Coordinate Geometry
RD Sharma
CBSE
Grade 10
Question:
Show that the points $(a, a), (-a, -a)$ and $(-\sqrt{3}a, \sqrt{3}a)$ are the vertices of an equilateral triangle.
Step-by-Step Solution
Key Concept: $d_1^2 = (a+a)^2 + (a+a)^2 = 4a^2 + 4a^2 = 8a^2 \Rightarrow d_1 = 2\sqrt{2}a$.<br>$d_2^2 = (-\sqrt{3}a+a)^2 + (\sqrt{3}a+a)^2 = a^2(3+1-2\sqrt{3} + 3+1+2\sqrt{3}) = 8a^2 \Rightarrow d_2 = 2\sqrt{2}a$.<br>$d_3^2 = (a+\sqrt{3}a)^2 + (a-\sqrt{3}a)^2 = 8a^2 \Rightarrow d_3 = 2\sqrt{2}a$. All 3 sides equal $\Rightarrow$ Equilateral triangle.
$AB = \sqrt{(2a)^2 + (2a)^2} = \sqrt{8a^2} = 2\sqrt{2}a$. [1.0 Mark]
$BC = \sqrt{(a - \sqrt{3}a)^2 + (a + \sqrt{3}a)^2} = \sqrt{a^2(1 - 2\sqrt{3} + 3 + 1 + 2\sqrt{3} + 3)} = \sqrt{8a^2} = 2\sqrt{2}a$. [1.0 Mark]
$CA = \sqrt{(a + \sqrt{3}a)^2 + (a - \sqrt{3}a)^2} = \sqrt{8a^2} = 2\sqrt{2}a$. Since $AB = BC = CA = 2\sqrt{2}a$, the triangle is equilateral. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Calculating $AB = 2\sqrt{2}a$: 1.0 Mark
Calculating $BC = 2\sqrt{2}a$: 1.0 Mark
Calculating $CA = 2\sqrt{2}a$ and concluding equilateral triangle: 1.0 Mark
Correct Answer:
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