Question

Show that the line5x-2y-1=0 is mid-parallel to the lines 5x-2y-9=0 and 5x-2y+7=0

Answer 1

we have to show that line 5x - 2y - 1 = 0 is mid - parallel to lines 5x - 2y - 9 = 0 and 5x - 2y + 7 = 0
means, we have to show that distance between line 5x - 2y - 1 = 0 and line 5x - 2y - 9 = 0 is same as distance between line 5x - 2y - 1 = 0 and line 5x - 2y + 7 = 0.

we know, formula of distance between two
parallel lines .it is given by $d=\frac{c_1-c_2}{\sqrt{a^2+b^2}}$

so, distance between 5x - 2y - 1 = 0 and 5x - 2y - 9 = 0 is $d_1=\frac{|-9-(-1)|}{\sqrt{5^2+2^2}}$
= $\frac{8}{\sqrt{29}}$

and distance between 5x - 2y - 1 = 0 and 5x - 2y + 7 = 0 is $d_2=\frac{7-(-1)}{\sqrt{5^2+2^2}}$
=$\frac{8}{\sqrt{29}}$

as you can see that, $d_1=d_2$
so, it is clear that line5x-2y-1=0 is mid-parallel to the lines 5x-2y-9=0 and 5x-2y+7=0

Answer 2

Given:

Line to be proved parallel: 5x - 2y - 1 = 0

Line 1: 5x - 2y - 9 = 0

Line 2: 5x - 2y + 7 = 0

To prove:

Same distance between the line and the other two parallel lines.

Proof:

To find the distance between two parallel lines,

c1 - c2 /√a^2 + b^2

So,

Distance between 5x - 2y - 1 = 0 and 5x - 2y - 9 = 0 is 8 / √29

Distance between 5x - 2y - 1 = 0 and 5x - 2y + 7 = 0 is 8 / √29

It is proven that the line 5x - 2y - 1 = 0 is mid-parallel to the lines 5x - 2y - 9 = 0 and 5x - 2y + 7 = 0 as their respective distance is 8 / √29

Hence, Proved.