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, (i
1 answer:
9514 1404 393
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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Answer:
yes
Step-by-step explanation:
You can always separate an equation into two parts and see where those graphs intersect.
Joel's method works well.
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<em>Additional comments</em>
Preston should know that the invention of logarithms makes it easy to solve equations like this. x = log₂(14) = log(14)/log(2) ≈ 3.8073549.
As for Joel's method, I prefer to subtract the right side to get the equation ...
2^x -14 = 0
Then graphing y = 2^x -14, I look for the x-intercept. Most graphing calculators make it easy to find x- and y-intercepts. Not all make it easy to find points of intersection between different curves.
