Sure, let's solve the problem step-by-step.
Part (a): AB is perpendicular to AC
- Given the points are A(-1, 6), B(3, -12), and C(k, 6).
- First, we find the slope of AB:
[tex]\[
\text{slope of AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-12 - 6}{3 - (-1)} = \frac{-18}{4} = -4.5
\][/tex]
- Next, we find the slope of AC. Given C has coordinates (k, 6) and A(-1, 6):
[tex]\[
\text{slope of AC} = \frac{6 - 6}{k - (-1)} = \frac{0}{k + 1} = 0
\][/tex]
- For the line AB to be perpendicular to AC, the product of their slopes should be -1. That is:
[tex]\[
(\text{slope of AB}) \times (\text{slope of AC}) = -1
\][/tex]
Substituting the values we have:
[tex]\[
(-4.5) \times 0 = 0
\][/tex]
Clearly, the product is not equal to -1. Hence, there is no value of 'k' that satisfies the condition where AB is perpendicular to AC.
Part (b): A, B, and C are collinear
- For the points to be collinear, the slopes of AB and AC must be equal.
We already have:
[tex]\[
\text{slope of AB} = -4.5
\][/tex]
[tex]\[
\text{slope of AC} = 0
\][/tex]
As seen previously, the slope of AC is 0 since:
[tex]\[
\text{slope of AC} = \frac{6 - 6}{k - (-1)} = 0
\][/tex]
- Since we need the slopes to be equal for collinearity:
[tex]\[
\text{slope of AB} = \text{slope of AC}
\][/tex]
-4.5 must equal 0 for the points to be collinear, which is not possible. Therefore, no value of ‘k’ will make the slopes of both AB and AC equal.
To find a value for [tex]\(k\)[/tex], we need to ensure that point C has to be specifically aligned with the given points A and B. Checking alignment for [tex]\((A, C)\)[/tex]:
Examining the X-coordinates of A(-1, 6) and C(k, 6):
- Points will be collinear when C is directly aligned vertically with A and B, hence k must be -1 (same X-coordinate).
Thus, the value of [tex]\(k\)[/tex] that makes A, B and C collinear is:
[tex]\[
k = -1
\][/tex]
So the answers are:
- For part (a): No value of 'k' satisfies the condition that AB is perpendicular to AC.
- For part (b): The value of 'k' is -1 to make A, B, and C collinear.
