# PROVE THE GIVEN POINTS ARE COLLINEAR USING THE CONCEPT SLOPE

The points which lie on the same line are known as collinear points.

In case we have three point and we need to prove that the given points are collinear, we may follow the steps given below.

## Working Rule

Step 1 :

Choose the points A and B and find the slope.

Step 2 :

Choose the points B and C and find the slope.

Step 3 :

If the slopes found in step 1 and step 2 are equal, then the points A, B and C are collinear.

Formula to find the slope of a line joining the two points (x1, y1) and (x2, y2) :

m  =  (y2 - y1) / (x- x1)

## Examples

Example 1 :

Using the concept of slope, prove that each of the following set of points are collinear.

(i) (2 , 3), (3 , -1) and (4 , -5)

(ii) (4 , 1), (-2 , -3) and (-5 , -5)

(iii) (4 , 4), (-2 , 6) and (1 , 5)

Solution :

(i) Let the given points be A (2 , 3)  B (3 , -1) and C (4 , -5)

If the points A, B and C  are collinear then,

Slope of AB  =  Slope of BC

 Slope of AB :m  =  (y2 - y1)/(x2 - x1)A (2 , 3)  B (3 , -1)m  =  (-1 - 3)/(3 - 2)m  =  -4/1m  =  -4  ---(1) Slope of BC :m  =  (y2 - y1)/(x2 - x1)B (3 , -1) C (4 , -5)m  =  (-5 + 1)/(4 - 3)m  =  -4/1m  =  -4  ---(2)

(1)  =  (2)

Hence the given points A, B and C are collinear.

(ii) (4 , 1), (-2 , -3) and (-5 , -5)

Solution :

Let the given points be A(4 , 1) B(-2 , -3) and C(-5 , -5)

If the points A, B and C  are collinear then,

Slope of AB  =  Slope of BC

 Slope of AB :m  =  (y2 - y1)/(x2 - x1)A (4, 1)  B (-2, -3)m  =  (-3 - 1)/(-2 - 4)m  =  -4/(-6)m  =  2/3 ---(1) Slope of BC :m  =  (y2 - y1)/(x2 - x1)B (-2, -3) C (-5, -5)m  =  (-5 + 3)/(-5 + 2)m  =  -2/(-3)m  =  2/3  ---(2)

(1)  =  (2)

Hence the given points are collinear.

(iii) (4 , 4), (-2 , 6) and (1 , 5)

Solution :

Let the given points be A(4 , 4) B(-2 , 6) and C(1 , 5)

If the points A, B and C  are collinear then,

Slope of AB  =  Slope of BC

 Slope of AB :m  =  (y2 - y1)/(x2 - x1)A (4, 4)  B (-2, 6)m  =  (6 - 4)/(-2 - 4)m  =  2/(-6)m  =  -1/3 ---(1) Slope of BC :m  =  (y2 - y1)/(x2 - x1)B (-2, 6) C (1, 5)m  =  (5 - 6)/(1 + 2)m  =  -1/3 ---(2)

(1)  =  (2)

So, the given points are collinear.

Example 2 :

If the points (a, 1), (1, 2) and (0, b+1) are collinear, then prove that (1/a) + (1/b) = 1

Solution :

Let the given points be A(a, 1) B(1, 2) and C(0, b+1). Since the given points are collinear, slope of AB is equal to slope of BC.

 Slope of AB :m  =  (y2 - y1)/(x2 - x1)A (a, 1)  B (1, 2)m  =  (2 - 1)/(1 - a)m  =  1/(1 - a) ---(1) Slope of BC :m  =  (y2 - y1)/(x2 - x1)B (1, 2) C (0, b + 1)m  =  (b + 1 - 2)/(0 - 1)m  =  (b - 1)/(-1) ---(2)

1/(1 - a)  =  -(b - 1)

1  =  (1 - b)(1 - a)

1  =  1 - a - b + ab

a + b = ab + 1 - 1

a + b  =  ab

Dividing by ab, we get

(a/ab) + (b/ab)  =  (ab/ab)

1/b + 1/a  =  1

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