The angle $\beta$ is simply the direction angle of that normal vector. The X-intercept $a$ is found by setting $y=0$, and $p$ is given by the standard distance formula. Once we compute these, the expression becomes straightforward arithmetic.

    ---

    Step-by-step solution:

    1. Find the X-intercept $a$. The X-intercept occurs where $y = 0$. Substitute into $x + y - 2 = 0$:

    $x + 0 - 2 = 0 \quad\Rightarrow\quad x = 2.$

    So $a = 2$.

    2. Find the perpendicular distance $p$ from the origin to the line. For a line $Ax + By + C = 0$, the distance from $(0,0)$ is

    $p = \frac{|C|}{\sqrt{A^2 + B^2}}.$

    Here $A = 1$, $B = 1$, $C = -2$, so

    $p = \frac{|-2|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}.$

    3. Find $\tan \beta$. The perpendicular from the origin to the line is along the normal vector of the line. The line $x + y - 2 = 0$ has normal vector $\vec{n} = (1, 1)$. The angle $\beta$ that this vector makes with the positive X-axis (anticlockwise) satisfies …

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12. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.If Q and R are the images of the point P(2,3) with respect to the lines $x-y+2=0$ and $2x+y-2=0$ respectively, then Q and R lie on (A) the same side of the line $2x+y-2=0$ (B) the opposite sides of the line $2x-y-2=0$ (C) the same side of the line $x+y+2=0$ (D) the opposite sides of the line $x-y+2=0$

    ›Reveal solutionSolution

    Reflect $P(2,3)$ across the two given lines to get $Q$ and $R$, then test each option by checking the sign of the corresponding line expression at $Q$ and $R$. Q and R turn out to lie on the same side of $x+y+2=0$ — option (C).

    We are given point $P(2,3)$ and two lines:

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

    We reflect $P$ across $L_1$ to get $Q$, and across $L_2$ to get $R$, then check the four options.

    ---

    Concept and Intuition

    The reflection of a point across a line is found using the standard foot-of-perpendicular formula. Once $Q$ and $R$ are known, their position relative to any line is found by substituting their coordinates into the line's equation: points on the same side give the same sign, points on opposite sides give opposite signs.

    ---

    Step-by-step solution

    1. Find $Q$, the reflection of $P(2,3)$ across $L_1: x - y + 2 = 0$.

       For a line $ax+by+c=0$, the reflection $(x',y')$ of $(x_1,y_1)$ satisfies

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

    For $L_1$: $a=1, b=-1, c=2$.

    $ax_1+by_1+c = 2-3+2=1, \qquad a^2+b^2=2, \qquad -2\cdot\frac{1}{2}=-1.$

    $x'-2 = 1\cdot(-1)=-1 \Rightarrow x'=1, \qquad y'-3=(-1)(-1)=1 \Rightarrow y'=4.$

    So $Q=(1,4)$.

    2. Find $R$, the reflection of $P(2,3)$ across $L_2: 2x+y-2=0$.

       For $L_2$: $a=2,b=1,c=-2$.

    $ax_1+by_1+c = 4+3-2=5, \qquad a^2+b^2=5, \qquad -2\cdot\frac{5}{5}=-2.$

    $x'-2=2(-2)=-4 \Rightarrow x'=-2, \qquad y'-3=1(-2)=-2 \Rightarrow y'=1.$

    So $R=(-2,1)$.

    3. Check each option by evaluating the relevant line expression at $Q$ and $R$.

       • (A) "same side of $2x+y-2=0$": $f(Q)=2(1)+4-2=4>0$, $f(R)=2(-2)+1-2=-5<0$ — opposite signs, so (A) is false. …

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