Q.The base of an equilateral triangle with side $2a$ lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

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:

   $2y - 6 = x + 4$

   Rearrange to general form ($Ax+By+C=0$):

   $x - 2y + 10 = 0$

   ✓Final answer

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

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.

   Using the two-point form of a line through $(a,0)$ and $(0,b)$: $\dfrac{y-0}{x-a} = \dfrac{b-0}{0-a}$, which simplifies to $-a y = b(x-a) \Rightarrow bx + ay = ab$. Dividing throughout by $ab$: $\dfrac{x}{a} + \dfrac{y}{b} = 1$.

   ✓Final answer

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

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.

   Any point P in the plane is uniquely specified by its x-coordinate (signed perpendicular distance from the y-axis) and its y-coordinate (signed perpendicular distance from the x-axis), written as the ordered pair $(x, y)$.

   ✓Final answer

   The co-ordinates of a point in the xy-plane are of the form $(x, y)$.

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 $x = 0$:

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

   So the $y$-intercept is $2$.

   ✓Final answer

   The $y$-intercept of $2x - 3y + 6 = 0$ matches option (a) 2.

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 $(2,1)$:

   $y - 1 = -1(x - 2)$

   $y - 1 = -x + 2$

   $x + y - 3 = 0$

   Check with the second point: $5 + (-2) - 3 = 0$ ✓.

   ✓Final answer

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

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 $m_2 = -3$.

   For the two lines to be parallel, $m_1 = m_2$:

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

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

   $1+2k = 3(3k-1) = 9k - 3$

   $1 + 3 = 9k - 2k$

   $4 = 7k$

   $k = \frac{4}{7}$

   ✓Final answer

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

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:

   $3x+4y=c$

   Since this line passes through $(4,3)$, substitute:

   $3(4)+4(3)=c \implies 12+12=c \implies c=24$

   So the equation is:

   $3x+4y=24$

   ✓Final answer

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

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:

   $y-3 = -4(x-(-2)) = -4(x+2)$

   $y-3 = -4x-8$

   $4x+y+5=0.$

   ✓Final answer

   $4x+y+5=0$.

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 $(3,-2)$:

   $y-(-2) = -\dfrac32(x-3)$

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

   $2y+4 = -3x+9$

   $3x+2y-5=0.$

   ✓Final answer

   $3x+2y-5=0$.

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

    Line through A with this slope: $y-1=4(x-3)\Rightarrow y-1=4x-12\Rightarrow 4x-y-11=0$.

    ✓Final answer

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

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

    Line through A$(3,1)$, parallel to SM (same slope): $y-1=-\dfrac65(x-3)$.

    Multiply by 5: $5y-5=-6x+18\Rightarrow 6x+5y-23=0$.

    ✓Final answer

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

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): $\dfrac{0-1}{-1/2-3}=\dfrac{-1}{-7/2}=\dfrac{2}{7}$.

    Line through A: $y-1=\dfrac27(x-3)\Rightarrow 7y-7=2x-6\Rightarrow -2x+7y-1=0\Rightarrow 2x-7y+1=0$.

    ✓Final answer

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