Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear

Question

Given that \( \cot \left( \frac{A+B}{2} \right) \cdot \tan \left( \frac{A-B}{2} \right) = \), and the equation \( \frac{x}{2} + \frac{y}{3} + \frac{2}{6} - 1 = 0 \), find the area of \( \Delta ABC = 2 \).

Answer 1

The vertices of the triangle are given as A=(1, 5), B=(2, 3), and C=(−2, −11).

The lengths of the sides are calculated as follows:

AB = \(\sqrt{(1-2)^2+(5-3)^2}=\sqrt{(-1)^2+2^2}=\sqrt{1+4}=\sqrt5\)

BC = \(\sqrt{(2-(-2))^2+(3-(-11))^2}=\sqrt{(2+2)^2+(3+11)^2}=\sqrt{4^2+14^2}=\sqrt{16+196}=\sqrt{212}\)

CA = \(\sqrt{(1-(-2))^2+(5-(-11))^2}=\sqrt{(1+2)^2+(5+11)^2}=\sqrt{3^2+16^2}=\sqrt{9+256}=\sqrt{265}\)

Since AB + BC \(eq\) CA, the points (1, 5), (2, 3), and (−2, −11) are not collinear.

Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear

Solution and Explanation

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