Point A is chosen on y-axis in such a way that ΔABC is an equilateral triangle. The base BC of the ΔABC is shown in the figure. Find the coordinates of (i) the mid-point of BC (ii) the area of the triangle (iii) the vertices of a triangle.
Answers
Answer:
O is the mid-point of the base BC.
Therefore, coordinates of point B are (0,3).
So, BC = 6 units
Let the coordinates of point A be (x,0).
Using distance formula,
AB=
(0−x)
2
+(3−0)
2
=
x
2
+9
BC=
(0−0)
2
+(−3−3)
2
=
36
Also, AB = BC
x
2
+9
=
36
x
2
+9=36
x
2
=27
x=±3
3
Coordinates of A = (3
3
,0).
Answer:
I am assuming O as (0, 0)
1) O is mid point /coodinates are (0, 0)
Point C lies on x axis so it would (3, 0) and point B will be(-3, 0) as O is mid point
(as 6/2=3)
Now we know AC = 6 units as triangle is Equilateral and OC =3 units
2) By Pythagoras theorem
OA^2 + OC^2 = AC^2
so OA =√AC^2 - OC^2
OA =√6^2 - 3^2
OA = √36-9
OA = 5
and point A lies on y axis so x coordinate is 0
i:e: point coordinates are (0, 5)
3) Area of triangle will be 1/2 base * height
=1/2 6*5
=15 unit^2
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