Question

equation of the line which is equidistant from 3x+4y-5=0 and 3x+4y-13=0​

Answer 1

Answer:

In order to determine the distance let’s first find a line perpendicular to 3x - 4y + 15 =0 which passes through the origin (to find te perpendicular distance from 3x - 4y + 15 =0 to the origin). In order to find the equation of that line we need its gradient and a point on the line.

To find the gradient we first need to find the gradient of the the line 3x - 4y + 15 =0. By rearranging the equation in the form y=mx+c — where “m” is the gradient and c is the y-intercept — we get:

4y = 3x + 15

y = 3/4 . x + 15/4

We hence now know that the gradient of the line 3x - 4y + 15 =0 is 3/4.

Any line perpendicular to 3x - 4y + 15 =0 will hence has as gradient: -4/3

Hence the gradient of our perpendicular line is -4/3.

Now we need a known point on the line. We know that our perpendicular line passes through the origin. We hence know that it passes through the point (0,0).

Answer 2

Answer:

3x+4y-9=0

Step-by-step explanation:

The line is equidistant from

1)3x+4y-5=0

2)3x+4y-13=0

then,

3x+4y+k =0

where k is the mean of -5 and -13(constant of (1) and (2)

k= -5-13/2

k=-18/2

k= -9

So the equation of line is

3x + 4y - 9 = 0