If (-5, 3) and (5, 3) are two vertices of an equilateral triangle, then find co-ordinates of the third vertex, given that origin lies inside the triangle. (Take $\sqrt{3} = 1.7$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Let the two given vertices be A(–5, 3) and B(5, 3), and let the third vertex be C(x, y).
Side length AB:
$$AB = \sqrt{(5-(-5))^2 + (3-3)^2} = \sqrt{100} = 10 \text{ units}$$
For an equilateral triangle, CA = CB = AB = 10.
From CA = CB:
$$\sqrt{(x+5)^2 + (y-3)^2} = \sqrt{(x-5)^2 + (y-3)^2}$$
Squaring: $(x+5)^2 = (x-5)^2 \Rightarrow x = 0$
From CA = 10 (with x = 0):
$$\sqrt{25 + (y-3)^2} = 10$$
$$(y-3)^2 = 75 \Rightarrow y - 3 = \pm 5\sqrt{3}$$
$$y = 3 + 5\sqrt{3} \quad \text{or} \quad y = 3 - 5\sqrt{3}$$
Since $\sqrt{3} = 1.7$: $\quad 5\sqrt{3} = 8.5$
- $y = 3 + 8.5 = 11.5$
- $y = 3 - 8.5 = -5.5$
Since the origin (0, 0) must lie inside the triangle, the third vertex must be below the line y = 3 (on the opposite side from y = 11.5).
∴ The coordinates of the third vertex are (0, –5.5).
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Explanation
- The two given vertices both have y = 3, so AB is horizontal. By symmetry, the third vertex lies on x = 0 (the perpendicular bisector of AB).
- Two values of y arise; the condition "origin lies inside" picks the one below y = 3 (i.e., y = –5.5), because the origin (0, 0) would be outside the triangle if the apex were at (0, 11.5) which is far above both given vertices and the origin.
- Examiners award marks for: finding AB, applying CA = CB to get x = 0, solving for y, and correctly applying the "origin inside" condition to choose y = –5.5.