Let P=(23−1), Q=(−101), R=(21−4) and S=(53−2). Let l1 be the line passing through P and Q,

Question

and let l2 be the line passing through R and S.?

 

(i) The distance between R and l1 is....
(ii) The distance between l1 and l2 is....

 

Answer 1

(1)

The Points are P(23, -1) , Q (-10, 1) , R (21, -4) and S (53, -2)

L1 is the line Equation due to Points P and Q then

Points P (23, -1) and Q (-10, 1)

Slope of the line

m = (1-(-1))/(-10-23)

m = -2/33

Therefore line equation L1 is

Point be (-10,1)

y - 1 = (-2/33) ( x + 10)

33 y - 33 = -2x - 20

2x + 33y = 13.

Therefore Line Equation L1 is 2x + 33y = 13.

Now distance between R (21, -4) and L1 2x + 33y -13 = 0 is

Distance =

= 3.1155

Therefore distance between R (21, -4) and L1 2x + 33y -13 = 0 is 3.1155.

Answer 2

(2)

The Points are P(23, -1) , Q (-10, 1) , R (21, -4) and S (53, -2)

L1 is the line Equation due to Points P and Q then

Points P (23, -1) and Q (-10, 1)

Slope of the line

m = (1-(-1))/(-10-23)

m = -2/33

Therefore line equation L1 is

Point be (-10,1)

y - 1 = (-2/33) ( x + 10)

33 y - 33 = -2x - 20

2x + 33y = 13.

Therefore Line Equation L1 is 2x + 33y = 13.

L2 is the line Equation due to Points R and S then

PointsR (21, -4) and S (53, -2)

Slope of the line

m = (-2-(-4))/(53-21)

m = 2/32

Therefore line equation L1 is

Point be (21, -4)

y + 4 = (2/32) ( x - 21)

32 y + 128 = 2x - 42

2x - 32y = 170.

Therefore Line Equation L2 is 2x - 32y = 170.

The Equations L1 and L2 are not parallel. So they wil cross each other at some point then Distance is 0.