# Using distance formula show that the points L(4, 1), M(1,-3) and N(-2, -5) are collinear.

the coordinates of the given points are
let us calculate the distances between the given points:
$LM=\sqrt{{\left(1-4\right)}^{2}+{\left(-3-1\right)}^{2}}=\sqrt{9+16}=\sqrt{25}\phantom{\rule{0ex}{0ex}}LM=5\phantom{\rule{0ex}{0ex}}LN=\sqrt{{\left(-2-4\right)}^{2}+{\left(-5-1\right)}^{2}}=\sqrt{36+36}=\sqrt{72}\phantom{\rule{0ex}{0ex}}LN=6\sqrt{2}$
$MN=\sqrt{{\left(-2-1\right)}^{2}+{\left(-5+3\right)}^{2}}=\sqrt{9+4}\phantom{\rule{0ex}{0ex}}MN=\sqrt{13}$
here sum of two distances is not equal to the third one, therefore the given points are not collinear.

hope this helps you

• 0
LM, MN,NL
therefore, L=(4,1) (x1,y1) M=(1,-3) (x2, y2)
By using Distance Formula,
(x2-x1)2 + (y2-y1)2
(1-4)2 + (-3,1)2
(-3)2 + (-2)2
9 + 4
13 Units.
therefore, the distance brtween the point L M is 13 Units

M= (1-3)(x1,y1) N= (-2,-5) (x2,y2)
By using Distance Formula,
(x2-x1)2 + (y2-y1)2
[-2-(-3)]2 + [-5-(-3)]2
(-2+3)2 + (-5+3)2
(1)2 + (-2)2
1 + 4
5 Units.
therefore, the distance between the points M N is 5 Units.

N= (-2,-5) (x1,y1) , L=(4,1) (x2, y2)
By using Distance Formula,
(x2-x1)2 + (y2-y1)2
[4 - (-2)]2 + [1- (-2)]2
[4 + 2]2 + [1+ 2]2
[7]2 + [3]2
49 + 9
58 units.
therefore, the distance between the points N L is 58 Units

Therefore, Mast. Gaurav Mane The points L,M N is not Colinear.

• -2
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