195o
135o
225o
045o
Correct answer is B
tan\(\theta\) = \(\frac{100}{100}\) = 1
\(\theta\) = tan-1(1) = 45o
The bearing of x from z is N45oE or 135o
\(\frac{2}{3}\)
\(\frac{3}{5}\)
\(\frac{1}{5}\)
\(\frac{4}{5}\)
Correct answer is C
tan\(\theta\) = \(\frac{3}{4}\)
from Pythagoras tippet, the hypotenus is T
i.e. 3, 4, 5.
then sin \(\theta\) = \(\frac{3}{5}\) and cos\(\theta\) = \(\frac{4}{5}\)
cos\(\theta\) - sin\(\theta\)
\(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\)
Find the equation of a line perpendicular to line 2y = 5x + 4 which passes through (4, 2).
5y - 2x -18 = 0
5y + 2x - 18 = 0
5y - 2x + 18 = 0
5y + 2x - 2 = 0
Correct answer is B
2y = 5x + 4 (4, 2)
y = \(\frac{5x}{2}\) + 4 comparing with
y = mx + e
m = \(\frac{5}{2}\)
Since they are perpendicular
m1m2 = -1
m2 = \(\frac{-1}{m_1}\) = -1
\(\frac{5}{2}\) = -1 x \(\frac{2}{5}\)
The equator of the line is thus
y = mn + c (4, 2)
2 = -\(\frac{2}{5}\)(4) + c
\(\frac{2}{1}\) + \(\frac{8}{5}\) = c
c = \(\frac{18}{5}\)
y = -\(\frac{2}{5}\)x + \(\frac{18}{5}\)
5y = -2x + 18
or 5y + 2x - 18 = 0
The midpoint of P(x, y) and Q(8, 6) is (5, 8). Find x and y.
(2, 10)
(2, 8)
(2, 12)
(2, 6)
Correct answer is A
P(x, y) Q(8, 6)
midpoint = (5, 8)
x + 8 = 5
\(\frac{y + 6}{2}\) = 8
x + 8 = 10
x = 10 - 8 = 2
y + 6 = 16
y + 16 - 6 = 10
therefore, P(2, 10)
The perpendicular bisector of a line XY is the locus of a point
whose distance from X is always twice its distance from Y
whose distance from Y is always twice its distance from X.
which moves on the line XY
which is equidistant from the points X and Y
Correct answer is D
No explanation has been provided for this answer.