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.
48 cm3
33 cm3
60 cm3
27 cm3
Correct answer is B
Volume of cube = L3
33 = 27cm3
volume of rectangular tank = L x B X h
= 3 x 4 x 5
= 60cm3
volume of H2O the tank can now hold
= volume of rectangular tank - volume of cube
= 60 - 27
= 33cm3
4√6 cm
3√6 cm
6√6 cm
2√6 cm
Correct answer is A
From Pythagoras theorem
|OA|2 = |AN|2 + |ON|2
72 = |AN|2 + (5)2
49 = |AN|2 + 25
|AN|2 = 49 - 25 = 24
|AN| = \(\sqrt {24}\)
= \(\sqrt {4 \times 6}\)
= 2√6 cm
|AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts)
|AN| + |NB| = |AB|
2√6 + 2√6 = |AB|
|AB| = 4√6 cm