MHB Analytic Geometry: Lines Problem

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In triangle ABC with vertices A(-2,7) and C(7,-5), the altitude from AC is 5 units, and the altitude from BC is √45. The slope of line AC is calculated as -4/3, leading to the equation of line AC. The coordinates of point B, which lies below line AC, can be determined using the area of the triangle and the relationship between the lengths of the sides and altitudes. Two possible coordinates for point B are found: B1(-3,0) and B2(9,-16), with B1 being the valid solution as it meets the condition of being below line AC.
Yankel
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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.
 
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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)
 
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.
 

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