Question

Find the distance of the point (–1, 1) from the line 12( x + 6) = 5( y – 2).

Answer 1

The equation of line is 12( x+6 )=5( y−2 ) .

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

Ax+By+C=0 (1)

Rearrange the terms of equation of line.

12x+72=5y−10 12x−5y+72+10=0 12x−5y+82=0

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

A=12, B=−5, C=82 (2)

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

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

Substitute the value of ( x 1 , y 1 ) as ( −1,1 ) and the values of A , B ,and C from equation (2) to equation (3).

d= | 12×−1+−5×1+82 | 12 2 + ( −5 ) 2 d= | −12−5+82 | 144+25 d= | −17+82 | 169 d= | 65 | 13

Further simplify the above equation.

d= 65 13 =5 units

Thus, the distance of the point ( −1,1 ) from the line 12( x+6 )=5( y−2 ) is 5 units .