The family of lines,forming an isosceles tria | vedclass

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(C) The family of lines forming an isosceles triangle with two given lines must be parallel to the angle bisector of the angle between the two lines.F

Vedclass · Accepted Answer

$9x - 7y + c = 0$

Vedclass · Answer

(C) The family of lines forming an isosceles triangle with two given lines must be parallel to the angle bisector of the angle between the two lines.First,we find the angle bisectors of the lines $L_1: 3x - 4y - 2 = 0$ and $L_2: 12x - 5y + 6 = 0$.The equation of the bisectors is given by $\frac{3x - 4y - 2}{\sqrt{3^2 + (-4)^2}} = \pm \frac{12x - 5y + 6}{\sqrt{12^2 + (-5)^2}}$.$\frac{3x - 4y - 2}{5} = \pm \frac{12x - 5y + 6}{13}$.Taking the negative sign (to get the bisector of the angle containing the origin or specific orientation):$13(3x - 4y - 2) = -5(12x - 5y + 6)$$39x - 52y - 26 = -60x + 25y - 30$$99x - 77y + 4 = 0$.Dividing by $11$,we get $9x - 7y + \frac{4}{11} = 0$.Thus,the family of lines parallel to this bisector is $9x - 7y + c = 0$.

The family of lines,forming an isosceles triangle with the lines $3x - 4y - 2 = 0$ and $12x - 5y + 6 = 0$,is

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