The point-slope form of a line that has a slope of 1/4 and passes through the point (3, 0) is shown. y-0-r-3) What is the equati
on in slope-intercept form?
1 answer:
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Answer:
Part A
The bearing of the point 'R' from 'S' is 225°
Part B
The bearing from R to Q is approximately 293.2°
Step-by-step explanation:
The location of the point 'Q' = 35 km due East of P
The location of the point 'S' = 15 km due West of P
The location of the 'R' = 15 km due south of 'P'
Part A
To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that =
, the right triangle ΔRPS is an isosceles right triangle
∴ ∠PRS = ∠PSR = 45°
The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°
Part B
∠PRQ = arctan(35/15) ≈ 66.8°
Therefore the bearing from R to Q = 270 + 90 - 66.8 ≈ 293.2°
