Q. 74.1( 7 Votes )

# 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.

Answer :

Let ABC be the given equilateral triangle with side 2a

⇒ AB= BC = AC = 2a

Assuming that the base AB lies on the y axis such that the mid-point of AB is at the origin

i.e.BO = OA = a and O is the origin

**⇒ Co-ordinates of point A are (0,a) and that of B are (0,-a)**

Since the line joining a vertex of an equilateral ∆ with the mid-point of its opposite side is perpendicular

⇒ Vertex of A lies on the y –axis

On applying Pythagoras theorem

(AC)^{2} = OA^{2} + OC^{2}

⇒(2a)^{2}= OA^{2} + a^{2}

⇒ 4a^{2} – a^{2} = OA^{2}

⇒ 3a^{2}= OA^{2}

⇒ OA =

∴ Co-ordinates of point C =

Thus, the vertices of the given equilateral triangle are (0, a) , (0, -a) , (

Or (0, a), (0, -a) and (

Rate this question :

Find all the points on the line x + y = 4 that lie at a unit distance from the line 4x+3y=10.

RS Aggarwal - MathematicsThe distance between the orthocenter and circumcentre of the triangle with vertices (1, 2) (2, 1) and is

RD Sharma - MathematicsWhat are the points on the x-axis whose perpendicular distance from the line is 4 units?

RS Aggarwal - MathematicsFind the equation of a line for which

p = 8, α = 225°

RD Sharma - MathematicsFind the equation of a line for which

p = 4, α = 150°

RD Sharma - MathematicsArea of the triangle formed by the points ((a + 3)(a + 4), a + 3), ((a + 2)(a + 3), (a + 2)) and ((a + 1)(a + 2), (a + 1)) is

RD Sharma - MathematicsFind the equation of straight line passing through ( – 2, – 7) and having an intercept of length 3 between the straight lines 4x + 3y = 12 and 4x + 3y = 3.

RD Sharma - MathematicsFind the distance of the line 2x + y = 3 from the point ( – 1, – 3) in the direction of the line whose slope is 1.

RD Sharma - Mathematics