Question

Are the points a (2,3) b ( 3,-1) and c (4,-5) collinear? justify​

Answer 1

Given:

Point a = (2, 3)

Point b = (3, -1)

Point c = (4, -5)

To find:

Justify whether the three points are collinear or not.

Solution:

There are many ways to prove that whether the three points are collinear or not, here we will use the method in which the slopes are calculated between two points and checked whether they are equal or not:

Slop of line joining point a nd b = y₂- y₁/ x₂- x₁

= -1 -3/ 3-2

= -4

Slope of the line joining the points b and c = y₂- y₁/ x₂- x₁

= -5+1 / 4-3

= -4

Since the two slopes are equal, therefore the points are collinear.

Answer 2

SOLUTION

TO CHECK

The points A (2,3) B ( 3,-1) and C (4,-5) collinear

CONCEPT TO BE IMPLEMENTED

1. The area of the triangle formed by the points

$\sf{(x_1,y_1), (x_2,y_2),(x_3,y_3)} \: is$

$\displaystyle \sf{ = \frac{1}{2} \bigg| x_1(y_2 - y_3) + x_2(y_3 - y_1) +x_3(y_1 - y_2) \bigg| } \: \: sq \: unit$

2. If three points are collinear then the triangle formed by the three points is zero

EVALUATION

Here the given three points are A (2,3) B ( 3,-1) and C (4,-5)

Then the area of the triangle ABC is

$\displaystyle \sf{ = \frac{1}{2} \bigg| 2( - 1 + 5) + 3( - 5 - 3) +4(3 + 1) \bigg| } \: \: sq \: unit$

$\displaystyle \sf{ = \frac{1}{2} \bigg| 8 - 24 +16 \bigg| } \: \: sq \: unit$

$\displaystyle \sf{ = \frac{1}{2} \times 0 \: \: \: \: sq \: unit }$

$= 0$

Hence the points A (2,3) B ( 3,-1) and C (4,-5) collinear

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