Question

The equation y - 5 = m(x - 2) represents a line, no matter what value m has. For what value of m does this line form a triangle of area 36 with the positive axes?

Answer 1

The "positive axes" include the positive x- and y-axes.
I'd suggest you choose one or two arbitrary values for the slope, m, and then graph a representative line for each. Can you identify the point through which these lines will pass?
Once you have a first sketch of this situation, think about how you might find the area of the triangle whose boundaries are the x- and y-axes and the line(s) you have drawn.
What is the formula for the areea of a triangle?

Answer 2

I guess I can use the x- and y-intercepts.
x- int : (2- (5/m), 0)
y- int: (0, -2m + 5)

Answer 3

area = (b x h)/2

Answer 4

What would be your next step? You'd need to find the value of the slope, m, for which the triangle has a n area of 36 square units.

Yes, \[A=\frac{ 1 }{ 2 }bh, or \frac{ bh }{ 2 }\]

Answer 5

and we'd need to set this equal to 36 square units.

Answer 6

The base would cover the distance from 0 until the x-int and the height would cover the distance from 0 to the y-int

Answer 7

so we could do
\[\frac{ (-2m + 5) \times (2 - \frac{5}{m}) }{ 2 } = 36\]

Answer 8

Thus, bh=2(36)=72 square units. This is your constraint on the values of b and h.
Yes, I agree with your statement.
Can you think of any other possible relationship between b and h?
Have you sketched a possible triangle yet? To do that, you'd need to select a value for m.

Answer 9

I think this would help you.

I hope so!

Answer 10

Your\[\frac{ (-2m + 5) \times (2 - \frac{5}{m}) }{ 2 } = 36\] is very promising because it has only one variable, m, which is the variable whose value you want to find.

Answer 11

Want to try solving your equation for m?
I see you'll end up with a quadratic equation in m.

Answer 12

Yes I am solving it right now

Answer 13

I get the quadratic:
\(0 = -4m^2 + 20m - 97\)

Answer 14

No wait I made a mistake...

Answer 15

My result, for comparison purposes, is\[-4m^2+20m-25=72m\]

Answer 16

What is 20m-72m?

Answer 17

I am sorry but I still didn't get that quadratic

Answer 18

Please share what you have obtained.

Answer 19

Can we agree on Can we agree on -4m^2+20m-25 =72m?

Answer 20

I see my mistake. I forgot to multiply m to the other side of the equation (meaning with 72).

Answer 21

Why would it be appropriate to mult the other side of the eq'n by 72?

Answer 22

I meant to multiply both sides of the equation by m.

Answer 23

\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 24

I got \(0 = -4m^2 - 52m - 25\)

Answer 25

A step or two back, I did that (multiplied both sides of the eq'n by m).
I agree with your result. How would we go about finding the values of m? (There are 2.)

Answer 26

0 = -4m^2 - 2m - 50m - 25
0 = -2m(2m +1) and -25(2m + 1)
0 = (-2m + 25)(2m + 1)

Answer 27

Looks promising. m=?

Answer 28

m = 25/2, m = -1/2
but m = 25/2 seems more reasonable

Answer 29

Why do you see it that way? Think: That ONE line, passing through (2,5), creates a triangle whose other boundaries are the x- and y-axes.

Answer 30

Would that "work" if m=+25/2?

Answer 31

Wait.. if the slope is positive and it passes through the point 2,5 then it won't go through both of the axes. so if the slope is negative (which means -1/2) it can pass through both of the positive axes

Answer 32

On the other hand, supposing that m=-1/2, then the equation of that line would be

\[y-5=\frac{ -1 }{ 2 }(x-2)\]

Answer 33

This would be easy to graph, should you want to do so.

Answer 34

Again, I agree with your reasoning. Very good!

Answer 35

Thank you! Can you help me with another question which stems off of "The equation y - 5 = m(x - 2) represents a line, no matter what value m has"

Answer 36

What remains to be done? Seems to me that you now should find the base and the height of the triangle we have created.

Answer 37

I've enjoyed this dialogue tremendously. You've been right on target.
Sure, go ahead and post your related question.

Answer 38

Should I make a new post?

Answer 39

since we'll likely finish this problem between us, just present the problem there. Thx

Answer 40

Alright so here it is:

Show that the area of a first-quadrant triangle formed by this line must be at least 20 units.

Answer 41

You have an equatiion for "this line," and thus you can express "y" as a function of x.
Why not come up with a formula for the area of our triangle that depends upon x, and then graph this? I'd bet the graph would show quickly that the min. area would be around 20 square units.

Answer 42

I assume you know differential calculus. Were you to come up with this general formula for the area of the triangle in terms of x, you could differentiate this formula and set it equal to zero in the usual way. One of the resulting roots may be the x-value at which the area of the triangle is at a min., assuming that \[0<x\]

Answer 43

I am sorry but I do not know differential calculus. My username is sort of misleading in that sense. I am just a freshmen in high school

Answer 44

OK. Come up with a general formula for the area of this triangle in terms of x alone.
Graph this formula. Look for the x-value(s) at which the area is at a minimum.

Answer 45

Are you allowed to use a graphing calculator?

Answer 46

Yes

Answer 47

I think you already have a formula for the area of the triangle. Mind going back to look for one?

Only thing is, it must be dependent upon x only.

Answer 48

I am kind of confused with how I should solve the area in terms of x?

Answer 49

Upon 2nd thought, that's not right; m would be a variable also.

\[A=\frac{ 1 }{ 2 }bh, or \frac{ bh }{ 2 }\]

Write b and h in terms of both m and x.
You may have to treat m as a parameter, e. g., m=-0.5, m=-1, and so on.

I think you already have developed formulas for the base and the height of your triangle, so just substitutte them into \[A=\frac{ 1 }{ 2 }bh, or \frac{ bh }{ 2 }\]

Answer 50

Earlier, you wrote:

"I guess I can use the x- and y-intercepts.
x- int : (2- (5/m), 0)
y- int: (0, -2m + 5)"

Agreed.

So, what is the x-intercept in terms of m alone?
What is the y intercept?