Find the equation of median without finding its vertices There is a triangle ABC. Equation of AB is $x + y = 2$, Equation of AC is $2x + 3y = 5$ and Equation of BC is $5x - y = 7$.
Given above, how do I find the equation of median $AD$ without finding any vertices of triangle ABC?
 A: It is easy to check that if points $B$ and $C$ had the same $x$ coordinate, then the slope of median $AD$ would be the average of the slopes of lines $AB$ and $AC$: 
$$
m_{AD}={y_D-y_A\over x_D-x_A}={(y_B+y_C)/2-y_A\over x_D-x_A}
={1\over2}{y_B-y_A\over x_B-x_A}+{1\over2}{y_C-y_A\over x_C-x_A}
={1\over2}(m_{AB}+m_{AC}).
$$
As $m_{BC}=5$, we can have line $BC$ parallel to the $y$-axis if we perform a counterclockwise rotation of an angle $\phi={\pi\over2}-\arctan5=\arctan{1\over5}$. Under such a rotation, the slope $m=\tan\theta$ of a generic line becomes $m'=\tan(\theta+\phi)$, that is:
$$
m'={m+1/5\over 1-m/5}.
$$
It is then easy to compute the slopes of lines $AB$ and $AC$ after the rotation:
$$
m'_{AB}=-{2\over3},\quad m'_{AC}=-{7\over 17},
$$
and then find the slope of median $AD$ after the rotation:
$$
m'_{AD}={1\over2}(m'_{AB}+m'_{AC})=-{55\over102}.
$$
We can then rotate back by the same angle to find $m_{AD}$:
$$
m_{AD}={m'_{AD}-1/5\over 1+m'_{AD}/5}=-{29\over35}.
$$
To find the equation of line $AD$ we can then consider a linear combination of the equations of lines $AB$ and $AC$:
$$
\alpha(x+y-2)+\beta(2x+3y-5)=0.
$$
Such an equation represents the pencil of all lines passing through $A$ and we just have to choose now coefficients $\alpha$ and $\beta$ such that the slope of the corresponding line is $m_{AD}$. This leads to the equation
$$
-{\alpha+2\beta\over\alpha+3\beta}=-{29\over35},
\quad\hbox{that is}\quad
{\alpha\over\beta}={17\over6},
$$
which can be satisfied for instance by $\alpha=17$ and $\beta=6$. Substituting these values into the above equation finally gives the equation of line $AD$:
$$
29x+35y-64=0.
$$
A: I'd first considered something along the lines of @Aretino's slope-averaging argument. Since that's been done, I'll use an area-halving argument. Write
$$\begin{align}
\overleftrightarrow{BC}: &\quad h_1 x+k_1y+p_1=0 \\
\overleftrightarrow{CA}: &\quad h_2 x+k_2y+p_2=0 \\
\overleftrightarrow{AB}: &\quad h_3 x+k_3y+p_3=0
\end{align} \tag{1}$$
The equation of the median through $A$ must be a combination of the other lines through $A$, say, 
$$\overleftrightarrow{AD}=m\overleftrightarrow{AB}+\overleftrightarrow{CA}: \quad (h_2+m h_3)x+(k_2+mk_3)y+(p_2+mp_3) = 0 \tag{2}$$
Now, we'll refer to a nifty formula, first brought to my attention in this question, that gives the area of a triangle from the equations of its edge-lines. (This might be out-of-scope for OP's class, but, as @G-man mentioned in that question, the formula deserves greater familiarity, so I'm doing my part to promote it.)

$$|\triangle ABC| = \frac{\left|\begin{array}{ccc}
h_1 & k_1 & p_1 \\
h_2 & k_2 & p_2 \\
h_3 & k_3 & p_3
\end{array}\right|^2}{2
\left|\begin{array}{cc} h_1 & k_1 \\ h_2 & k_2 \end{array}\right| 
\left|\begin{array}{cc} h_2 & k_2 \\ h_3 & k_3 \end{array}\right| 
\left|\begin{array}{cc} h_3 & k_3 \\ h_1 & k_1 \end{array}\right|} 
\tag{$\star$}$$

Technically, the area is the absolute value of the expression above, but we can assume that the lines are ordered appropriately so as to give a non-negative result. In writing an expression for $\triangle ADC$, with $D$ the midpoint of $\overline{BC}$, we need to make sure that the bounding lines are ordered in the same way; namely, $\overleftrightarrow{DC}$ (that is, $\overleftrightarrow{BC}$), $\overleftrightarrow{CA}$, and $\overleftrightarrow{AD}$, we also have
$$|\triangle ADC| = \frac{\left|\begin{array}{ccc}
h_1 & k_1 & p_1 \\
h_2 & k_2 & p_2 \\
h_2+mh_3 & k_2+mk_3 & p_2+mp_3
\end{array}\right|^2}{2
\left|\begin{array}{cc} h_1 & k_1 \\ h_2 & k_2 \end{array}\right| 
\left|\begin{array}{cc} h_2 & k_2 \\ h_2+mh_3 & k_2+mk_3 \end{array}\right| 
\left|\begin{array}{cc} h_2+m h_3 & k_2+mk_3 \\ h_1 & k_1 \end{array}\right|} 
\tag{3}$$
Recall that determinants are unchanged if we replace a row by a linear combination of that row with another. Also, we can "factor-out" a common factor in a row. Thus,
$$\begin{align}|\triangle ADC| &= \frac{\left|\begin{array}{ccc}
h_1 & k_1 & p_1 \\
h_2 & k_2 & p_2 \\
mh_3 & mk_3 & mp_3
\end{array}\right|^2}{2
\left|\begin{array}{cc} h_1 & k_1 \\ h_2 & k_2 \end{array}\right| 
\left|\begin{array}{cc} h_2 & k_2 \\ mh_3 & mk_3 \end{array}\right| 
\left|\begin{array}{cc} h_2+m h_3 & k_2+mk_3 \\ h_1 & k_1 \end{array}\right|}  \\
&= \frac{m\left|\begin{array}{ccc}
h_1 & k_1 & p_1 \\
h_2 & k_2 & p_2 \\
h_3 & k_3 & p_3
\end{array}\right|^2}{2
\left|\begin{array}{cc} h_1 & k_1 \\ h_2 & k_2 \end{array}\right| 
\left|\begin{array}{cc} h_2 & k_2 \\ h_3 & k_3 \end{array}\right| 
\left|\begin{array}{cc} h_2+m h_3 & k_2+mk_3 \\ h_1 & k_1 \end{array}\right|} 
\end{align} \tag{4}$$
Notice that the expressions for $|\triangle ABC|$ and $|\triangle ADC|$ have a lot in common, and that we can write
$$\frac{|\triangle ADC|}{|\triangle ABC|} = \frac{m\left|\begin{array}{cc} h_3 & k_3 \\ h_1 & k_1 \end{array}\right|}{\left|\begin{array}{cc} h_2+mh_3 & k_2+mk_3 \\ h_1 & k_1 \end{array}\right|} =
\frac{m(h_3 k_1 - h_1k_3)}{h_2k_1-h_1k_2+m(h_3k_1-h_1 k_3)} \tag{5}$$
Finally, we invoke the fact that a median bisects a triangle's area: the ratio in $(5)$ is $1/2$, which tells us

$$m = \frac{h_1 k_2-h_2 k_1}{h_1k_3-h_3 k_1} \tag{$\star\star$}$$ 

For the problem at hand, $m = 17/6$, so that

$$\overleftrightarrow{AD}: \quad 29x + 35 y - 64 = 0$$

as in @Aretino's solution. $\square$

Note. As in that other answer, assuming our original equations are in normal form provides a geometric insight; namely 
$$m = \frac{\sin C}{\sin B}$$
so that
$$\overleftrightarrow{AD} \;=\; \overleftrightarrow{AB}_{\text{n}}\;\sin C + \overleftrightarrow{AC}_{\text{n}}\;\sin B$$
where "n" indicates that we're to use the normal form in the calculation. (The resulting equation will not necessarily be in normal form.) Nifty!
A: 1) Draw two parallels $B_1C_1$ and $B_2C_2$ to the line $BC: 5x-y=7$, for example $y=5x-10$ and $y=5x-\dfrac{25}{2}$.
2) calculate the intersections of $B_1C_1$ with the lines $AB$ and $AC$ and do the same with the line $B_2C_2$ so you get four points $B_1,C_1,B_2,C_2$
3) You have now two points of the required median which are $P=\dfrac{B_1+C_1}{2}$ and $Q=\dfrac{B_2+C_2}{2}$

Calculation gives $$B_1=(2,0),C_1=(\dfrac{35}{17},\dfrac{5}{17}),B_2=(\dfrac{29}{12},\dfrac{-5}{12}),C_2=(\dfrac52,0)\\P=(\frac{69}{34},\frac{5}{34}),Q=(\frac{59}{24},\frac{-5}{24})$$ Thus the equation $$\color{red}{29x+35y=64}$$
A: 
Equation of AB is $x + y = 2$, equation of AC is $2x + 3y = 5$

Cheating somewhat, it is obvious by inspection that both lines pass through $\,(1,1)\,$. Then, choosing a complex plane with the origin at $\,(1,1)\,$, the conditions can be written (with $\,\lambda, \mu, \nu \in \mathbb{R}\,$) as:
$$
\begin{align}
b &= \lambda (1-i) \\
c &= \mu (3-2i) \\
b - c &= \nu (1+5i)
\end{align}
$$
The latter equation implies $\,\lambda (1-i) - \mu (3-2i) = \nu (1+5i) \iff \begin{cases}\lambda - 3 \mu = \nu \\ -\lambda+ 2 \mu = 5 \nu \end{cases}\,$. Eliminating $\,\nu\,$ between the two gives $\,-\lambda+2\mu=5(\lambda-3\mu) \iff6 \lambda = 17 \mu\,$. It follows that:
$$
2d = b+c = \lambda (1-i) + \mu (3-2i) = \frac{\mu}{6}\big(17(1-i)+6(3-2i)\big) = \frac{\mu}{6} (35-29i)
$$
Therefore, the median line is $\,z = (35-29i)\,t \,\mid\, t \in \mathbb{R}\,$. Translating back to cartesian:
$$
x-1 + (y-1)i = 35 t -29 t\,i
 \iff \begin{cases} x-1 = 35 t \\y - 1 = -29 t\end{cases}$$
Eliminating $\,t\,$ between the last two equations gives in the end $\,29x+35y-64=0\,$.
