Here are some more questions I'm having trouble with. Please walk me through them so I can fully understand them.

#2: Let A be (5,9), B be (-3,-5), and C be (1,1). The median of a triangle connects a vertex of a triangle to the midpoint of the opposite side. *Another note: would you be able to explain the median of a triangle more fully? I don't get what my teacher meant by the midpoint on the other side* For example, the median of triangle ABC from vertex A connects A to the midpoint of line segment BC (I still don't understand it).

a. Find an equation describing the line that contains the median from A to the midpoint of line segment BC.
b. Find an equation describing the line that contains the median from B to the midpointof line segment AC.
c. What point (x,y) is on both of the lines you found in parts a and b (I might be able to figure this out if you explain a and b well).
d. Find an equation describing the line that contains the median A to the midpoint of line segment AB.

#3: A line with slope 3 intersects with slope 5 at the point (10,15). What is the distance between the x-intercepts of these two lines?

#4: Jon begins jogging at a steady 3 m/sec down the middle of lane one in one of a public track. Laura starts even with him in the center of lane two but moves at 4m/sec. At the instant they begin, Ellis is located 100 meters down the track in lane four and is heading towards them in his lane at 6m/sec. After how many seconds will the runners lie in a straight line?
a. Let t be the number of seconds the three have been running. Write expressions for the number of meters each has run after t seconds.
b. Consider the location of each runner as a point on a graph. What quantity might you use as the x-coordinate? What quantity might you use as the y-coordinate?
c. How can you tell if the three points are on a line? Use this to solve the problem.

Please explain this very thoroughly so I can completely understand it. I am only in 7th grade algebra so there are a lot of terms I still don't understand.

Thank you so much!

look at your triangle and find BC.

Mark the middle of that line, or the midpoint.
Join that point to point A with a line.
This line is called the median.
In any triangle there are three such medians, one from each vertex

We have just "connected vertex A to the midpoint of the opposite side BC", and called it the "median"

Now let's do this algebraically
the midpoint between any two given points is found by adding the x's, then dividing by 2,
and adding the y's , then dividing by 2
so the midpoint of BC is ((-3+1)/2,(-5+1)/2)
or (-1,-2), let's call it point D

in a) you are asked to find the equation of that line AD
slope AD = (-2-9)/(-1-5) = 11/6

so the line AD is y = (11/6)x + b
sub in the point (5,9), it lies on the line
9 = (11/6)(5) + b
multiply each side by 6
54= 55 + 6b
b = -1/6

so the median AD has equation y = (11/6)x - 1/6

now repeat the same steps for b)
for c) solve the two equations.
in d), I think you have a typo, probably meant to say "median C to the midpoint of line segment AB"

sounds really confusing, sorry but i can't help you! i think this is too much for someone to explain online.

WOWOWOWOW REINY YOU ARE A GENIUS!!!!! ARE YOU A MATHEMATICIAN?? HOW DO YOU KNOW ALL THIS STUFF?? YOU ROCK GIRL:)

You're being so helpful, thank you! Just wondering, my triangle turned out to be very skinny. It's supposed to be this way, right?

And for part d, that is how my teacher typed it, but in the other day's homework he also did a typo.

Thank you so much! You were actually MUCH more help than my teacher. My teacher just stuck the homework in front of our faces and expected us to understand it.

how do i set up three proportions with measurements of 3x3x3 that show the side splitter theorem or similar triangles

What is the midpoint of the line segment that connects the points (1, 2) and (6, 7)?

No problem! I'll walk you through each of the questions step by step, and I'll make sure to explain everything thoroughly. Let's start with question 2.

#2: Let A be (5,9), B be (-3,-5), and C be (1,1). The median of a triangle connects a vertex of a triangle to the midpoint of the opposite side. In this case, the median of triangle ABC from vertex A connects A to the midpoint of line segment BC.

a. To find the equation describing the line that contains the median from A to the midpoint of BC, we need to first find the midpoint of BC. The midpoint formula is ( (x1+x2)/2, (y1+y2)/2).

Let's calculate the midpoint of BC:
BC has endpoints B(-3, -5) and C(1, 1).
Midpoint = ( (x1+x2)/2, (y1+y2)/2) = ( (-3+1)/2, (-5+1)/2) = (-1, -2)

Now, we have the midpoint of BC as (-1, -2). The median from A to the midpoint of BC will pass through A(5, 9) and the midpoint (-1, -2).

Let's find the equation of the line passing through these two points using the point-slope form: y - y1 = m(x - x1), where m is the slope of the line.

Slope (m) = (y2 - y1) / (x2 - x1) = (-2 - 9) / (-1 - 5) = -11 / -6 = 11 / 6

Now, we can plug in the values of one of the points (let's use A) and the slope into the point-slope form:

y - 9 = (11/6)(x - 5)

Simplifying, we get: y - 9 = (11/6)x - 55/6

Rearranging the equation to the standard form, we get: (11/6)x - y = 55/6 - 54/6

Final equation: (11/6)x - y = 1/6

b. Similarly, to find the equation describing the line that contains the median from B to the midpoint of AC, we first need to find the midpoint of AC.

AC has endpoints A(5, 9) and C(1, 1).
Midpoint = ( (x1+x2)/2, (y1+y2)/2) = ( (5+1)/2, (9+1)/2) = (3, 5)

Now, we have the midpoint of AC as (3, 5). The median from B to the midpoint of AC will pass through B(-3, -5) and the midpoint (3, 5).

Let's find the equation of the line passing through these two points using the point-slope form again:

Slope (m) = (y2 - y1) / (x2 - x1) = (5 - (-5)) / (3 - (-3)) = 10 / 6 = 5 / 3

Plugging in the values of one of the points (let's use B) and the slope into the point-slope form:

y - (-5) = (5/3)(x - (-3))

Simplifying, we get: y + 5 = (5/3)x + 5

Rearranging the equation to the standard form, we get: (5/3)x - y = 0

Final equation: (5/3)x - y = 0

c. To find the point (x, y) that is on both of the lines found in parts a and b, we need to solve the system of equations formed by the two equations from parts a and b.

(11/6)x - y = 1/6 --> Equation from part a
(5/3)x - y = 0 --> Equation from part b

We can solve this system of equations using any preferred method, such as substitution or elimination. Let's use the elimination method.

By multiplying the first equation by 2 and the second equation by 3, we can eliminate the y term:

(22/6)x - 2y = 2/6
(15/6)x - 3y = 0

Now, subtracting the second equation from the first:

(22/6)x - 2y - (15/6)x + 3y = 2/6

Simplifying, we get:

(7/6)x + y = 2/6

Rearranging the equation to the standard form, we get:

7x + 6y = 2

Final equation: 7x + 6y = 2

d. To find the equation describing the line that contains the median from A to the midpoint of AB, we first need to find the midpoint of AB.

AB has endpoints A(5, 9) and B(-3, -5).
Midpoint = ( (x1+x2)/2, (y1+y2)/2) = ( (5+(-3))/2, (9+(-5))/2) = (1, 2)

Now, we have the midpoint of AB as (1, 2). The median from A to the midpoint of AB will pass through A(5, 9) and the midpoint (1, 2).

Let's find the equation of the line passing through these two points using the point-slope form:

Slope (m) = (y2 - y1) / (x2 - x1) = (2 - 9) / (1 - 5) = (-7) / (-4) = 7 / 4

Plugging in the values of one of the points (let's use A) and the slope into the point-slope form:

y - 9 = (7/4)(x - 5)

Simplifying, we get: y - 9 = (7/4)x - 35/4

Rearranging the equation to the standard form, we get: (7/4)x - y = 35/4 - 36/4

Final equation: (7/4)x - y = -1/4

Now, let's move on to question 3.