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Mathematics

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

Coordinate Geometry

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Answer

Let the points be A(1, 5), B(2, 3) and C(-2, -11).

By formula,

Distance between two points (D) = (y2−y1)2+(x2−x1)2\sqrt{(y2 - y1)^2 + (x2 - x1)^2}

Substituting values we get :

AB=(3−5)2+(2−1)2=(−2)2+(1)2=4+1=5.BC=(−11−3)2+(−2−2)2=(−14)2+(−4)2=196+16=212.AC=(−11−5)2+(−2−1)2=(−16)2+(−3)2=256+9=265.AB = \sqrt{(3 - 5)^2 + (2 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (1)^2} \\[1em] = \sqrt{4 + 1} \\[1em] = \sqrt{5}. \\[1em] BC = \sqrt{(-11 - 3)^2 + (-2 - 2)^2} \\[1em] = \sqrt{(-14)^2 + (-4)^2} \\[1em] = \sqrt{196 + 16} \\[1em] = \sqrt{212}. \\[1em] AC = \sqrt{(-11 - 5)^2 + (-2 - 1)^2} \\[1em] = \sqrt{(-16)^2 + (-3)^2} \\[1em] = \sqrt{256 + 9} \\[1em] = \sqrt{265}. \\[1em]

Calculating,

⇒ AB + BC = 5+212\sqrt{5} + \sqrt{212}

⇒ BC + AC = 212+265\sqrt{212} + \sqrt{265}

⇒ AB + AC = 5+265\sqrt{5} + \sqrt{265}

Since, AB + AC ≠ BC and BC + AC ≠ AB and AB + BC ≠ AC.

Hence, points (1, 5), (2, 3) and (-2, -11) are not collinear.

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