Finding K for Collinear Points in Coordinate Geometry

Three points are said to be collinear if they lie on the same straight line. We can determine if points are collinear by checking if the slope between the first two points is equal to the slope between the second and third points.

Given Points

Let the three given points be:

• $P_1 = (K, 2 - 2K)$
• $P_2 = (-K + 1, 2K)$
• $P_3 = (-4 - K, 6 - 2K)$

Method: Using Slopes

The slope formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.

For the points to be collinear, the slope between $P_1$ and $P_2$ must be equal to the slope between $P_2$ and $P_3$. Let's denote these slopes as $m_{12}$ and $m_{23}$ respectively.

Calculating Slopes

Slope between $P_1$ and $P_2$ ($m_{12}$):

Using the slope formula:

$m_{12} = \frac{(2K) - (2 - 2K)}{(-K + 1) - K}$

$m_{12} = \frac{2K - 2 + 2K}{-K + 1 - K}$

$m_{12} = \frac{4K - 2}{1 - 2K}$

Slope between $P_2$ and $P_3$ ($m_{23}$):

Using the slope formula:

$m_{23} = \frac{(6 - 2K) - (2K)}{(-4 - K) - (-K + 1)}$

$m_{23} = \frac{6 - 4K}{-4 - K + K - 1}$

$m_{23} = \frac{6 - 4K}{-5}$

Condition for Collinearity

For the points to be collinear, $m_{12} = m_{23}$.

$\frac{4K - 2}{1 - 2K} = \frac{6 - 4K}{-5}$

Solving for K

Now, we solve this equation for $K$. Cross-multiply:

$-5(4K - 2) = (1 - 2K)(6 - 4K)$

Expand both sides:

$-20K + 10 = 1(6 - 4K) - 2K(6 - 4K)$

$-20K + 10 = 6 - 4K - 12K + 8K^2$

$-20K + 10 = 8K^2 - 16K + 6$

Rearrange the terms to form a quadratic equation:

$8K^2 - 16K + 20K + 6 - 10 = 0$

$8K^2 + 4K - 4 = 0$

Divide the entire equation by 4 to simplify:

$2K^2 + K - 1 = 0$

Factoring the Quadratic Equation

We can factor this quadratic equation. We look for two numbers that multiply to $(2 \times -1) = -2$ and add up to $1$. These numbers are $2$ and $-1$.

$2K^2 + 2K - K - 1 = 0$

Factor by grouping:

$2K(K + 1) - 1(K + 1) = 0$

$(2K - 1)(K + 1) = 0$

Determining the Values of K

Setting each factor to zero gives the possible values for $K$:

• $2K - 1 = 0 \implies 2K = 1 \implies K = \frac{1}{2}$
• $K + 1 = 0 \implies K = -1$

Conclusion

The points are collinear when $K = \frac{1}{2}$ or $K = -1$. This corresponds to options (A) and (D).