1
$\begingroup$

A triangle $\triangle ABC$ one of his vertex is the point $C(4,3)$.

The bisector line equation is $x+2y-5=0$ and the median line equation is $4x+3y-10=0$ drawn from the same vertex.

Find the coordinates of the vertices A and B.

What i have done:

I started using the fact that the two equations starts on the same vertex, but is not $C$. So, let $A$ be that vertex, i calculate the intersection point by the system: $$y_A=-1/2\cdot x_A+5/2$$ $$y_A=-4/3\cdot x_A+10/3$$

Which implies that $A(1,2)$

But i dont know how to find $B$.

$\endgroup$
  • $\begingroup$ Hint: use the fact that $\vec{AC}+\vec{AB}$ is parallel to the director vector of the median... $\endgroup$ – Martigan Oct 27 '14 at 17:02
1
$\begingroup$

The segment connecting $A$ in $(1,2)$ with point $C$ in $(4,3)$ lies on the line $y=\frac{x}{3}+\frac{5}{3}$ and has length $\sqrt{3^2+1^2}=\sqrt{10}$.

Note that the bisector crosses the $x$-axis in point $(5,0)$. Let us call $D$ this point. The segment $CD$ lies on the line $y=-3x+15$ and has length $\sqrt{3^2+1^2}=\sqrt{10}$.

Because segments $AC$ and $CD$ have equal length and lie on lines whose slopes are one the negative inverse of the other, we get that $ACD$ is an isosceles right triangle. Thus, the bisector $AD$ forms with $AC$ a $45^o$ degree angle, which means that the angle $CAB$ is right. As a result, the leg $AB$ of the right triangle $ABC$ has slope equal to $CD$, and then lies on the line $y=-3x+5$.

Now let us call $s$ the slope of the segment $BC$ and $E$ its middle point. This segment lies on the line $y=sx+3-4s$, which crosses the median $y=\frac{10-4x}{3}$ in $(\frac{12s+1}{3s+4}, \frac{12-6s}{3s+4})$. So the length of segment $CE$ is

$$\sqrt{(4-\frac{12s+1}{3s+4})^2+(3-\frac{12-6s}{3s+4})^2} \\ = \frac{\sqrt{15^2+(15s)^2}}{3s+4}= \frac{15\sqrt{s^2+1}}{3s+4}$$

On the other hand, the segment $BC$ crosses the leg $AB$ in $(\frac{4s+2}{s+3}, \frac{9-7s}{s+3})$. So the length of segment $BE$ is

$$\sqrt{( \frac{4s+2}{s+3} -\frac{12s+1}{3s+4})^2+( \frac{9-7s}{s+3} -\frac{12-6s}{3s+4})^2 } \\ = \frac{\sqrt{(5-15s)^2+(5s-15s^2)^2}}{(s+3)(3s+4)}= \frac{5(3s-1)\sqrt{s^2+1}}{(s+3)(3s+4)}$$

Since $BE=CE$ we get

$$\frac{5(3s-1)\sqrt{s^2+1}}{(s+3)(3s+4)}=\frac{15\sqrt{s^2+1}}{3s+4})$$

which reduces to

$$\frac{3s-1}{s+3}=3$$

The last equation has no real solutions, except for $s \rightarrow \infty$. This means that the segment $BC$ is vertical, i.e., it lies on a line of the form $x=k$. Since it includes point $C$, we get that segment $BC$ lies on the line $x=4$. Knowing this, it is not difficult to find its intersection with line $AB$, allowing to conclude that point $B$ is in $(4,-7)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.