Find the equation of the bisectors of the angles between the lines $4x+3y-12=0$ and $y=3x$.

THE DISTANCE

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[The angle bisectors are the locus of a point which is equidistance from the two lines, and this provides a method of finding their equations.] Let $P(X, Y)$ be a point on the locus, then the distance of $P$ from the lines $4x+3y-12=0$ and $y-3x=0$ are $\pm\frac{4X-3Y-12}{\sqrt{(4^2+3^2)}}$ and $\pm\frac{Y-3X}{\sqrt{(3^2+1^2)}}$

But $P$ is equidistant from the two lines, therefore

$\frac{4X-3Y-12}{5}=\pm\frac{Y-3X}{\sqrt{10}}$

[One $\pm$ sign has been dropped, since there are only two distinct equations: one given by the same sign each side, the other by different signs.]

Simplifying these equations we obtain

$4\sqrt{10X}+3\sqrt{10Y}-12\sqrt{10}=5Y-15X$

And

$4\sqrt{10X}+3\sqrt{10Y}-12\sqrt{10}=-5Y+15X$

Therefore the equations of the angle bisectors of the lines are

$(4\sqrt{10}+15)x+(3\sqrt{10}-5)y-12\sqrt{10}=0$

And

$(4\sqrt{10}-15)x+(3\sqrt{10}+5)y-12\sqrt{10}=0$

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Asked: 1 month ago