a) To find the gradient of the line AB, we use the formula:
gradient = (y2 - y1) / (x2 - x1)
where (x1, y1) = (-1, -2) and (x2, y2) = (7, 2)
Gradient of AB = (2 - (-2)) / (7 - (-1))
= 4 / 8
= 0.5
b) Since ABC is a right angle, the gradient of AC * gradient of BC = -1
Let m be the gradient of AC (from A to C), then the gradient of BC would be -1/m
The gradient of AC = (4 - (-2)) / (k - (-1)) = 6 / (k+1)
Therefore, the gradient of BC = -1 / (6 / (k + 1)) = - (k + 1) / 6
Since AB is perpendicular to BC, their gradients are negative reciprocals of each other.
Therefore gradient of AB = a = -1 / m, the gradient of AC
0.5 = -1 / (6 / (k + 1))
0.5 = - (k + 1) / 6
0.5 = (1 - k) / 6
3 = 1 - k
k = -2
c) First, we find the gradient of the line passing through B and C:
gradient = (4 - 2) / (k - 7) = 2 / (7 - k) = -2/(k-7)
The equation of the line passing through B (7, 2) with gradient -2/(k-7) is given by:
y - 2 = -2 / (k - 7) (x - 7)
y - 2 = -2 / (k - 7) x + 2
y = -2 / (k - 7) x + 4
The points A(-1,-2),B(7,2) and C(k, 4) where k is constant, are the verticles of angle ABC. ABC is a right angle.
a) Find the gradient of AB
b) Calculate the value of k
c) Find the equation of the line passing through B and C
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