P is the point (16,9) and Q is the point (22. 24). (a) Find the equation of the line perpendicular to PQ that passes through the point (5, 1).
Give your answer in the form y=mx+c.
(b) N is the point on PQ such that PN=2NQ. Find the co-ordinates of N.

see explanation

Step-by-step explanation:

(a)

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m =

with (x₁, y₁ ) = (P(16, 9) and (x₂, y₂ ) = Q(22, 24)

= = =

Given a line with slope m then the slope of a line perpendicular to it is

= - = - = - , then

y = - x + c ← is the partial equation

To find c substitute (5, 1) into the partial equation

1 = - 2 + c ⇒ c = 1 + 2 = 3

y = - x + 3 ← equation of perpendicular line

(b)

Given PN = 2NQ , then N sections PQ in the ratio 2 : 1

Using the Section formula to find coordinates of N

N = ( , )

   = ( , )

    = ( , )

    = (20, 19 )

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29

arc length = radius * x ( in radians)

6 * pi = 18 * x

divide by 18 on each side

6 pi / 18 = x

pi/3 = x

the angle is pi / 3 radians

or   60 degrees