Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear
Let the points (1, 5), (2, 3), and (−2, −11) be representing the vertices A, B, and C of the given triangle respectively.
Let, A=(1,5); B=(2, 3); C=(−2, −11)
\(\therefore\) AB=\(\sqrt{(1-2)^2+(5-3)^2}=\sqrt5\)
BC=\(\sqrt{(2-(-2))^2+(3-(-11))^2}=\sqrt{4^2+14^2}=\sqrt{16+196}=\sqrt{212}\)
CA=\(\sqrt{(1-(-2))^2+(5-(-11))^2}=\sqrt{3^2+16^2}=\sqrt{9+256}=\sqrt{265}\)
Since, AB+BC\(\neq\)CA
Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.
What is the angle between the hour and minute hands at 4:30?
Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
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Frequency | 11 | 8 | 15 | 7 | 10 | 9 |