Find the value of k for which the points (3k-1,k-2)(k,k-7)&(k-1,-k+2)are collinear
Answers
Given:
- (3k - 1, k - 2)
- (k, k - 7)
- (k - 1, -k + 2)
To find: The value of k.
Answer:
If the given points were collinear, then,
From the given points,
Using them in the formula,
Let's use the quadratic formula to find the value of x.
From the equation,
- a = 1
- b = -6
- c = 1
Using them in the formula,
Given:
The points (3k - 1, k - 2), (k, k - 7) and (k - 1, - k + 2) are collinear.
To find: We have to find the value of k.
Solution:
Let us know how to find the area of a triangle in Cartesian coordinates:
Let (x₁, y₁), (x₂, y₂) and (x₃, y₃) are the coordinates of the vertices of a triangle.
The area of the triangle is given by
= 1/2 * | {x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)} | unit²
We find the area of a triangle formed by the given points in order to show that the points are collinear:
The coordinates of the vertices are (3k - 1, k - 2), (k, k - 7), (k - 1, - k + 2).
Thus the area of the triangle is given by
1/2 * {(3k - 1) (k - 7 + k - 2) + k (- k + 2 - k + 2) + (k - 1) (k - 2 - k + 7)} unit²
= 1/2 * {(3k - 1) (2k - 9) + k (- 2k + 4) + (k - 1) (5)} unit²
= 1/2 * (6k² - 27k - 2k + 9 - 2k² + 4k + 5k - 5) unit²
= 1/2 * (4k² - 20k + 4) unit²
= 2 (k² - 5k + 1) unit²
Now we use the condition for collinear points where the area of the triangle must be zero:
ATQ:
k² - 5k + 1 = 0
or, k² - 2 * k * 5/2 + (5/2)² + 1 - 25/4 = 0
or, (k - 5/2)² - 21/4 = 0
or, (k - 5/2)² - {(√21)/2}² = 0
or, {k - 5/2 + (√21)/2} {k - 5/2 - (√21)/2} = 0
This gives
k = 5/2 - (√21)/2, 5/2 + (√21)/2
or, k = (5 - √21)/2, (5 + √21)/2
or, k = (5 ± √21)/2