Q.Find the equation of a line drawn perpendicular to the line $\dfrac{x}{4} + \dfrac{y}{6} = 1$ through the point, where it meets the y-axis.

Showing the 12 most recent of 13 on this concept.

1. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.If the foot of the perpendicular drawn from the origin to the line $y = mx + c$ is $(1,1)$ then the value of $m$ and $c$ are, respectively, (A) 1 and 2 (B) 1 and -2 (C) -1 and 2 (D) 2 and 2 (E) -1 and -2

   ›Reveal solutionSolution

   Use two conditions: the foot lies on the line, and the origin-to-foot segment is perpendicular to the line.

   The foot of the perpendicular $(1,1)$ lies on $y=mx+c$: $1=m+c$. …

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2. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.The line $y=5x+7$ is perpendicular to the line joining the points $(2,12)$ and $(12,k)$. Then the value of $k$ is equal to (A) $12$ (B) $-12$ (C) $8$ (D) $-8$ (E) $10$

   ›Reveal solutionSolution

   Perpendicular slopes multiply to $-1$.

   The line $y=5x+7$ has slope $5$. The segment joining $(2,12)$ and $(12,k)$ has slope $\dfrac{k-12}{10}$. Perpendicularity requires …

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3. KEAM 2024Set eng-2024-06074 marksMCQ
   Q.The equation of the line perpendicular to the line $7x-5y=11$ and passing through $(7,-9)$ is (A) $5x+7y+28=0$ (B) $5x+7y-28=0$ (C) $5x+7y+38=0$ (D) $5x+7y-38=0$ (E) $5x-7y+28=0$

   ›Reveal solutionSolution

   A line perpendicular to $7x-5y=11$ has the form $5x+7y=c$. Passing through $(7,-9)$: $5(7)+7(-9)=35-63=-28$, so $5x+7y+28=0$.

   The given line $7x-5y=11$ has normal-swapped perpendicular family $5x+7y=c$ (swap coefficients and change one sign). …

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4. KEAM 2024Set eng-2024-06074 marksMCQ
   Q.If $(a,-6)$ lies on the perpendicular bisector of the line segment joining $(-2,-1)$ and $(4,-13)$, then the value of $a$ is equal to (A) $1$ (B) $-2$ (C) $2$ (D) $-3$ (E) $3$

   ›Reveal solutionSolution

   A point on the perpendicular bisector is equidistant from the endpoints. Setting distances from $(a,-6)$ to $(-2,-1)$ and $(4,-13)$ equal: $(a+2)^2+25=(a-4)^2+49\Rightarrow 12a=36\Rightarrow a=3$.

   The perpendicular bisector consists of points equidistant from $(-2,-1)$ and $(4,-13)$. So

   $(a+2)^2+(-6+1)^2=(a-4)^2+(-6+13)^2.$ …

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5. KEAM 2024Set eng-2024-06084 marksMCQ
   Q.If $P(-3,4)$ and $Q(3,1)$ are points on a straight line, then the slope of the straight line perpendicular to PQ is (A) $1$ (B) $-2$ (C) $2$ (D) $-1$ (E) $\sqrt{3}$

   ›Reveal solutionSolution

   Find the slope of $PQ$, then take its negative reciprocal.

   For $P(-3,4)$ and $Q(3,1)$:

   $m_{PQ}=\frac{1-4}{3-(-3)}=\frac{-3}{6}=-\frac12.$ …

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6. KEAM 2024Set eng-2024-06084 marksMCQ
   Q.The equation of the straight line passing through the point $(1,1)$ and perpendicular to the line $x+y=5$, is (A) $x-y=2$ (B) $x-y=0$ (C) $x-y=-2$ (D) $x+y=2$ (E) $x+y=0$

   ›Reveal solutionSolution

   Perpendicular to slope $-1$ has slope $1$; use point-slope through $(1,1)$.

   The line $x+y=5$ has slope $-1$. A line perpendicular to it has slope $+1$. …

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7. KEAM 2024Set eng-2024-06094 marksMCQ
   Q.The equation of the line passing through the point $(1,2)$ and perpendicular to the line $x + y + 1 = 0$ is (A) $x + y + 1 = 0$ (B) $y - x + 1 = 0$ (C) $y - x - 1 = 0$ (D) $y - x + 2 = 0$ (E) $y - x - 2 = 0$

   ›Reveal solutionSolution

   Perpendicular slope is the negative reciprocal, then use point-slope.

   The line $x+y+1=0$ has slope $-1$, so a perpendicular line has slope $+1$.

   Through $(1,2)$: …

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8. KEAM 2023Set eng-2023-P2-B24 marksMCQ
   Q.If $(2,-6),(5,2)$ and $(-2,2)$ constitute the vertices of a triangle, then the line joining the origin and its orthocentre is (A) $x + 4y = 0$ (B) $x - 4y = 0$ (C) $4x - y = 0$ (D) $4x + y = 0$ (E) $x - y = 0$

   ›Reveal solutionSolution

   The orthocentre is $(2,\tfrac12)$, so the line through it and the origin is $x-4y=0$.

   Concept and Intuition

   The orthocentre is the intersection of the altitudes. Since side $BC$ is horizontal, one altitude is vertical, which simplifies the work.

   Step-by-Step Solution

   1. $A(2,-6),B(5,2),C(-2,2)$. $BC$ is horizontal ($y=2$), so the altitude from $A$ is the vertical line $x=2$.
   2. Slope of $AC=\dfrac{2-(-6)}{-2-2}=-2$, so the altitude from $B$ has slope $\tfrac12$: $y-2=\tfrac12(x-5)$. …

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9. KEAM 2023Set eng-2023-P2-B24 marksMCQ
   Q.The line perpendicular to $4x - 5y + 1 = 0$ and passing through the point of intersection of the straight lines $x + 2y - 10 = 0$ and $2x + y + 5 = 0$ is (A) $5x + 4y = 0$ (B) $y + \dfrac{5}{4}x = \dfrac{50}{3}$ (C) $5x + 4y = 1$ (D) $y + \dfrac{5}{4}x = -\dfrac{50}{3}$ (E) $4x + 5y = 0$

   ›Reveal solutionSolution

   The required line is $5x+4y=0$.

   Concept and Intuition

   Find the intersection point of the two given lines, then use the perpendicular-slope condition to the line $4x-5y+1=0$.

   Step-by-Step Solution

   1. Solve $x+2y=10$ and $2x+y=-5$: $x=-\tfrac{20}{3},\ y=\tfrac{25}{3}$.
   2. $4x-5y+1=0$ has slope $\tfrac45$; the perpendicular slope is $-\tfrac54$.
   3. Line: $y-\tfrac{25}{3}=-\tfrac54\left(x+\tfrac{20}{3}\right)$. …

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10. KEAM 2022Set eng-2022-P2-B14 marksMCQ
    Q.If the line $y = mx + c$ is perpendicular to $y = 1 + x$ and passes through the point $(1, 2)$, then the value of $c$ is equal to (A) $1$ (B) $-1$ (C) $-3$ (D) $3$ (E) $0$

    ›Reveal solutionSolution

    $c=3$.

    Concept and Intuition

    Perpendicular slopes multiply to $-1$; the given line has slope 1, so $m=-1$. Then substitute the point to find $c$.

    Step-by-Step Solution

    1. Slope of $y=1+x$ is 1, so $m=-1$.
    2. Line: $y=-x+c$.
    3. Through $(1,2)$: $2=-1+c\Rightarrow c=3$.

    Common Mistakes …

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11. KEAM 2022Set eng-2022-P2-B14 marksMCQ
    Q.The normal to the curve $y = \sqrt{x}$ at the point $(25, 5)$ intersects the $y$-axis at (A) $(0, 245)$ (B) $(0, 255)$ (C) $(255, 0)$ (D) $(245, 0)$ (E) $(0, 100)$

    ›Reveal solutionSolution

    The normal meets the $y$-axis at $(0,255)$.

    Concept and Intuition

    The normal's slope is the negative reciprocal of the tangent slope; find where it crosses $x=0$.

    Step-by-Step Solution

    1. $y=\sqrt{x}\Rightarrow y'=\frac{1}{2\sqrt{x}}$; at $(25,5)$, $y'=\frac{1}{10}$.
    2. Normal slope $=-10$.
    3. Normal line: $y-5=-10(x-25)$.
    4. Set $x=0$: $y=5+250=255$.
    5. Intersection: $(0,255)$.

    Common Mistakes …

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12. KEAM 2022Set eng-2022-P2-B14 marksMCQ
    Q.If $x + 13y = 40$ is normal to the curve $y = 5x^2 + \alpha x + \beta$ at the point $(1, 3)$, then the value of $\alpha\beta$ is equal to (A) $15$ (B) $-6$ (C) $6$ (D) $13$ (E) $-15$

    ›Reveal solutionSolution

    $\alpha=3$, $\beta=-5$, so $\alpha\beta=-15$.

    Concept and Intuition

    The normal's slope is the negative reciprocal of the tangent's slope. Use the tangent slope from the derivative and the fact the point lies on the curve.

    Step-by-Step Solution

    1. Normal line: $x+13y=40\Rightarrow y=\frac{40-x}{13}$, slope $=-\frac1{13}$.
    2. Tangent slope $=$ negative reciprocal $=13$.
    3. $y=5x^2+\alpha x+\beta\Rightarrow y'=10x+\alpha$; at $x=1$: $10+\alpha=13\Rightarrow\alpha=3$.
    4. Point $(1,3)$ on curve: $3=5(1)+\alpha+\beta=5+3+\beta\Rightarrow\beta=-5$. …

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