Given: O ∈ AB , ON is ∠ bisector of ∠AOC, OF is ∠ bisector of ∠COB, m∠FOB=70°. Find: m∠AON
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Answer:
The coordinates of HF are (1, 4)
Step-by-step explanation:
The parameters of the line are;
The coordinate of the end points are H = (-11, 7), and J = (5, 3)
The ratio by which the point F divides the line = 3:1
The segments in the line are HF, and FJ
Therefore;
The fraction of the length of HJ that is represented by HF = 3/(3 + 1) × HJ = 3/4 × HJ
HF = 3/4 × HJ
Which gives the coordinates of the point F as follows;
Coordinate of F = (-11 +(5 - (-11))×3/4, 7 + (3 - 7)×3/4) = (1, 4)
The coordinates of F are (1, 4)
We check the length of HF, from the equation for the length to of a line to get;
Similarly, we check the length of HJ, to get;
The length of HF = 3·√(17)
The length of HJ = 4·√(17)
Therefore, from HF = 3/4× HJ, we have;
HF = 3/4 × 4·√(17) = 3·√(17)
Therefore, the coordinates of HF are (1, 4)