Q.Case study - based question. Villages of Shanu and Arun are 40 km apart and are situated on Imphal-Moirang highway as shown in the following picture. Another highway YY' crosses Moirang-Imphal highway at O (0,0). A small local road PQ crosses both the highways at points A and B such that OA = 10Km and OB = 12Km. Also the villages of Barun and Jeetu are on the similar highway YY'. Barun's village B is 12Km from O and that of Jeetu is 15Km from O. Based on the above information, answer the following questions: Find

Showing the 12 most recent of 23 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Equation of the line passing through the points $(-4, 3)$ with slope $\dfrac{1}{2}$ is

   (a) $x - 2y + 10 = 0$

   (b) $x - 2y - 10 = 0$

   (c) $2x - y + 10 = 0$

   (d) $x + 2y + 10 = 0$

   ›Reveal solutionSolution

   Apply the point-slope equation $y-y_1 = m(x-x_1)$ and simplify to general form.

   Point-slope form: $y - y_1 = m(x - x_1)$ with $(x_1,y_1)=(-4,3)$ and $m=\dfrac12$:

   $y - 3 = \dfrac12(x-(-4)) = \dfrac12(x+4)$

   Multiply both sides by 2: …

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2. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: Equation of a line making intercepts $a$ and $b$ on the x and y axis respectively is ______.

   ›Reveal solutionSolution

   A line cutting intercepts $a$ and $b$ on the x- and y-axis respectively has equation $\dfrac{x}{a}+\dfrac{y}{b}=1$.

   The line passes through $(a, 0)$ on the x-axis and $(0, b)$ on the y-axis.

   …

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3. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: The co-ordinates of points in the xy-plane are in the form of ______.

   ›Reveal solutionSolution

   Every point in the Cartesian xy-plane is represented as an ordered pair $(x, y)$.

   In the Cartesian coordinate system, two perpendicular number lines (the x-axis and y-axis) intersect at the origin.

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Match the column: Column A entry 'y-intercept of the line $2x - 3y + 6 = 0$' — find the matching value from Column B.

   (a) 2

   (b) 8

   (c) $\dfrac{2}{3}$

   (d) $\dfrac{2\tan x}{1 - \tan^2 x}$

   (e) $\sin 2x$

   (f) 10

   (g) 20

   (h) $\dfrac{2\tan x}{1 + \tan^2 x}$

   (i) 4

   ›Reveal solutionSolution

   Substituting $x = 0$ into the line equation $2x - 3y + 6 = 0$ gives $y = 2$, the $y$-intercept.

   For the line $2x - 3y + 6 = 0$, the $y$-intercept is found by setting $x = 0$:

   $2(0) - 3y + 6 = 0 \Rightarrow -3y = -6 \Rightarrow y = 2$. …

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5. CBSE 2024Set ANNUAL1 markMCQ
   Q.The equation of the line passing through the points $(2,1)$ and $(5,-2)$ is

   (a) $x+y-3=0$

   (b) $x+y+3=0$

   (c) $5x+3y+2=0$

   (d) None of these

   ›Reveal solutionSolution

   Find the slope from the two given points, then use the point-slope form of a line.

   Slope through $(2,1)$ and $(5,-2)$:

   $m = \frac{-2-1}{5-2} = \frac{-3}{3} = -1$

   Using point-slope form $y - y_1 = m(x-x_1)$ with point $(2,1)$:

   $y - 1 = -1(x - 2)$ …

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6. CBSE 2024Set ANNUAL1 markMCQ
   Q.If the line $(x-y+2)+k(2x+3y+5)=0$ is parallel to the line $3x+y=0$, then the value of $k$ will be

   (a) $k=1$

   (b) $k=\dfrac{4}{7}$

   (c) $k=\dfrac{7}{4}$

   (d) None of these

   ›Reveal solutionSolution

   Expand the family of lines to standard form $Ax+By+C=0$, write its slope $-A/B$, and set it equal to the slope of the given parallel line.

   Expand $(x-y+2)+k(2x+3y+5)=0$:

   $x - y + 2 + 2kx + 3ky + 5k = 0$

   $(1+2k)x + (3k-1)y + (2+5k) = 0$

   The slope of a line $Ax+By+C=0$ is $-A/B$, so this line's slope is:

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

   The line $3x+y=0$ has slope $m_2 = -3$.

   …

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7. CBSE 2023Set ANNUAL1 markMCQ
   Q.The equation of the straight line which passes through the point $(4, 3)$ and parallel to the line $3x+4y=12$ is

   (a) $3x+4y=10$

   (b) $3x+4y=24$

   (c) $3x+4y-20=0$

   (d) $3x-4y+24=0$

   ›Reveal solutionSolution

   Parallel lines share the same $x,y$ coefficients; substituting the given point fixes the constant as 24.

   A line parallel to $3x+4y=12$ has the same left-hand-side coefficients, differing only in the constant:

   $3x+4y=c$

   …

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8. CBSE 2023Set ANNUAL1 mark
   Q.Find the equation of the line through $(-2, 3)$ with slope $-4$.

   ›Reveal solutionSolution

   The equation of the line is $4x+y+5=0$.

   Using point-slope form:

   $y-3 = -4(x-(-2)) = -4(x+2)$ …

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9. CBSE 2023Set ANNUAL1 mark
   Q.Find the equation of the line through the points $(3, -2)$ and $(-1, 4)$.

   ›Reveal solutionSolution

   The equation of the line through $(3,-2)$ and $(-1,4)$ is $3x+2y-5=0$.

   Slope: $m=\dfrac{4-(-2)}{-1-3}=\dfrac6{-4}=-\dfrac32$.

   Using point-slope form through $(3,-2)$: …

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10. CBSE 2023Set ANNUAL1 markMCQ
    Q.Case study: Due to Covid-19 pandemic situation, people are maintaining social distances. Three friends Along, Mharemo and Senti are sitting on the vertices of a triangle whose coordinates are A(3, 1), M(2, -3) and S(-3, 3). The equation of the line AM is

    (a) $4x+y-11=0$

    (b) $4x-y-11=0$

    (c) $4x+y+11=0$

    (d) $4x-y+11=0$

    ›Reveal solutionSolution

    Find the slope of AM, then use point-slope form through A.

    A$(3,1)$, M$(2,-3)$.

    Slope of AM $=\dfrac{-3-1}{2-3}=\dfrac{-4}{-1}=4$.

    …

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11. CBSE 2023Set ANNUAL1 markMCQ
    Q.Case study (continued — triangle with vertices A(3,1), M(2,-3), S(-3,3)): The equation of a line passing through A and parallel to SM is

    (a) $6x+5y-23=0$

    (b) $6x-5y-23=0$

    (c) $6x-5y+23=0$

    (d) $6x+5y+23=0$

    ›Reveal solutionSolution

    Find slope of SM, then write the parallel line through A with that same slope.

    S$(-3,3)$, M$(2,-3)$.

    Slope of SM $=\dfrac{-3-3}{2-(-3)}=\dfrac{-6}{5}$.

    …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.Case study (continued — triangle with vertices A(3,1), M(2,-3), S(-3,3)): Equation of median through A is

    (a) $2x+7y+1=0$

    (b) $2x-7y+1=0$

    (c) $2x+7y-1=0$

    (d) $2x-7y-1=0$

    ›Reveal solutionSolution

    The median from A passes through the midpoint of the opposite side SM.

    Midpoint of S$(-3,3)$ and M$(2,-3)$: $\left(\dfrac{-3+2}{2},\dfrac{3-3}{2}\right)=\left(-\dfrac12,\,0\right)$.

    Slope of the median (through A$(3,1)$ and this midpoint): $\dfrac{0-1}{-1/2-3}=\dfrac{-1}{-7/2}=\dfrac{2}{7}$.

    …

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