Q. 23.6( 8 Votes )

# Prove that the po

Answer :

Let the points be

A = (x_{1}, y_{1}) = (3, -2) and

B = (x_{2}, y_{2}) = (-5, 4) and

C = (x_{3}, y_{3}) = (-1, 1)

Now if the points A, B and C are collinear then the area formed by the triangle by joining these three points would be 0

Area of triangle is given by formula

Area = × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

Where (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are vertices of triangle

So, if we prove that area of ΔABC = 0 then points A, B and C are collinear

⇒ Area = × [3(4 – 1) + (-5)(1 – (-2)) + (-1)(-2 – 4)]

⇒ Area = × [9 + (-5)(3) + 6]

⇒ Area = × [15 + (-15)]

⇒ Area = 0

Hence points (3, –2), (–5, 4) and (–1, 1) are collinear.

__Method 2__

Another method is that you can find distances between every pair of two points and if any distance is equal to sum of other two distances then points are collinear. Here AC + BC = AB.

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