BITSAT 2011 Paper

The given line is 12 (x + 6) = 5(y – 2)
⇒ 12x + 72 = 5y – 10
or 12x – 5y + 72 + 10 = 0
⇒ 12x – 5y + 82 = 0
The perpendicular distance from (x1,y1) to
the line ax + by + c = 0 is

(ax1+by1+c)

√a2+b2

The point (x1,y1) is (–1, 1), therefore, perpendicular distance from (–1, 1) to the line 12x - 5y + 82 = 0
=

|−12−5+82|

√122+(−5)2

=

65

√144+25

=

65

√169

= 5