Show that the perpendiculars falling from any point of the st.line 2x+11y=5 upon the two st. lines 24x+7y=20 and 4x-3y=2 are equal to each other.

solution:
given
L1 :24x+7y=20 ; slope M1 = -24/7
L2 ;4x-3y=2 ; slope M2= 4/3
and L3 :2x+11y=5 ; slope M3= -2/11

ifperpendiculars falling from any point from the line L3 to L1 and L2 areequal to each other.... then it means.. line L3 is angle bisector of then intersecting lines L1 and L2... that means ..
the angle between L1 and L3 = angle b/w L2 and L3

use slope formula..
angle b/w L1 and L3 = (M1 – M3)/(1 + M1*M3) …......(1)

angle b/w L2 and L3 = (M2 – M3)/(1 + M2*M3)................(2)

eqn (1) = eqn(2)




thanks and regards