Question

Check whether the points (-6,10), (-4,6) and (3,-8) are collinear

Answer 1

Answer:

if the area of triangle formed by the points (x,y) ,(x,y),and (x,y) is zero then the poins are collinear.

Step-by-step explanation:

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Answer 2

Question:

$\longrightarrow$ Check whether the points (-6,10),(-4,6) and (3,-8) are collinear.

To Find:

$\longrightarrow$ To check that the points are collinear.

Solution:

We know that,

$\: \: \: \: \: \: \: \: \: \: \bullet$The distance between the points $\sf{(x_1,y_1),(x_2,y_2)}$ is$\sqrt{\rm{(x_2 - x_1)^{2} + (y_2,y_1)}^{2}}$

Let A(-6,10),B(-4,6) and C(3,-8) are given points.

Now,AB:

$\: \: \: \: \: \: \implies \sf{AB = \sqrt{\big( - 4 - ( -6) \big)^{2} + \big(6 - 10 \big)^{2}}}$

$\: \: \: \: \: \: \: \implies \sf{AB = \sqrt{( - 4 - 6)^{2} + (6 - 10)^{2}}}$

$\: \: \: \: \: \: \: \: \: \: \implies \sf{AB = \sqrt{(-10)^{2} + (-4)^{2}}}$

$\: \: \: \: \: \: \: \: \: \implies \sf{AB = \sqrt{100 \: + 16}}$

$\: \: \: \: \: \implies \sf{AB = \sqrt{116}} \:units$

Now,BC:

$\: \: \: \: \implies \sf{BC = \sqrt{ \big(3 - ( - 4) \big)^{2} + \big( - 8 - 6 \big)^{2}}}$

$\: \: \: \: \implies \sf{ BC = \sqrt{( - 3 + 4)^{2} + ( - 8 - 6)^{2}}}$

$\: \: \: \: \: \: \implies \sf{BC = \sqrt{(1)^{2} +( - 14)^{2}}}$

$\: \: \: \: \: \: \implies \sf{BC = \sqrt{1 + 196}}$

$\: \: \: \: \: \: \implies \sf{BC = \sqrt{197}} \:units$

Now,AC:

$\: \: \: \: \implies \sf{AC = \sqrt{\big(3 -( - 6) \big)^{2} + \big( - 8 - 10 \big)^{2}}}$

$\: \: \: \: \: \: \: \: \: \: \implies \sf{AC = \sqrt{(3 + 6)^{2} + ( - 8 - 10)^{2}}}$

$\: \: \: \: \: \: \: \implies \sf{AC = \sqrt{(9)^{2} + ( - 18)^{2} }}$

$\: \: \: \: \: \: \: \implies \sf{AC = \sqrt{81 + 324}}$

$\: \: \: \: \: \implies \sf{ AC = \sqrt{405}} \:units$

• No,The points are not collinear.Because AB+BC is not equal to AC.

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More to Know:-

Collinear points: The points which lie on the same line are called collinear points.