Question

a straight line is parallel to the line 3x-y-3=0 and 3x-y+5=0 and lies between them find equation of the line if it's distance from these lines are in tha ratio 3:5​

Answer 1

Answer:

Required line is 3x - y = 0

Step-by-step explanation:

Let the parallel line be

3x - y + k = 0 ..... (1)

Then the distance of the line (1) from the parallel line 3x - y - 3 = 0 is

= |k + 3|/√(3² + 1²) units

= |k + 3|/√10 units

and the distance from the line 3x - y + 5 = 0 is

= |k - 5|/√(3² + 1²) units

= |k - 5|/√10 units

By the given condition,

|k + 3|/√10 : |k - 5|/√10 = 3 : 5

or, |k + 3| / |k - 5| = 3/5

or, 5 |k + 3| = 3 |k - 5|

or, 25 (k² + 6k + 9) = 9 (k² - 10k + 25)

or, 25k² + 150k + 225 = 9k² - 90k + 225

or, 16k² + 240k = 0

or, k (k + 15) = 0

Either k = 0 or, k + 15 = 0

i.e., k = 0, - 15

Though the line 3x - y - 15 = 0 [ by (1) ] is parallel to the given lines, but doesn't lie between them; only 3x - y = 0 lies between them as their parallel.

Therefore the required parallel line is 3x - y = 0

Remark:

However if we alter the ratio to be 5 : 3, we get another line 3x - y + 2 = 0 parallel to the given lines and lying between them. Give it a try.

Answer 2

Answer:

3x - y + 2 = 0

3x - y = 0

Step-by-step explanation:

a straight line is parallel to the line 3x-y-3=0 and 3x-y+5=0

so line would be

3x - y + k  = 0

As lines are parallel

Distance between given lines  = 5 -(-3) = 8

Distance of new line is in ratio of 3: 5

=> 3  & 5

Let say Distance 3  from 3x-y+5=0

Then k  = 5 ± 3   = 8 or 2

as line lies between given lines

Hence k = 2

3x - y + 2 = 0

Let say Distance 5  from 3x-y+5=0

Then k  = 5 ± 5   = 10 or 0

as line lies between given lines

Hence k = 0

3x - y + 0 = 0

=> 3x - y = 0