# Need Help With Math. How to do line Slope?

Indicate the equation of the given line in standard form.

The line containing the midpoints of the legs of right triangle ABC where A(-5, 5), B(1, 1), and C(3, 4) are the vertices.

Indicate the equation of the given line in standard form.

The line containing the hypotenuse of right triangle ABC where A(-5, 5), B(1, 1), and C(3, 4) are the vertices.

I have about 30 of these to do but i cant figure them out? Any help is appreciated.

I think you should plot the triangles in the Cartesian co-ordinate plane first of all. That will give you a good idea of how to proceed.

Take the first example.

Step (1): Plot the triangle using pencil and paper.

If you plot the triangle through A(-5, 5), B(1, 1), and C(3, 4), you'll find that the legs of the right triangle are the sides BC and BA. The right angle is formed at vertex B.

Step (2):

Find out mid-points of the legs.

The line-segment BC is formed by points B(1,1) and C(3,4). The mid-point, say X, is the point

X ( (1+3)/2, (1+4)/2) i.e. X(2, 5/2).

Similarly the midpoint of the line-segment BA, formed by points B(1,1) and A(-5,5), is

Y ( (1-5)/2, (1+5)/2) i.e. Y(-2, 3)

Step (3):

Now that you know the co-ordinates of the points X and Y, you can easily find the line through them.

Let the line be y = mx + c

It passes through X => 5/2 = 2m + c

It also passes through Y => 3 = -2m + c

Adding the respective sides of these two equations, we have

(5/2) + 3 = 2m + c - 2m + c

=> 11/2 = 2c

=> c = 11/4

Substituting in the 1st equation:

5/2 = 2m + 11/4

=> 2m = 5/2 - 11/4

=> 2m = -1/4

=> m = -1/8

So, the equation of the required line is: y = -x/8 + 11/4

For standard form:

y = -x/8 + 11/4

=> 8y = -x + 22

=> 8y + x - 22 = 0

Problem # 2:

The vertices are the same as Problem # 1, so if you plot the triangle first, you'll find that the hypotenuse is the line-segment AC that passes through the points A(-5,5) and C(3,4).

Let the line passing through A and C have the equation: y = mx + c

It passes through A(-5,5) => 5 = -5m + c .............(1)

It also passes through C(3,4) => 4 = 3m + c ..........(2)

Subtracting the respective sides of equation (2) from equation (1), we have:

1 = -8m => m = -1/8

Substituting this value of m in (1), we have:

5 = -5(-1/8) + c

=> 5 = 5/8 + c

=> c = 35/8

So the equation of the line through the hypotenuse is: y = -x/8 + 35/8.

For standard form:

y = -x/8 + 35/8

=> 8y = -x + 35

=> 8y + x - 35 = 0

May 28 at 21:22

First find the slope of each side to determine the right-angle:

Side AB slope (y1-y2)/(x1-x2)=(5-1)/(-5-1)= -4/6 = -2/3

Do same for other sides to determine which two slopes multiplied together equal "-1". These contain the right angle.

Next halve the Y and X distances between ends of each 'leg' of the right triangle i.e. not the hypotenuse side.

When you have these coordinates put into equation (y-y1)/(x-x1)=(y-y2)/(x-x2) and then rearrange to get standard form.

Do same for the coords representing the ends of the hypotenuse

May 29 at 1:8