ABC is an equilateral triangle the coordinates of vertices B and C are (3,0) and (- 3,0) respectively find the coordinates vertex A
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Since ABC is an equilateral triangle, therefore
AB = AC = BC
OB = 3 units ( along + ve direction of x-axis)
OC = -3 units ( Along - ve direction of x-axis)
But side CB = OB - OC
=> CB = 3 - ( -3)
=> CB = 6 units
=> BC = 6 units
=> AC = BC = 6 units
In right traingle AOC, we have
ac {}^{2} = ao^{2} + oc {}^{2}ac
2
=ao
2
+oc
2
(6) {}^{2} = oa {}^{2} + (3) {}^{2}(6)
2
=oa
2
+(3)
2
36 = oa {}^{2} + 936=oa
2
+9
OA {}^{2} = 36 - 9OA
2
=36−9
OA {}^{2} = 27OA
2
=27
OA = 3 \sqrt{3} \: unitsOA=3
3
units
The point A is at
3 \sqrt{3}3
3
and x co-ordinates of A is O
Hence,
The co-ordinates of vertex A are (0, 3✓3)
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