Question

Find the distance between parallel lines (i) 15 x + 8 y – 34 = 0 and 15 x + 8 y + 31 = 0 (ii) l ( x + y ) + p = 0 and l ( x + y ) – r = 0

Answer 1

(i)

The parallel lines are 15x+8y−34=0 and 15x+8y+31=0 .

The general form of the equation of line is given by,

Ax+By+C=0 (1)

Let A 1 , B 1 , C 1 be the values for the line 15x+8y−34=0 and A 2 , B 2 , C 2 be the values for the line 15x+8y+31=0 .

Compare the above expression with the general form of equation of line from equation (1).

A 1 =15,  B 1 =8,  C 1 =−34 A 2 =15,  B 2 =8,  C 2 =31

It is observed from the above expression

A 1 = A 2 ,  B 1 = B 2 C 1 =−34,  C 2 =31

The formula for the distance d between two parallel lines Ax+By+ C 1 =0 and Ax+By+ C 2 =0 is given by,

d= | C 1 − C 2 | A 2 + B 2 (2)

Substitute the values of A,B, C 1 , C 2 as 15,8,−34,31 in equation (2).

d= | −34−31 | 15 2 + 8 2 = | −65 | 225+64 = | −65 | 289 = | −65 | 17

Further simplify the above expression.

d= −( −65 ) 17 = 65 17  units

Thus, the distance between the parallel lines 15x+8y−34=0 and 15x+8y+31=0 are 65 17  units .

(ii)

The parallel lines are l( x+y )+p=0 and l( x+y )−r=0 .

Rearrange the terms of equation of lines.

lx+ly+p=0 lx+ly−r=0

Let A 1 , B 1 , C 1 be the values for the line lx+ly+p=0 and A 2 , B 2 , C 2 be the values for the line lx+ly−r=0 .

Compare the above expression with the general form of equation of line from equation (1).

A 1 =l,  B 1 =l,  C 1 =p A 2 =l,  B 2 =l,  C 2 =−r

It is observed from the above expression

A 1 = A 2 ,  B 1 = B 2 C 1 =p,  C 2 =−r

Substitute the values of A,B, C 1 , C 2 as l,l,p,−r in equation (2).

d= | p−( −r ) | l 2 + l 2 = | p+r | 2 l 2 = | p+r | 2 l = 1 2 | p+r | l

Thus, the distance between the parallel lines l( x+y )+p=0 and l( x+y )−r=0 is 1 2 | p+r | l .