The coordinates of three points are A(- 1, - 3) , B(2, 3) and C(6, k) . If AB is perpendicular to BC find (i) the value of k, (ii) the gradient of AC (iii) the acute angle that AC makes with the x-axis.​

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Answer:

  (i) k = 1

  (ii) 4/7

  (iii) arctan(4/7) ≈ 29.7°

Step-by-step explanation:

(i) A graph of the given points is helpful. It shows us the slope of AB is ...

  mAB = rise/run = 2/1 = 2

so the slope of BC must be the opposite reciprocal, -1/2.

Point C is 6-2 = 4 units to the right of point B, so will be (-1/2)(4) = -2 units from point B in the vertical direction. That is, ...

  k = 3 -2

  k = 1

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(ii) The gradient of AC is found from the slope formula:

  m = (y2 -y1)/(x2 -x1)

  m = (1 -(-3))/(6 -(-1))

  mAC = 4/7

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(iii) The angle that line AC makes with the +x axis is the arctangent of the slope:

  arctan(4/7) ≈ 29.7° . . . angle between AC and +x axis

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