Q.Find the foot of perpendicular from the point $(2, 3, -8)$ to the line $\dfrac{4-x}{2} = \dfrac{y}{6} = \dfrac{1-z}{3}$. Also, find the perpendicular distance from the given point to the line.

Showing the 12 most recent of 18 on this concept.

1. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.If $L_1$ is a line passing through the point $P(4, -3)$ and perpendicular to the line $3x - 4y + k = 0$, then the distance of $P$ from the line $5x - 3y - 2 = 0$ measured along the line $L_1$ is (A) $5$ (B) $\sqrt{13}$ (C) $\sqrt{41}$ (D) $13$

   ›Reveal solutionSolution

   The key idea is to find the intersection of the given line with the line through P perpendicular to the first line, then compute the distance between P and that intersection. The answer is 5.

   We are asked: the distance of P from the line $5x - 3y - 2 = 0$ measured along the line $L_1$.

   This means: start at P, travel along $L_1$ until you hit the line $5x - 3y - 2 = 0$; the distance you travel is what we want. So we need the intersection point of $L_1$ with that line, then the distance from P to that point.

   ---

   1. Find the equation of $L_1$.

   $L_1$ is perpendicular to $3x - 4y + k = 0$. The slope of that line is $\frac{3}{4}$ (rewrite as $y = \frac{3}{4}x + \frac{k}{4}$).

   A line perpendicular to it has slope $-\frac{4}{3}$ (negative reciprocal).

   $L_1$ passes through $P(4, -3)$, so its equation:

   $y + 3 = -\frac{4}{3}(x - 4)$

   Multiply: $3y + 9 = -4x + 16$

   So $4x + 3y - 7 = 0$.

   Tip

   The constant $k$ in the original line doesn't affect the slope, so it doesn't affect $L_1$'s direction — only its position. That's why we could find $L_1$ without knowing $k$.

   2. Find where $L_1$ meets the line $5x - 3y - 2 = 0$.

   Solve the system:

   $\begin{cases} 4x + 3y = 7 \\ 5x - 3y = 2 \end{cases}$

   Add the equations: $9x = 9 \Rightarrow x = 1$.

   Substitute into $4(1) + 3y = 7 \Rightarrow 3y = 3 \Rightarrow y = 1$. …

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2. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.A plane $\pi_1$ passing through the point $3\vec{i} - 7\vec{j} + 5\vec{k}$ is perpendicular to the vector $\vec{i} + 2\vec{j} - 2\vec{k}$ and another plane $\pi_2$ passing through the point $2\vec{i} + 7\vec{j} - 8\vec{k}$ is perpendicular to the vector $3\vec{i} + 2\vec{j} + 6\vec{k}$. If $p_1$ and $p_2$ are the perpendicular distances from the origin to the planes $\pi_1$ and $\pi_2$ respectively, then $p_1 - p_2 =$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   The key idea is to write each plane in normal form using a point and a normal vector, then compute the perpendicular distance from the origin using the formula $p = \frac{|\vec{a} \cdot \vec{n}|}{|\vec{n}|}$. The difference $p_1 - p_2$ simplifies to $2$, so the correct option is (B).

   Concept and Intuition

   A plane can be defined by a point on it and a normal vector. The distance from the origin to a plane is the absolute value of the scalar projection of any point’s position vector onto the unit normal. Since the normal vectors are given, we can directly compute the distances without finding the full Cartesian equations.

   Step-by-step solution

   1. Equation of plane $\pi_1$ The plane passes through $A(3, -7, 5)$ and has normal $\vec{n}_1 = \vec{i} + 2\vec{j} - 2\vec{k}$. The equation is $\vec{r} \cdot \vec{n}_1 = \vec{a} \cdot \vec{n}_1$, where $\vec{a} = 3\vec{i} - 7\vec{j} + 5\vec{k}$. Compute:

   $\vec{a} \cdot \vec{n}_1 = 3(1) + (-7)(2) + 5(-2) = 3 - 14 - 10 = -21.$

   So $\pi_1: \vec{r} \cdot (\vec{i} + 2\vec{j} - 2\vec{k}) = -21$.

   2. Distance $p_1$ from origin to $\pi_1$ The distance from origin $(0,0,0)$ to plane $\vec{r} \cdot \vec{n} = d$ is $\frac{|d|}{|\vec{n}|}$. Here $d = -21$, $|\vec{n}_1| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$. Hence

   $p_1 = \frac{|-21|}{3} = \frac{21}{3} = 7.$

   3. Equation of plane $\pi_2$ The plane passes through $B(2, 7, -8)$ and has normal $\vec{n}_2 = 3\vec{i} + 2\vec{j} + 6\vec{k}$. Compute $\vec{b} \cdot \vec{n}_2$: $2(3) + 7(2) + (-8)(6) = 6 + 14 - 48 = -28.$ …

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3. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.Let $6x - 3y + 2z - 6 = 0$ be the given plane. If $a, b, c$ are the intercepts made by the plane on $X, Y, Z$ - axes respectively; $l, m, n$ are the direction cosines of a normal drawn to the plane and $p$ is the perpendicular distance from the origin to the plane, then $|al + bm + cn| =$ (A) $p$ (B) $2p$ (C) $3p$ (D) $4p$

   ›Reveal solutionSolution

   The expression $|al + bm + cn|$ equals the perpendicular distance from the origin to the plane, i.e. $p$. The correct option is (A).

   The key idea here is that $a, b, c$ are the intercepts of the plane on the axes, and $l, m, n$ are the direction cosines of the normal. The product $al$ is the projection of the intercept vector along the normal direction — and summing these gives the distance from the origin to the plane.

   Let’s work through it step by step.

   1. Find the intercepts $a, b, c$.

      The plane is $6x - 3y + 2z - 6 = 0$.

      To get the $x$-intercept, set $y = 0, z = 0$:

      $6x - 6 = 0 \implies x = 1$, so $a = 1$.

      For the $y$-intercept, set $x = 0, z = 0$:

      $-3y - 6 = 0 \implies y = -2$, so $b = -2$.

      For the $z$-intercept, set $x = 0, y = 0$:

      $2z - 6 = 0 \implies z = 3$, so $c = 3$.

   2. Find the direction cosines $l, m, n$ of the normal.

      The normal vector to the plane $6x - 3y + 2z - 6 = 0$ is $\vec{n} = (6, -3, 2)$.

      Its magnitude is $|\vec{n}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.

      Hence the direction cosines are:

      $l = \frac{6}{7}, \quad m = \frac{-3}{7}, \quad n = \frac{2}{7}$.

   3. Compute $al + bm + cn$.

      $al = 1 \cdot \frac{6}{7} = \frac{6}{7}$

      $bm = (-2) \cdot \frac{-3}{7} = \frac{6}{7}$

      $cn = 3 \cdot \frac{2}{7} = \frac{6}{7}$

      Sum: $al + bm + cn = \frac{6}{7} + \frac{6}{7} + \frac{6}{7} = \frac{18}{7}$.

   4. Find $p$, the perpendicular distance from the origin to the plane.

      For a plane $Ax + By + Cz + D = 0$, the distance from $(0,0,0)$ is $\frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.

      Here $A = 6, B = -3, C = 2, D = -6$, so

      $p = \frac{|-6|}{\sqrt{36 + 9 + 4}} = \frac{6}{7}$.

   5. Compare $|al + bm + cn|$ with $p$.

      $|al + bm + cn| = \frac{18}{7} = 3 \times \frac{6}{7} = 3p$. …

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4. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If the point P$(x_1, y_1)$ lying on the curve $y = x^2 - x + 1$ is the closest point to the line $y = x - 3$ then the perpendicular distance from P to the line $3x + 4y - 2 = 0$ is (A) $1$ (B) $\dfrac{7}{5}$ (C) $\dfrac{16}{5}$ (D) $4$

   ›Reveal solutionSolution

   The closest point on a curve to a line is where the tangent is parallel to the line; solving gives P(1,1), and its distance to the given line is 1, so option (A).

   We need the point on $y = x^2 - x + 1$ that is closest to the line $y = x - 3$. The key idea: the shortest distance from a curve to a line occurs at a point where the tangent to the curve is parallel to the line. Why? Because if you imagine sliding a line parallel to the given one until it just touches the curve, the point of tangency is the closest point. This is a standard optimization trick — it avoids calculus with distances directly.

   1. Find the slope of the given line.

      The line is $y = x - 3$, so its slope is $1$.

   2. Find the slope of the tangent to the curve.

      The curve is $y = x^2 - x + 1$. Differentiate:

   $\frac{dy}{dx} = 2x - 1.$

   3. Set the tangent slope equal to the line’s slope. For the closest point,

   $2x - 1 = 1 \quad \Rightarrow \quad 2x = 2 \quad \Rightarrow \quad x = 1.$

   4. Find the corresponding y-coordinate. Substitute $x = 1$ into the curve:

   $y = 1^2 - 1 + 1 = 1.$

   So the point is $P(1, 1)$.

   Tip

   Always check that this point actually lies on the curve — it does. Also, the line $y = x - 3$ does not intersect the curve (try solving $x^2 - x + 1 = x - 3$ gives $x^2 - 2x + 4 = 0$, no real roots), so the closest point is indeed a tangency point, not an intersection.

   5. Now find the perpendicular distance from P to the line $3x + 4y - 2 = 0$. …

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5. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The tangent drawn at a point P on the circle $x^2 + y^2 + 6x + 6y - 2 = 0$ cuts the line $5x - 2y + 6 = 0$ at a point Q. If PQ = 5, then a point Q having integral coordinates is (A) $(0, 3)$ (B) $(2, 8)$ (C) $(-2, -2)$ (D) $(-4, -7)$

   ›Reveal solutionSolution

   The key idea is to use the length of the tangent from an external point Q to the circle: $PQ^2 = \text{(power of Q)}$. We find the circle’s center and radius, set up the power condition, and test which given point satisfies $PQ = 5$.

   Concept & Intuition

   When a tangent is drawn from an external point Q to a circle, the segment from Q to the point of tangency P is perpendicular to the radius at P. A classic result (the tangent-secant power theorem) tells us that the square of the length of the tangent from Q equals the power of Q with respect to the circle:

   $PQ^2 = QO^2 - r^2$

   where O is the center and r the radius. So instead of finding P explicitly, we can directly test each candidate Q by checking whether its distance to the center, minus the square of the radius, equals $5^2 = 25$.

   Step-by-step solution

   1. Rewrite the circle equation in standard form

      Given: $x^2 + y^2 + 6x + 6y - 2 = 0$

      Complete the square:

      $(x^2 + 6x) + (y^2 + 6y) = 2$

      $(x^2 + 6x + 9) + (y^2 + 6y + 9) = 2 + 9 + 9$

      $(x + 3)^2 + (y + 3)^2 = 20$

      So center $O = (-3, -3)$ and radius $r = \sqrt{20} = 2\sqrt{5}$.

   2. Apply the tangent length formula

      For any point Q = $(x, y)$, the length of the tangent from Q to the circle is

   $PQ = \sqrt{ (x+3)^2 + (y+3)^2 - 20 }$

   We are told $PQ = 5$, so

   $(x+3)^2 + (y+3)^2 - 20 = 25$

   $(x+3)^2 + (y+3)^2 = 45$

   This is the condition Q must satisfy.

   3. Test each option
      • (A) $(0, 3)$: $(0+3)^2 + (3+3)^2 = 9 + 36 = 45$ ✓
      • (B) $(2, 8)$: $(2+3)^2 + (8+3)^2 = 25 + 121 = 146 \neq 45$ ✗
      • (C) $(-2, -2)$: $(-2+3)^2 + (-2+3)^2 = 1 + 1 = 2 \neq 45$ ✗
      • (D) $(-4, -7)$: …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If a line L passing through the point $A(-2, 4)$ makes an angle of $60^\circ$ with the positive direction of X-axis in anti-clockwise direction and $B(p, q)$ lying in the 3rd quadrant is a point on L at the distance of 6 units from the point A, then $\sqrt{p^2 + q^2 - 8q} =$ (A) $8$ (B) $7$ (C) $9$ (D) $6$

   ›Reveal solutionSolution

   The line through A at 60° has slope √3; using parametric form, point B is 6 units away in the 3rd quadrant; substituting into the expression gives 8.

   We are given a point $A(-2,4)$ and a line $L$ through it making a $60^\circ$ angle with the positive X-axis (counterclockwise). That means the slope is $\tan 60^\circ = \sqrt{3}$. The line also passes through a point $B(p,q)$ in the 3rd quadrant (so $p<0, q<0$), exactly 6 units from $A$. We need $\sqrt{p^2 + q^2 - 8q}$.

   Concept & Intuition

   When a line’s direction is known, the easiest way to locate a point at a given distance along it is to use the parametric form:

   $(x, y) = (x_0 + r\cos\theta,\; y_0 + r\sin\theta)$

   where $r$ is the signed distance from $(x_0, y_0)$. Here $\theta = 60^\circ$, so $\cos 60^\circ = \frac12$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$. The distance is 6, but the sign of $r$ determines which side of A we go. Since B is in the 3rd quadrant (both coordinates negative), we must go in the direction that makes both coordinates decrease from $A(-2,4)$. That means moving opposite to the positive direction of the line — so we take $r = -6$.

   Step-by-step

   1. Parametric coordinates of B Starting from $A(-2,4)$, with $r = -6$:

   $p = -2 + (-6)\cos 60^\circ = -2 - 6\cdot\frac12 = -2 - 3 = -5$

   $q = 4 + (-6)\sin 60^\circ = 4 - 6\cdot\frac{\sqrt{3}}{2} = 4 - 3\sqrt{3}$

   2. Check quadrant

      $p = -5 < 0$.

      $q = 4 - 3\sqrt{3} \approx 4 - 5.196 = -1.196 < 0$.

      So B is indeed in the 3rd quadrant. Good.

   3. Compute the expression

      We need $\sqrt{p^2 + q^2 - 8q}$.

      First, $p^2 = (-5)^2 = 25$. …

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7. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.The distance between the points of concurrency of the two families of straight lines given by $x + (5\lambda + 1)y + 1 - 3\lambda = 0$ and $(5\mu + 2)x - 3y + 3 + 6\mu = 0$ is (A) 4 (B) $\dfrac{2\sqrt{2}}{5}$ (C) $\dfrac{\sqrt{2}}{5}$ (D) 6

   ›Reveal solutionSolution

   Each family of lines passes through a fixed point (its point of concurrency). Find these two fixed points, then compute the distance between them. The distance is $\boxed{\frac{2\sqrt{2}}{5}}$.

   The key idea is that a family of straight lines depending on a parameter (here $\lambda$ or $\mu$) usually represents all lines through a common fixed point. That fixed point is the intersection of any two distinct lines from the family. Once we find the two fixed points, the distance between them is straightforward.

   1. First family: $x + (5\lambda + 1)y + 1 - 3\lambda = 0$ Rewrite it by grouping terms containing $\lambda$ and those that don't:

   $x + y + 1 + \lambda(5y - 3) = 0$

   This is of the form $L_1 + \lambda L_2 = 0$, where $L_1 = x + y + 1 = 0$ and $L_2 = 5y - 3 = 0$.

   For any $\lambda$, the line passes through the intersection of $L_1 = 0$ and $L_2 = 0$.

   Solve $5y - 3 = 0 \Rightarrow y = \frac{3}{5}$.

   Substitute into $x + y + 1 = 0$: $x + \frac{3}{5} + 1 = 0 \Rightarrow x = -\frac{8}{5}$.

   So the fixed point is $P_1 = \left(-\frac{8}{5}, \frac{3}{5}\right)$.

   2. Second family: $(5\mu + 2)x - 3y + 3 + 6\mu = 0$ Group terms: $2x - 3y + 3 + \mu(5x + 6) = 0$ Here $M_1 = 2x - 3y + 3 = 0$ and $M_2 = 5x + 6 = 0$. Solve $5x + 6 = 0 \Rightarrow x = -\frac{6}{5}$. Substitute into $2x - 3y + 3 = 0$: $2\left(-\frac{6}{5}\right) - 3y + 3 = 0 \Rightarrow -\frac{12}{5} + 3 - 3y = 0 \Rightarrow \frac{3}{5} - 3y = 0 \Rightarrow y = \frac{1}{5}$. …

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8. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If Q is the image of the point P(1,1) with respect to the straight line $x + y + 1 = 0$, then the length of the perpendicular drawn from Q to the line $3x - 4y + 3 = 0$ is (A) $\dfrac{5}{2}$ (B) $2$ (C) $1$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   To find the length of the perpendicular from point Q, we first determine the coordinates of Q by reflecting point P(1,1) across the line $x+y+1=0$. Once Q is found, we use the perpendicular distance formula to calculate the distance from Q to the line $3x-4y+3=0$. The final distance is $\boxed{1}$.

   The problem asks us to perform two main tasks: first, find the image of a given point with respect to a line, and second, calculate the perpendicular distance from this image point to another line.

   Concept and Intuition

   1. Image of a point with respect to a line:

      Geometrically, the image of a point P with respect to a line L is the point Q such that L is the perpendicular bisector of the line segment PQ. This means:

      • The line segment PQ is perpendicular to the line L.
      • The midpoint of PQ lies on the line L. These two conditions are used to derive the formula for the image of a point. If P is $(x_1, y_1)$ and the line L is $ax + by + c = 0$, and its image Q is $(x', y')$, then the formula is:

      The image $(x', y')$ of a point $(x_1, y_1)$ with respect to the line $ax + by + c = 0$ is given by:

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

   2. Perpendicular distance from a point to a line:

      The perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is the shortest distance between the point and the line. It is calculated using a standard formula derived from geometry.

      The perpendicular distance $d$ from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by:

      $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

   Now, let's apply these concepts step-by-step.

   Step-by-Step Solution

   1. Find the coordinates of point Q, the image of P(1,1) with respect to the line $x + y + 1 = 0$. Here, the given point is $P(x_1, y_1) = (1,1)$. The line of reflection is $x + y + 1 = 0$. Comparing this with $ax + by + c = 0$, we have $a=1$, $b=1$, and $c=1$. Let the image point be $Q(x', y')$. Using the image formula:

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

   Substitute the values:
   

   $\frac{x' - 1}{1} = \frac{y' - 1}{1} = -2 \frac{(1)(1) + (1)(1) + 1}{1^2 + 1^2}$

   $$ \frac{x' - 1}{1} = \frac{y' - 1}{1} = -2 \frac{1 + 1 + 1}{1 + 1} $$ …
   

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9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If the points $(1,1,\lambda)$ and $(-3,0,1)$ are equidistant from the plane $3x+4y-12z+13=0$, then the values of $\lambda$ are (A) $-1, \dfrac{7}{3}$ (B) $1, -\dfrac{7}{3}$ (C) $-1, -\dfrac{7}{3}$ (D) $1, \dfrac{7}{3}$

   ›Reveal solutionSolution

   Setting the two perpendicular distances equal gives $|20-12\lambda|=8$, whose solutions are $\lambda=1$ and $\lambda=\tfrac{7}{3}$ — option (D).

   Distance formula. The distance from $(x_0,y_0,z_0)$ to $3x+4y-12z+13=0$ is

   $\frac{|3x_0+4y_0-12z_0+13|}{\sqrt{3^2+4^2+12^2}}=\frac{|3x_0+4y_0-12z_0+13|}{13}.$

   Evaluate at each point.

   $\text{At }(1,1,\lambda):\quad 3+4-12\lambda+13=20-12\lambda.$

   $\text{At }(-3,0,1):\quad -9+0-12+13=-8.$

   Equidistant condition (equal denominators): …

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10. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.Let $A\ (4, 3, 5),\ B\ (1, -2, 1),\ C\ (3, 2, 1)$ be the vertices of a triangle $ABC$. If the internal bisector of $\angle BAC$ meet the side $BC$ at $D$, then $CD =$ (A) $\dfrac{\sqrt{5}}{4}$ (B) $\dfrac{3\sqrt{5}}{4}$ (C) $2\sqrt{5}$ (D) $\dfrac{5\sqrt{5}}{2}$

    ›Reveal solutionSolution

    The internal angle bisector theorem gives the ratio in which D divides BC as AB:AC. After computing side lengths, CD is a fraction of BC; the result is $\frac{3\sqrt{5}}{4}$, so option (B).

    We have triangle $ABC$ with vertices $A(4,3,5)$, $B(1,-2,1)$, $C(3,2,1)$. The internal bisector of $\angle BAC$ meets $BC$ at $D$. We need $CD$.

    Concept and intuition

    The key idea is the Angle Bisector Theorem: In any triangle, the internal bisector of an angle divides the opposite side into segments proportional to the adjacent sides.

    So for $\angle BAC$, the point $D$ on $BC$ satisfies

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

    Thus, if we compute the lengths $AB$ and $AC$, we can find the ratio $BD:DC$, and then $CD$ as a fraction of the whole side $BC$.

    Step-by-step solution

    1. Compute $AB$ $A(4,3,5)$, $B(1,-2,1)$

    $AB = \sqrt{(4-1)^2 + (3-(-2))^2 + (5-1)^2} = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}.$

    2. Compute $AC$ $A(4,3,5)$, $C(3,2,1)$

    $AC = \sqrt{(4-3)^2 + (3-2)^2 + (5-1)^2} = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}.$

    3. Apply the Angle Bisector Theorem

    $\frac{BD}{DC} = \frac{AB}{AC} = \frac{5\sqrt{2}}{3\sqrt{2}} = \frac{5}{3}.$

    So $BD : DC = 5 : 3$.

    This means $BC$ is divided into $5+3=8$ equal parts, and $DC$ corresponds to $3$ of those parts.

    4. Compute $BC$ $B(1,-2,1)$, $C(3,2,1)$ $$$$

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11. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.A straight line passes through a point A(2,5) and makes an angle of $45^\circ$ with the positive X-axis when measured in the positive direction. If this straight line intersects the line passing through the points (1,–2) and (3,–4) at B, then AB = (A) $2\sqrt{2}$ (B) $5\sqrt{2}$ (C) $4\sqrt{2}$ (D) $\sqrt{82}$

    ›Reveal solutionSolution

    The line through $A(2,5)$ with slope $\tan45^\circ=1$ is $y=x+3$; the line through $(1,-2),(3,-4)$ is $y=-x-1$. They meet at $B(-2,1)$, giving $AB=\sqrt{32}=4\sqrt2$ — option (C).

    Line through $A$. Angle $45^\circ$ with the positive X-axis gives slope $m_1=\tan45^\circ=1$. Through $A(2,5)$:

    $y-5=1\cdot(x-2)\;\Rightarrow\;y=x+3.$

    Line through the two given points. Slope

    $m_2=\frac{-4-(-2)}{3-1}=\frac{-2}{2}=-1.$

    Through $(1,-2)$:

    $y+2=-1\cdot(x-1)\;\Rightarrow\;y=-x-1.$ …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.$L = x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive, $\alpha$ lies in the fourth quadrant and perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units then $p =$ (A) $5$ (B) $\dfrac{5}{2}$ (C) $10$ (D) $\dfrac{15}{2}$

    ›Reveal solutionSolution

    The line $L$ is perpendicular to $x+y+1=0$, so its normal vector is parallel to $(1,1)$. Using the given distance from $(\sqrt2,\sqrt2)$ and the quadrant condition for $\alpha$, we find $p = 5$, which corresponds to option (A).

    ---

    We are told that $L = x\cos\alpha + y\sin\alpha - p = 0$ is a line, and that it is perpendicular to the line $x + y + 1 = 0$.

    The key idea: the normal vector of $L$ is $(\cos\alpha,\sin\alpha)$, and the normal vector of $x+y+1=0$ is $(1,1)$. The key idea: the normal vector of $L$ is $(\cos\alpha,\sin\alpha)$, and the normal vector of $x+y+1=0$ is $(1,1)$. Two lines are perpendicular exactly when their normal vectors are perpendicular — if $\mathbf{d}_1,\mathbf{d}_2$ are the direction vectors and $\mathbf{n}_1,\mathbf{n}_2$ the normals of the two lines, then $\mathbf{d}_1\cdot\mathbf{n}_1=0$ and $\mathbf{d}_2\cdot\mathbf{n}_2=0$; the lines being perpendicular means $\mathbf{d}_1\cdot\mathbf{d}_2=0$, which in turn forces $\mathbf{n}_1\cdot\mathbf{n}_2=0$. (Quick check: $L_1:x=0$ has normal $(1,0)$ and $L_2:y=0$ has normal $(0,1)$ — both the lines and their normals are perpendicular.)

    Thus for our lines:

    • Normal of $L$: $(\cos\alpha,\sin\alpha)$
    • Normal of $x+y+1=0$: $(1,1)$

    Perpendicular condition:

    $(\cos\alpha,\sin\alpha) \cdot (1,1) = 0 \quad\Rightarrow\quad \cos\alpha + \sin\alpha = 0$

    So $\sin\alpha = -\cos\alpha$, i.e. $\tan\alpha = -1$.

    Now $\alpha$ lies in the fourth quadrant. In the fourth quadrant, $\cos\alpha > 0$, $\sin\alpha < 0$, and $\tan\alpha = -1$ gives $\alpha = -\frac{\pi}{4}$ (or $315^\circ$). So:

    $\cos\alpha = \frac{1}{\sqrt{2}},\quad \sin\alpha = -\frac{1}{\sqrt{2}}$

    Thus the line $L$ becomes:

    $x\cdot\frac{1}{\sqrt{2}} + y\cdot\left(-\frac{1}{\sqrt{2}}\right) - p = 0$

    Multiply through by $\sqrt{2}$:

    $x - y - p\sqrt{2} = 0$

    Now we are told the perpendicular distance from $(\sqrt{2},\sqrt{2})$ to this line is $5$ units. The distance from a point $(x_1,y_1)$ to line $ax+by+c=0$ is:

    $$$$

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