Question

Find equation of the line which is equidistant from parallel lines 9 x + 6 y – 7 = 0 and 3 x + 2 y + 6 = 0.

Answer 1

The given equations of parallel lines are,

9x+6y−7=0(1)

3x+2y+6=0(2)

Let, ( p,q ) be the point equidistant from both the lines.

The general form of equation of line is given by,

Ax+By+C=0

Let, A 1 , B 1 , C 1 and A 2 , B 2 , C 2 be the values for the line having equations (1) and (2).

Comparing equations (1) and (2) with the general form of equation of line.

A 1 =9,  B 1 =6,  C 1 =−7 A 2 =3,  B 2 =2,  C 2 =6

The formula for the perpendicular distance d of a point ( x 1 , y 1 ) from the line Ax+By+C=0 is given by,

d= | A x 1 +B y 1 +C | A 2 +B 2 (3)

Let d 1 and d 2 be the perpendicular distance of the point ( p,q ) from the lines 9x+6y−7=0 and 3x+2y+6=0 respectively.

Substitute the values of A 1 ,  B 1 , and  C 1 in equation (3).

d 1 = | 9p+6q−7 | 9 2 + 6 2 = | 9p+6q−7 | 117 = | 9p+6q−7 | 3 13

Substitute the values of A 2 ,  B 2 , and  C 2 in equation (3).

d 2 = | 3p+2q+6 | 3 2 +2 2 = | 3p+2q+6 | 9+4 = | 3p+2q+6 | 13

The point ( p,q ) is equidistant from both the lines. So,

d 1 = d 2 (4)

Substitute the values of d 1 and d 2 in equation (4).

| 9p+6q−7 | 3 13 = | 3p+2q+6 | 13 | 9p+6q−7 | 3 = | 3p+2q+6 | 1 | 9p+6q−7 |=3×| 3p+2q+6 | ( 9p+6q−7 )=±3×( 3p+2q+6 )

Taking positive value,

9p+6q−7=3×( 3p+2q+6 ) 9p+6q−7=9p+6q+18 −7=18

This is not a possible solution.

Taking negative value,

9p+6q−7=−3×( 3p+2q+6 ) 9p+6q−7=−9p−6q−18 18p+12q+11=0

As ( p,q ) are coordinates along the x axis and y axis, respectively, rewrite the above equation in terms of x and y.

18x+12y+11=0

Thus, the equation of line is 18x+12y+11=0.