The equations of bisectors of two lines `L_1 & L_2` are `2x-16y-5=0` and `64x+ 8y+35=0`. lf the line `L_1` passes through `(-11, 4)`, the equation of acute angle bisector of `L_1` & `L_2` is:
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From the figure, `d_(1) = (|2(-11)-16(4)-5|)/(sqrt(4+256)) = (91)/(2sqrt(65)) ~= 5.6 " " (1)` `d_(2) = (|64(-11)+8(4)+35|)/(sqrt((64)^(2)+8^(2))) = (637)/(64.49) ~= 9.87 " " (2)` `d_(1) lt d_(2)` Therefore, 2x-16y-5=0 is acute angle bisector and 64x+8y+35=0 is obtuse angle bisector.
Find the equation of the bisectors of the anglebetween the lines 4x+3y=5 and x +2y+3=0.
theta_1 and theta_2 are the inclination of lines L_1a n dL_2 with the x-axis. If L_1a n dL_2 pass through P(x_1,y_1) , then the equation of one of the angle bisector of these lines is (a) (x-x_1)/(cos((theta_1+theta_2)/2))=(y-y_1)/(sin((theta_1+theta_2)/2)) (b) (x-x_1)/(-sin((theta_1+theta_2)/2))=(y-y_1)/(cos((theta_1+theta_2)/2)) (c) (x-x_1)/(sin((theta_1+theta_2)/2))=(y-y_1)/(cos((theta_1+theta_2)/2)) (d) (x-x_1)/(-sin((theta_1+theta_2)/2))=(y-y_1)/(cos((theta_1+theta_2)/2))
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