If (1, 2) is the centroid of an equilateral triangle and (– 2,

Engineering

Mathematics

Family of Straight lines

Question

If (1, 2) is the centroid of an equilateral triangle and (– 2, – 2) is one of the vertices, then

inradius of triangle is $\frac{5}{2}$ .

area of triangle is $\frac{225}{4 \sqrt{3}}$ .

area of triangle is $\frac{225}{4}$ .

equation of side opposite to vertex (– 2, – 2) is 6x + 8y = 47.

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$\sin 30^{\circ} = \frac{ID}{\sqrt{9 + 16}} \Rightarrow ID (inradius) = \frac{5}{2}$

$Also, AG = \sqrt{9 + 16} = 5$

$∴ BD (length of altitude) = 5 + \frac{5}{2} = \frac{15}{2}$

$∴ Area (△ ABC) = \frac{h^{2}}{\sqrt{3}} = \frac{225}{4 \sqrt{3}}$

Let co-ordinates of E(x, y)

So, $1 = \frac{2 (x) + 1 (- 2)}{3} \Rightarrow x = \frac{5}{2}$

Also, $2 = \frac{2 (y) + 1 (- 2)}{3} \Rightarrow y = 4$

So, $E (\frac{5}{2}, 4) \Rightarrow m_{BC} = \frac{- 1}{m_{AE}} = \frac{- 3}{4}$

⇒ Equation of BC is (y – 4) = $(\frac{- 3}{4}) (x - \frac{5}{2})$
⇒ 6x + 8y = 47

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