USING THE EQUATION OF LINE TO PROVE THREE POINTS ARE COLLINEAR

By using the concept of equation of line, prove that the three points are collinear.

Example 1 :

(4, 2), (7, 5), (9, 7)

Solution :

Formula for slope of a line joining two points :

m = (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = (4, 2) (x₂, y₂) = (7, 5).

m = (5 - 2)/(7 - 4)

= 3/3

= 1

Equation of line in slope-intercept form :

y = mx + b

Substitute m = 1.

y = x + b ----(1)

The line is passing through the point (4, 2).

So, substitute x = 4 and y = 2.

2 = 4 + b

-2 = b

Substitute b = -2 in (1).

y = x - 2

Now, substitute the third point (9, 7) into the equation of line and check whether the point satisfies the equation.

9 = 7 - 2 ?

7 = 7 ✓

The above result is true. So the point (9, 7) satisfies the equation of line.

Hence, the three points (4, 2), (7, 5) and (9, 7) are collinear.

Example 2 :

(1, 4), (3, -2), (-3, 16)

Solution :

Formula for slope of a line joining two points :

m = (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = (1, 4) (x₂, y₂) = (3, -2).

m = (-2 - 4)/(3 - 1)

= -6/2

= -3

Equation of line in slope-intercept form :

y = mx + b

Substitute m = -3.

y = -3x + b ----(1)

The line is passing through the point (1, 4).

So, substitute x = 1 and y = 4.

4 = -3(1) + b

4 = -3 + b

Substitute b = 7 in (1).

y = -3x + 7

Now, substitute the third point (-3, 16) into the equation of line and check whether the point satisfies the equation.

16 = -3(-3) + 7 ?

16 = 9 + 7 ?

16 = 16 ✓

The above result is true. So the point (-3, 16) satisfies the equation of line.

Hence, the three points (1, 4), (3, -2) and (-3, 16) are collinear.

Example 3 :

Check whether points are collinear or not A(1, -1), B(5, 2) and C(9, 5)

Solution :

Slope of the line joining the points A and B :

Slope = (y₂ - y₁) / (x₂ - x₁)

= (2 + 1) / (5 - 1)

= 3/4

Slope of the line joining the points A and C :

Slope = (y₂ - y₁) / (x₂ - x₁)

= (5 + 1) / (9 - 1)

= 6/8

= 3/4

Since the slopes of are equal, they must lie on the same line. So, the given points are collinear.

Example 4 :

The line joining the points (2, 1) and (5, 8) is trisected at the points P and Q. If point P lies on the line 2x – y + k = 0, find the value of k.

Solution :

Since P and Q are the points which are trisecting the line joining the points (2, 1) and (5, -8).

P is dividing the line segment in the ratio 1 : 2

= (lx₂ + mx₁)/(l + m), (ly₂ + my₁)/(l + m)

= [1(5) + 2(2)]/(1 + 2), [1(-8) + 2(1)]/(1 + 2)

= (5 + 4)/3, (-8 + 2)/3

= (9/3, -6/3)

= (3, -2)

Since the point P lies on the line 2x – y + k = 0

2(3) - (-2) + k = 0

6 + 2 + k = 0

8 + k = 0

k = -8

So, the value of k is -8.

Example 5 :

For what value of k, the points (k, –1), (5, 7) and (8, 11) are collinear? 

Solution :

Let the given points be A(k, –1), B(5, 7) and C(8, 11)

Slope of the line joining the points A and B :

Slope = (y₂ - y₁) / (x₂ - x₁)

= (7 + 1) / (5 - k)

= 6/(5 - k) -------(1)

Slope of the line joining the points A and C :

Slope = (y₂ - y₁) / (x₂ - x₁)

= (11 + 1) / (8 - k)

= 12/(8 - k) -------(2)

(1) = (2)

6/(5 - k) = 12/(8 - k)

1/(5 - k) = 2/(8 - k)

8 - k = 2(5 - k)

8 - k = 10 - 2k

-k + 2k = 10 - 8

k = 2

Sim the value of k is 2.

Example 6 :

For what value of k are the points (k, 2 – 2k), (–k + 1, 2k) and (–4 – k, 6 – 2k) are collinear?

Solution :

Since the points are collinear their slopes will be equal. Let the given points be  A (k, 2 – 2k), B (–k + 1, 2k) and C(–4 – k, 6 – 2k)

Slope of the line joining the points A and B :

Slope = (2k - (2 - 2k)) / (-k + 1 - k)

= (2k - 2 + 2k)/(-2k + 1)

= (4k - 2)/(-2k + 1) ---------(1)

Slope of the line joining the points A and C :

Slope = (6 – 2k - (2 - 2k)) / ((-4 - k)- k)

= (6 – 2k - 2 + 2k) / (-4 - k - k)

= 4 / (-4 - 2k) ---------(2)

(1) = (2)

(4k - 2)/(-2k + 1) = 4 / (-4 - 2k)

(4k - 2) (-4 - 2k) = 4(-2k + 1)

-16k - 8k² + 8 + 4k = -8k + 4

- 8k² + 8 - 12k = -8k + 4

- 8k² - 12k + 8k + 8 - 4 = 0

- 8k² - 4k + 4 = 0

Dividing by -4

==> 2k² + k - 1 = 0

2k² + k - 1 = 0

(2k - 1)(k + 1) = 0

k = 1/2 and k = -1

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