I am trying to find an equation using the give pair of points (1/4,-1/2) and (3/4,3)

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I am trying to find an equation using the give pair of points (1/4,-1/2) and (3/4,3)

Mathematics
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You need to use the two-point formula for a line. It says, if you have two points,\[(x_1,y_1)\]and\[(x_2,y_2)\]then the equation of the line containing those points is\[y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\]You can make any one of your two points be (x1,y1). The other will have to be the (x2,y2). So here, you'd have\[y-(-\frac{1}{2})=\frac{3-(-1/2)}{3/4-1}(x-1/4)\]that is\[y+1/2=\frac{7/2}{-1/4}(x-1/4) \rightarrow y = -14(x-1/4)-1/2\]
You can expand to get\[y=-14x+7/2-1/2=-14x+3\]So\[y=-14x+3\]
we are suppossed to use the mx+b formula is what it says. Is that possible?

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Sure. You should have said so! :) I'll do it differently.
Well, actually, if you notice when I gave you the two-point formula, outside the \[(x-x_1)\]part there is\[\frac{y_2-y_1}{x_2-x_1}\]...that's the slope, m. So you can calculate that first, then you pick any one of the two points to do the rest\[y-y_1=m(x-x_1)\]
They're actually the same formula...you get m from the two points.
You're not actually told the intercept in your question, so can't read off 'b'. You would need to have one of your points, (0,b).
So, if you were given two points, (1,2) and (0,3), you could use y=mx+b as: m= (3-2)/(0-1)=-1 b=3 so y=-x+3
Your question's slightly different; you can't jump to y=mx+b...you need to find b from the points like I've done.
the only points i have are (1/4,-1/2) and (3/4,3) sorry i did not mention it before. At the bottom it says what is the equation of the line? y= (Simplify your answer.Type the answer in the form y=mx+b using intergers or fractions) this is why I am having so much problem trying to figure out what they want me to do because everytime I put in an answer they say it is wrong
Okay...they're saying "Find the equation of the line ANY way you like (from ways you're taught) and THEN put it in the form y=mx+b...then type that in." I'll do it on paper to double-check.
okay- I made a mistake before - when I type on the fly, I sometimes stuff up. The equation of your line in the form you need is\[y=7x-\frac{9}{4}\]
ok that was right that time. thanks
np
ok now I have to write an equation of the line containing the given point and parallel to the given line. express your answer in the form of y=mx+b (6,7);x+7y=2
Well, the second line needs to be parallel to the first, so it must have the same slope. You should arrange the first equation in the form y=mx+b and then 'read off' the slope. That will be the slope of the second line. Now, since you have a point that the second line must have, and the slope, you can use the point-slope formula I gave above y - y1 = m(x - x1) You expand and rewrite it as a last step in the form you need.
ok so the slope would be 1 then right?
No...you have to put the equation in the right form. x+7y=2 is NOT in the form y=mx+_b
\[y=-\frac{1}{7}x+\frac{2}{7}\]
So the slope is -1/7
Your new line will be\[y-y_1=m(x-x_1) \rightarrow y-7=-\frac{1}{7}(x-6)\]
\[y=-\frac{1}{7}x+\frac{55}{7}\]when you rearrange and put it into the form you need.
so would I do the same thig with these points (-2.5);5x=7y+8
What are you being asked to do?

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