# Question #c42fa

Mar 12, 2015

You want to start by graphing the triangle.

Then you want to identify the midpoints of the top 2 legs of the ∆.
Using the midpoint formula $\left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$ we know that the midpoints are (-2.5, 3.5) and (5,1)

So our graph is this now:

In order to create the midsegment, we simply need to find the line which connects the 2 midpoints. We can do this by using the point-slope from of a line ($y - {y}_{1} = m \left(x - {x}_{1}\right)$). The equation of the line is $y - 3.5 = \left(\frac{2.5}{-} 7.5\right) \left(x + 2.5\right)$

So our graph is this now:

Now, switching gears, we need to find the altitude. We know that the altitude is a line which connects a vertex to the base at a 90º angle. In this case the vertex would be (3,5) and the base line would be the one formed by (-8,2) and (7,-3).

Since we know that a ⊥ line has a negative reciprocal slope, we need to find the equation of the base line. Again, we will use the point-slope from of a line.

The equation of the base line is $y - 2 = \left(\frac{5}{-} 15\right) \left(x + 8\right)$ which can be simplified to $y - 2 = \left(- \frac{1}{3}\right) \left(x + 8\right)$. Since the slope of the base line is $\left(- \frac{1}{3}\right)$, the slope of the ⊥ line will be 3.

So far, we know that the altitude equation will have the form y=3x+k, where k is the y intercept.

Since we know that the altitude must go through (3,5), we can solve for k by plugging in 3 for x and 5 for y.

Therefore: 5=3(3)+k, so k=-4.

Therefore, the altitude equation is y=3x-4.

So our graph is now this:

From here, all we need to determine is whether the altitude is being bisected by the midsegment. For me, the easiest way to do this is to measure the two segments. There may be an easier method, but I don't know it :)

So, we need to first determine the point of intersection between the altitude and the midsegment. We can do this by setting the 2 equations equal to each other like so:

Equation 1 (Midsegment) in mx+b form: $y = \left(\frac{2.5}{-} 7.5\right) \left(x + 2.5\right) + 3.5$

Equation 2 (Altitude) in mx+b form: $y = 3 x - 4$

Setting the 2 equal (since they both equal y): $\left(\frac{2.5}{-} 7.5\right) \left(x + 2.5\right) + 3.5 = 3 x - 4$

Solving, x=2. Substituting into an equation, we find that y=2.

Therefore, the intersection point is (2,2)

So our graph looks like this now:

Using the distance formula $\left(\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}\right)$, we can determine the length of the 2 segments.

The first segment (from (3,5) to (2,2) has a length of $\sqrt{10}$

To find the length of the second segment, we need to find the intersection point of the altitude and the base line. To do this, we will set the 2 equations equal, just like what we did with the altitude.

Equation 1 (Altitude) in mx+b form: $y = 3 x - 4$

Equation 2 (Base line) in mx+b form: $y = \left(- \frac{1}{3}\right) \left(x + 8\right) + 2$

Setting them equal: $3 x - 4 = \left(- \frac{1}{3}\right) \left(x + 8\right) + 2$

Solving, x=1. Substituting into an equation, we find that y=-1.
Therefore, the intersection point is (1,-1).

Now, solving for the length of the second segment (from (2,2) to (1,-1)), we find that the length is also $\sqrt{10}$.

Therefore, the midsegment bisects the altitude

Sorry for the super lengthy answer, but to accommodate your request for graphs and steps it was necessary.

Best of luck with all your future problems!

BTW, mad props to Desmos Graphing Calculator for these sweet graphs (https://www.desmos.com/calculator). It's free and you should check them out! (This is me giving credit for the images btw, so don't remove this pls)