Question

Find the value of k for which the points (3k-1,k-2)(k,k-7)&(k-1,-k+2)are collinear

Answer 1

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,

$\tt \dfrac{1}{2}\ \times\ \bigg[x_1(y_2\ -\ y_3)\ +\ x_2(y_3\ -\ y_1)\ +\ x_3(y_1\ -\ y_2)\bigg]\ =\ 0$

From the given points,

$\tt x_1\ =\ 3k\ -\ 1\\\\x_2\ =\ k\\\\x_3\ =\ k\ -\ 1\\\\y_1\ =\ k\ -\ 2\\\\y_2\ =\ k\ -\ 7\\\\y_3\ =\ -k\ +\ 2$

Using them in the formula,

$\tt \dfrac{1}{2}\ \times\ \bigg[3k\ -\ 1(k\ -\ 7\ +\ k\ -\ 2)\ +\ k(-k\ +\ 2\ -\ k\ +\ 2)\ +\ k\ -\ 1(k\ -\ 2\ -\ k\ +\ 7)\bigg]\ =\ 0$

$\tt \bigg[3k\ -\ 1(2k\ -\ 9)\ +\ k(-2k\ +\ 4)\ +\ k\ -\ 1(5)\bigg]\ =\ 0\\\\\\6k^2\ -\ 27k\ -\ 2k\ +\ 9\ -\ 2k^2\ +\ 4k\ +\ k\ -\ 5\ =\ 0\\\\\\4k^2\ -24k\ +\ 4\ =\ 0\\\\\\k^2\ -\ 6k\ +\ 1\ =\ 0$

Let's use the quadratic formula to find the value of x.

$\tt k\ =\ \dfrac{-b\ \pm\ \sqrt{p^2\ -\ 4ac}}{2a}$

From the equation,

• a = 1
• b = -6
• c = 1

Using them in the formula,

$\tt k\ =\ \dfrac{6\ \pm\ \sqrt{(6)^2\ -\ 4\ \times\ 1\ \times\ 1}}{2\ \times\ 1}\\\\\\k\ =\ \dfrac{6\ \pm\ \sqrt{36\ -\ 4}}{2}\\\\\\k\ =\ \dfrac{6\ \pm\ \sqrt{32}}{2}\\\\\\k\ =\ \dfrac{6\ \pm\ 4\sqrt{2}}{2}\\\\\\k\ =\ \dfrac{2(3\ \pm\ 2\sqrt{2})}{2}\\\\\\k\ =\ 3\ \pm\ 2\sqrt{2}\\\\\\\therefore\ \bf\ k\ =\ 3\ +\ 2\sqrt{2}\ \tt\ or\ \bf\ k\ =\ 3\ -\ 2\sqrt{2}.$

Answer 2

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

∴ k = (5 ± √21)/2