The base BC of an equilateral triangle ABC lies on y axis The coordinates of pt C are - Maths - Coordinate Geometry

Question

The base BC of an equilateral triangle ABC lies on y axis The
coordinates of pt C are (0,-3) If the origin is the mid pt of the
base BC, find the coordinates of the pt A and B and hance find the
area of the triangle

Ajanta Trivedi · Accepted Answer

given: ABC is an equilateral triangle BC lies on y-axis and origin is the midpoint of BC let the coordinates of B are (0,y), since it lies on y-axis therefore the coordinates of B are y-3/2=0 i e y=3 coordinates are (0,3) therefore BC = 3-(-3)=3+3=6 units OA is the perpendicular bisector of BC therefore the vertex A will lie on OA i e x-axis let the coordinates of A are (x,0) since ABC is an equilateral triangle AB = BC $A B^{2} = B C^{2} (x - 0)^{2} + (0 - 3)^{2} = 6^{2} x^{2} + 9 = 36 x^{2} = 36 - 9 = 27 x = \pm \sqrt{27} = \pm 3 \sqrt{3}$ thus the coordinates of A are $(3 \sqrt{3}, 0) o r (- 3 \sqrt{3}, 0)$ area of the equilateral triangle ABC = $\frac{\sqrt{3}}{4} B C^{2} = \frac{\sqrt{3}}{4} * 6^{2} = \frac{\sqrt{3}}{4} * 36 = 9 \sqrt{3} s q u n i t s$ hope this helps you

The base BC of an equilateral triangle ABC lies on y axis. The coordinates of pt. C are (0,-3). If the origin is the mid pt. of the base BC, find the coordinates of the pt. A and B and hance find the area of the triangle.