point A(-1,-2) B(7,2) C(k,4) K is a constant. The vertices of ABC. Angle ABC is a right angle.
Ive worked out the gradient of AB to be 1/2. And the negative reciprocals is -2.
How do you calculate the value of k??
thanks for the help
To find the value of k, you can use the fact that the product of the gradients of perpendicular lines is -1.
First, let's find the gradient (or slope) of BC. The gradient of BC is equal to the change in y divided by the change in x, which is (4 - 2) / (k - 7) = 2 / (7 - k).
Since angle ABC is a right angle, the gradient of BC is the negative reciprocal of the gradient of AB. In other words, you can set -2 = -1 / (1/2).
Simplifying the equation, you get -2 = -2/1.
Therefore, the equation 2/(7 - k) = -2/1 can be formed.
Cross-multiplying, you get 2 = -2(7 - k).
Expanding, you have 2 = -14 + 2k.
Simplifying further, you get 16 = 2k.
Dividing both sides by 2, you find k = 8.
Therefore, the value of k that satisfies the conditions of a right angle at vertex B is 8.