Find the value of k if the points (k, 3), (6, - 2), (- 3, 4) align.

Correct option is C

Given:

Points: (k, 3), (6, -2), (-3, 4).
For the points to be collinear, the area of the triangle formed by these points must be zero.

Formula Used:

The area of a triangle formed by three points $$(x_1, y_1), (x_2, y_2), (x_3, y_3)$$​ is:
$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$​
If the points are collinear, then:
Area = 0

Solution:

Substitute $$(x_1, y_1) = (k, 3), (x_2, y_2)$$​ = (6, -2), and (x_3, y_3) = (-3, 4) into the formula:
$$\frac{1}{2} \left| k(-2 - 4) + 6(4 - 3) + (-3)(3 - (-2)) \right| = 0$$​

1. Simplify inside the determinant:
$$\frac{1}{2} \left| k(-6) + 6(1) + (-3)(5) \right| = 0$$​

2. Expand and simplify:
$$\frac{1}{2} \left| -6k + 6 - 15 \right| = 0$$​
$$\frac{1}{2} \left| -6k - 9 \right| = 0$$​

3. Eliminate $$\frac{1}{2}$$​ :
$$\left| -6k - 9 \right| = 0$$​

4. Solve the equation:
-6k - 9 = 0
-6k = 9
k = $$-\frac{3}{2}$$​

Final Answer:

The value of k is $$**-\frac{3}{2}**$$​.