- two vertices of a triangle are (1,2), (3,5), and the centeroid is at the origin. Find the third vertex.
- Find the coordinates (x,y) of a point whose distance from points (3,5) is 5 units & that from (0,1) is 10 units , given that 3x=2y
- If 3b is the length of a side of an equilateral triangle ABC ,base BC lies on x-axis & B lies of origin , find the coordinate of vertices of ABC
(a) let the vertices of the triangle be A(1,2), B(3,5) and .
therefore centroid of the triangle ABC is given by:
given: the centroid is the origin i.e. (0,0).
hence the coordinates of third vertex is (-4,-7).
distance of (x,y) from the point (3,5) is 5 units.
and distance of (x,y) from the point (0,1) is 10 units.
subtracting equation (1) from (2):
since 3x=2y put this in (3)
hence the desired point is (6,9).
the coordinates of B is (0,0).
since BC lies on x-axis. and length of the side is 3b.
since C is on x-axis and at a distance of 3b units.
the coordinates of C will be either (3b,0) or (-3b,0)
now draw perpendicular from A to BC, let it intersect BC at D.
D will be the mid-point of BC, BD=3b/2
the coordinates of A (when coordinates of C is (3b,0)) is
the coordinates of A (when coordinates of C is (-3b,0)) is