Another problem XY Graph problem

yohanson77

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Joined
Jan 8, 2007
Messages
41
Hi all,
I tried to seek help on this but have been left dry.

ok.

Q. select the option which you think gives the equation of the line in the picture below.

"Picture is a XY Graph with a straight line cutting through y=2.5 and x=2"

Options:

A. y = 5/4 x + 5/2
B. y = -5x + 5/2
C. y = -5/4 x + 2
D. 4y + 5x = 10
E. 4y + 5x = 5/2
F. 5y + 4x = 10

I Know the answer is D.

I've been trying to work out;

y = MX + C

M = Slope of line (gradient)
C = intersect on y axis

Am i on the right lines? (not literally)

yours

Yohanson77
 
You are given ONLY a point. There is no way to determine the equation of the line. You simply must substitute the given value of x and see if you get the given value of y.

I suppose you could manipulate each given equation to see if you can put it into a point slope form that includes the given point. That might be fun, but I suspect it will take quite a bit longer than the first suggested method.
 
I've spent all afternoon on it, i will not rest. It's so hard to try and describe the picture.

I've got microsoft math and it spits out this:

x = -4y/5 + 2

Does this make any sence?

Yohanson77
 
I understand. You actually have a picture.

Very quickly, if you've a positive slope, you had better mark "A", snice that's the only one with that characteristic.

Sadly, your answer still makes little sense as you ahve insufficient data to define a line. What did you put in? Is there a discernable y-intercept?
 
yohanson77 said:
Hi all,
I tried to seek help on this but have been left dry.

ok.

Q. select the option which you think gives the equation of the line in the picture below.

"Picture is a XY Graph with a straight line cutting through y=2.5 and x=2"

Options:

A. y = 5/4 x + 5/2
B. y = -5x + 5/2
C. y = -5/4 x + 2
D. 4y + 5x = 10
E. 4y + 5x = 5/2
F. 5y + 4x = 10

I Know the answer is D.

I've been trying to work out;

y = MX + C

M = Slope of line (gradient)
C = intersect on y axis

Am i on the right lines? (not literally)

yours

Yohanson77
3"

I may be taking a BIG LEAP here....

but, I wonder if the stipulation that "y = 2.5 and x = 2" means that the line in question crosses the y axis at (0, 2.5) and the x-axis at (2, 0).

IF this is the case, then the points (0, 2.5) and (2, 0) are on the line.

Find the slope:

slope = (0 - 2.5) / (2 - 0)

slope = (-2.5/)2

slope = -1.25, or -5/4

Now, use the point-slope form of the equation of the line with slope " m" passing through (x<SUB>1</SUB>, y<SUB>1</SUB>)

Y - y<SUB>1</SUB> = m(x - x<SUB>1</SUB>)

Let us choose (2, 0) as a point on the line. Then,


y - 0 = (5/4)(x - 2)

y = (-5/4)x - (5/2)

Now, this does not seem to match any of your answer choices. Let's multiply each side of the equation by 4 to eliminate the fractions:

4y = 4(-5/4)x - 4(5/2)

4y = -5x + 10

5x + 4y = 10

That looks like answer F to me....
 
yohanson77 said:
I've got microsoft math and it spits out this:
x = -4y/5 + 2
...and that, in terms of y: y = (-5/4)x + 5/2 : as per Mrspi
 
Ok, the slope crosses y axis at 2.5 and slopes down to x axis 2.

The answer in the book say it's answer D.

This one has really got me.

Why are the slopes minus?

slope = (-2.5/)2

slope = -1.25, or -5/4
 
Thanks all for your input, it's gratefully appreciated.

Dennis thanks for the link, i think this will help heaps.

Thanks again

yohanson77:)
 
yohanson77 said:
"Picture is a XY Graph with a straight line cutting through y=2.5 and x=2"
Okay, intercepts.

\(\displaystyle \L\;\frac{x}{2}\;+\;\frac{y}{(5/2)}\;=\;1\)

A little algebra and you're done.

Note: Please observe that "x = 2" defines a vertical line. "(2,0)" defines a point.
 
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