For what value of 'k' the points ( -k + 1, 2k), (k, 2 - 2k) and ( - 4 - k, 6 - 2k) are
collinear.
they are collinear when the differences are all in the same ratio. That is, when the slope from P1 to P2 is the same as the slope from P2 to P3:
((2-2k)-(2k))/((k)-(-k+1)) = ((6-2k)-(2-2k))/((-4-k)-(k))
(2-4k)/(2k-1) = (4)/(-4-2k)
...
k = -1
(2,-2),(-1,4),(-3,8): y=-2x+2
To determine the value of 'k' for which the given points are collinear, we can use the concept of collinearity.
Points are collinear if they lie on the same line. In other words, the slope of the line passing through two points should be the same as the slope of the line passing through any other pair of points on that line.
Let's find the slopes of the lines passing through pairs of points to identify the condition for collinearity.
First, let's find the slope of the line passing through points (-k + 1, 2k) and (k, 2 - 2k):
The slope, denoted as m₁, is given by:
m₁ = (change in y) / (change in x)
Using the coordinates of the given points, we have:
m₁ = (2 - 2k - 2k) / (k - (-k + 1))
m₁ = (2 - 4k) / (k + k - 1)
m₁ = (2 - 4k) / (2k - 1)
Similarly, let's find the slope of the line passing through points (-k + 1, 2k) and (-4 - k, 6 - 2k):
The slope, denoted as m₂, is given by:
m₂ = (change in y) / (change in x)
Using the coordinates of the given points, we have:
m₂ = (6 - 2k - 2k) / (-4 - k - (-k + 1))
m₂ = (6 - 4k) / (-5 - k)
m₂ = (6 - 4k) / (-k - 5)
Since the points are collinear, the slopes m₁ and m₂ should be equal:
m₁ = m₂
Substituting the expressions for m₁ and m₂, we have:
(2 - 4k) / (2k - 1) = (6 - 4k) / (-k - 5)
To solve this equation for 'k', we can cross-multiply and simplify:
(2 - 4k)(-k - 5) = (6 - 4k)(2k - 1)
(-2k - 10 + 4k^2 + 20k) = (12k - 6k^2 - 3 + 2k - 4k^2)
(-2k + 4k^2 + 12k) = (-12k^2 + 14k - 3)
(2k^2 + 2k - 3) = 0
Now, let's solve this quadratic equation for 'k':
2k^2 + 2k - 3 = 0
This quadratic equation can be factored as:
(2k - 1)(k + 3) = 0
Setting each factor equal to zero, we have:
2k - 1 = 0 or k + 3 = 0
Solving these equations, we get:
2k = 1 or k = -3
Therefore, for the points (-k + 1, 2k), (k, 2 - 2k), and (-4 - k, 6 - 2k) to be collinear, the values of 'k' can be either 1/2 (when 2k = 1) or -3.
Hence, when k = 1/2 or k = -3, the given points are collinear.