# Question #9f434

Oct 10, 2016

$C \left(- \frac{672}{103} , - \frac{1103}{103}\right)$

#### Explanation:

We are given two lines which intersect at $C$, the line containing the altitude from $C$, and the line containing the median from $C$. Additionally, we are given the equation of the line containing the median, meaning all that remains is to find the equation of the line containing the altitude, and find their point of intersection.

An altitude of a triangle is a line segment which intersects a vertex of the triangle and the side opposite that vertex, intersecting the side at a right angle.

As the altitude from $C$ to $\overline{A B}$ is perpendicular to $\overline{A B}$, that means its slope will be equal to the negative reciprocal of the slope of $\overline{A B}$. That is, if ${m}_{\overline{A B}}$ is the slope of $\overline{A B}$, then the slope of the altitude will be ${m}_{\text{altitude}} = - \frac{1}{m} _ \overline{A B}$.

The slope of a line containing two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$. With that, we can calculate the slope of $\overline{A B}$ as

${m}_{\overline{A B}} = \frac{- 6 - 2}{2 - \left(- 5\right)} = - \frac{8}{7}$

thus

${m}_{\text{altitude}} = - \frac{1}{m} _ \overline{A B} = \frac{7}{8}$

We are also given that the $y$-intercept of the altitude is $\left(0 , - 5\right)$. Thus, using the slope-intercept form of a line, we get the equation of the line containing the altitude as

$y = \frac{7}{8} x - 5$

As this intersects the line $y = \frac{26}{15} x + \frac{3}{5}$ containing the median from $C$ only at $C$, we are looking for the solution to the system of equations

$\left\{\begin{matrix}y = \frac{7}{8} x - 5 \\ y = \frac{26}{15} x + \frac{3}{5}\end{matrix}\right.$

$\implies \frac{7}{8} x - 5 = \frac{26}{15} x + \frac{3}{5}$

$\implies - \frac{103}{120} x = \frac{28}{5}$

$\implies x = - \frac{672}{103}$

$\implies y = - \frac{1103}{103}$

This gives us the final result $C \left(- \frac{672}{103} , - \frac{1103}{103}\right)$

The graph would look as follows, with green dashed line containing the median and the blue dashed line containing the altitude.

Oct 10, 2016

$C = \left(- \frac{672}{103} , - \frac{1103}{103}\right)$

#### Explanation:

Segment $\overline{A B}$ is described by

$S \to p = A + \lambda \left(B - A\right)$ with $p = \left(x , y\right)$ and $\lambda \in \left[0 , 1\right]$

the line that supports the height is given by

${L}_{h} \to p = {p}_{h} + {\lambda}_{h} {\vec{v}}_{h}$ with ${p}_{h} = \left(0 , - 5\right)$ and
${\vec{v}}_{h}$ orthogonal to $B - A$. Here $B - A = \left(7 , - 8\right)$ so, a choice for ${\vec{v}}_{h}$ is ${\vec{v}}_{h} = \left(8 , 7\right)$

The support line for the median is

${L}_{m} \to p = {p}_{m} + {\lambda}_{m} {\vec{v}}_{m}$. This line is given as
${L}_{m} \to 26 x - 15 y + 9 = 0$ and can be represented as
${L}_{m} \to {a}_{m} \left(x - {x}_{m}\right) + {b}_{m} \left(y - {y}_{m}\right) = 0$ or
${L}_{m} \to \left\langlep - {p}_{m} , \vec{n}\right\rangle = 0$. Then follows that ${\vec{v}}_{m}$ must be orthogonal to $\vec{n}$
${\vec{v}}_{m} = \left(- {b}_{m} , {a}_{m}\right) = \left(15 , 26\right)$ and
${a}_{m} {x}_{m} + {b}_{m} {y}_{m} = - 9$ so
${p}_{m} = \left({x}_{m} , {y}_{m}\right) = \left(0 , \frac{3}{5}\right)$

Now, the vertice $C$ is at the intersection $C = {L}_{h} \cap {L}_{m}$

which is obtained solving for ${\lambda}_{h} , {\lambda}_{m}$ the system

${p}_{h} + {\lambda}_{h} {\vec{v}}_{h} = {p}_{m} + {\lambda}_{m} {\vec{v}}_{m}$ which gives ${\lambda}_{h} = - \frac{84}{103} , {\lambda}_{m} = - \frac{224}{515}$ and finally

$C = \left(- \frac{672}{103} , - \frac{1103}{103}\right)$

Attached a situation plot