Question

Find the equation of the angle bisectors of the 3x+4y=5 and 12x+5y=5. Also find the bisector which bisect the acute angle.

Answer 1

Answer on Question #44317 - Math - Other

Find the equation of the angle bisectors of the $3x + 4y = 5$ and $12x + 5y = 5$ . Also find the bisector which bisect the acute angle.

Solution:

The bisector of an angle is the locus of points on the plane that are equidistant from the rays that form the angle.

$$\mathrm {d} (\mathrm {P}, \mathrm {r}) = \mathrm {d} (\mathrm {P}, \mathrm {s})$$ $$r: A _ {1} x + B _ {1} y + C _ {1} = 0$$ $$r: 3 x + 4 y - 5 = 0$$ $$s: A _ {2} x + B _ {2} y + C _ {2} = 0$$ $$s: 1 2 x + 5 y - 5 = 0$$ $$\mathrm {P} (\mathrm {x}, \mathrm {y})$$ $$\frac {\left| A _ {1} x + B _ {1} y + C _ {1} \right|}{\sqrt {A _ {1} ^ {2} + B _ {1} ^ {2}}} = \frac {\left| A _ {2} x + B _ {2} y + C _ {2} \right|}{\sqrt {A _ {2} ^ {2} + B _ {2} ^ {2}}}$$

Two angle bisectors are perpendicular between themselves.

$$\frac {\left| 3 x + 4 y - 5 \right|}{\sqrt {3 ^ {2} + 4 ^ {2}}} = \frac {\left| 1 2 x + 5 y - 5 \right|}{\sqrt {1 2 ^ {2} + 5 ^ {2}}}$$ $$5 (1 2 x + 5 y - 5) = 1 3 (3 x + 4 y - 5)$$ $$2 1 x + 4 0 = 2 7 y$$ $$y = \frac {7}{9} x + \frac {4 0}{2 7}$$ $$5 (1 2 x + 5 y - 5) = - 1 3 (3 x + 4 y - 5)$$ $$9 9 x + 7 7 y = 9 0$$ $$y = - \frac {9}{7} x + \frac {9 0}{7 7}$$

Now let's take one of the given lines and let its slope be $m_{1}$ and take one of the bisectors and let its slope be $m_{2}$ .

If $\theta$ be the acute angle between them, then find $\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$

If $\tan \theta > 1$ then the bisector taken is the bisector of the obtuse angle and the other one will be the bisector of the acute angle.

First line $3x + 4y - 5 = 0$ has a slope of $-\frac{3}{4}$

Then $m_1 = -\frac{3}{4}$

$y = -\frac{9}{7}x + \frac{90}{77}$ is bisector, that has a slope of $-\frac{9}{7}$

Then $m_2 = -\frac{9}{7}$

$$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-\frac{3}{4} - \left(-\frac{9}{7}\right)}{1 + \left(-\frac{3}{4}\right) \left(-\frac{9}{7}\right)} \right| = \frac{3}{11}$$ $$\frac{3}{11} < 1$$

Thus, $y = -\frac{9}{7}x + \frac{90}{77}$ is bisector which bisect this acute angle.

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