MHB Analytic Geometry: Lines Problem

[Yankel]

In a triangle ABC: A(-2,7) and C(7,-5).

The length of the altitude of AC is 5, and the length of the altitude of BC is the square root of 45. I wish to find the vertex B, given that it is below the line AC.

I need your help, I have no idea how to approach this.

Thank you in advance.

I have started by calculating that AC is -(4/3)x+(13/3)

that's more or less all I have done.

 

[MarkFL]

I think I would first use the information that the altitude from $\overline{AC}$ is 5 to determine a line that must contain $B$. We find the slope of $\overline{AC}$ is:

$$\frac{7+5}{-2-7}=-\frac{4}{3}$$

This line must contain a point $\left(x_1,y_1\right)$ whose slope with $A$ is $$\frac{3}{4}$$ and whose distance from $A$ is 5. So, we may write:

$$\frac{7-y_1}{-2-x_1}=\frac{3}{4}\implies 7-y_1=-\frac{3}{4}\left(2+x_1\right)$$

$$\left(2+x_1\right)^2+\left(7-y_1\right)^2=25$$

$$\left(2+x_1\right)^2+\frac{9}{16}\left(2+x_1\right)^2=25$$

$$\frac{25}{16}\left(2+x_1\right)^2=25$$

$$\left(2+x_1\right)^2=16$$

$$2+x_1=\pm4$$

$$x_1=-2\pm4$$

Since $B$ is said to lie below $\overline{AC}$, we take the solution:

$$x_1=-6\implies y_1=4$$

And so now armed with the slope and a point on the line which must contain $B$, using the point-slope formula, we find this line to be:

$$y=-\frac{4}{3}(x+6)+4=-\frac{4}{3}x-4=-\frac{4}{3}(x+3)$$

And so we may now label the coordinates of point $B$ as:

$$\left(x_B,-\frac{4}{3}\left(x_B+3\right)\right)$$

To continue, you need to find the $x_B$, such that the perpendicular distance from $A$ to the line through $B$ and $C$ is $\sqrt{45}$. (Thinking)

 

[Yankel]

Thank you for the assistance.

There is something wrong, or that I do not understand in your solution.

View attachment 6513

I agree that the slope of BG is 3/4 like you said. But why is the distance between G and A is 5 ? The distance between B and G is 5. I am slightly confused. The line AC is y=(-4/3)+(13/3). G is on this line and the distance from G to B is 5.

 

[MarkFL]

What I am saying is that $B$ must lie on the red line:

\begin{tikzpicture}[xscale=0.5,yscale=0.5]
\usetikzlibrary{shapes,positioning,intersections,quotes}
\def\x{3};
\draw[<->] (-10.3,0) -- (10.3,0) node

{$x$};
\draw[<->] (0,-10.3) -- (0,10.3) node[above] {$y$};
\draw[domain=4.4:-10.1, variable=\x, red, ultra thick] plot ({\x}, {-(4/3)*(\x+3)}) node[above] {$y=-\dfrac{4}{3}(x+3)$};
\node[circle,draw=blue, fill=blue, inner sep=0pt,minimum size=5pt] (b) at (-2,7) {};
\node[above=1pt of {(-2,7)}, blue, outer sep=2pt,fill=none] {$A(-2,7)$};
\node[circle,draw=blue, fill=blue, inner sep=0pt,minimum size=5pt] (b) at (7,-5) {};
\node[below=1pt of {(7,-5)}, blue, outer sep=2pt,fill=none] {$C(7,-5)$};
\end{tikzpicture}​

 

[MarkFL]

Perhaps a simpler way to determine the coordinates of point $B$ is to observe the that area $A$ of the triangle must be:

$$A=\frac{1}{2}bh=\frac{1}{2}\overline{AC}\cdot5=\frac{5}{2}\sqrt{9^2+12^2}=\frac{75}{2}$$

We may also write:

$$A=\frac{1}{2}bh=\frac{1}{2}\overline{BC}\cdot\sqrt{45}=\frac{\sqrt{45}}{2}\sqrt{\left(7-x_B\right)^2+\left(\frac{4}{3}\left(x_B+3\right)-5\right)^2}$$

This implies:

$$\left(7-x_B\right)^2+\left(\frac{4}{3}\left(x_B+3\right)-5\right)^2=125$$

Solving this, we get:

$$x_B\in\{-3,9\}$$

And so this gives us two possible locations for point $B$:

$$B_1(-3,0)$$

$$B_2(9,-16)$$

\begin{tikzpicture}[xscale=0.375,yscale=0.375]
\usetikzlibrary{shapes,positioning,intersections,quotes}
\def\x{3};
\draw[<->] (-20.3,0) -- (20.3,0) node

{$x$};
\draw[<->] (0,-20.3) -- (0,20.3) node[above] {$y$};
\node[circle,draw=blue, fill=blue, inner sep=0pt,minimum size=5pt] (b) at (-2,7) {};
\node[left=1pt of {(-2,7)}, blue, outer sep=2pt,fill=none] {$A(-2,7)$};
\node[circle,draw=blue, fill=blue, inner sep=0pt,minimum size=5pt] (b) at (7,-5) {};
\node[right=1pt of {(7,-5)}, blue, outer sep=2pt,fill=none] {$C(7,-5)$};
\node[circle,draw=black!30!green, fill=black!30!green, inner sep=0pt,minimum size=5pt] (b) at (-3,0) {};
\node[above left=1pt of {(-3,0)}, black!30!green, outer sep=2pt,fill=none] {$B_1(-3,0)$};
\node[circle,draw=red, fill=red, inner sep=0pt,minimum size=5pt] (b) at (9,-16) {};
\node[right=1pt of {(9,-16)}, red, outer sep=2pt,fill=none] {$B_2(9,-16)$};
\draw[-,blue] (-2,7) -- (7,-5);
\draw[-,black!30!green] (-2,7) -- (-3,0);
\draw[-,black!30!green] (7,-5) -- (-3,0);
\draw[-,red] (-2,7) -- (9,-16);
\draw[-,red] (7,-5) -- (9,-16);
\end{tikzpicture}​

 

[Yankel]

This is simpler. I like your new idea, thanks !