Find the Equation

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TheKeyboardist

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Alright, here is another part i dont remember from algebra 1. The question is (ex) "Find the equation of a line that passes through (6,1) and (8,-4). Write equation in slope intercept form."

If you can help me that is awesome!
 
Start by finding the slope of the line:

m = (y1 - y2)/(x1 - x2)

You get to pick which point you call (x1,y1) and which will be (x2,y2). It doesn't matter.

Now you know the "m" in the slope-intercept eqn:

y = mx + b

but you still need to find "b".

To do this, put the x and y values from one of the given points into the eqn:

y1 = m*x1 + b

Rearrange the eqn to solve for b.

That's it. Hope that helps.
 
Alright, so to confirm that i did it right, would the answer for my example be:

y=-5/2x-14 ?


Sorry i didnt get back to you earlier, i was to busy working on the graphical problems.
 
and how would you do the ones that say find the equation of a line that goes through (4,2) and that is paralell to y=2x-4, there are even ones that include the former and replaces the "that is parallel to" with "and is perpendicular to" and also one with "paralell to a refrence line that passes through (3,3) and (0,6)! How would you go on solving those kinds of problems?
 
TheKeyboardist said:
How would you go on solving those kinds of problems?
Click to expand...
Since we cannot teach lessons here, please try online resources for lessons on this topic.

Eliz.
 
Find the equation of a line that passes through (6,1) and (8,-4). Write equation in slope intercept form.

y=-5/2x-14 ?
Click to expand...

Always check your final answer by plugging the original points back in. Do they work in your equation? (Your slope is correct. Check your b value calculations.)

Some hints on other problems:

Parallel lines have the same slope.

For perpendicular lines, if the slope of one line is (a/b), the line perpendicular to it will have slope (-b/a).
 
oh, i just saw my mistake, when i multiplied -5/2 and 6(x), i wrote 15 instead of -15!

And thanks alot wjm11! I guess ill be able to finish these problems now!
 
i was working on this problem:

"Find the equation of a line tat goes through (-2,0) and is perpendicular to y=-3x+7."

And so, in resoponse, i drew up this to begin: y=1/3x+b

Then i continued the equation to: 0=1/3(-2)+b

Then: 0=-2/3+b

Then: 2/3=b

And so, the answer would be y=1/3x+2/3 but when i graph it, the lines dont intersect at right angles! Can someone point out something that went wrong in my steps?
 
I found another problem that was confusing.

Find the equation of a line that is perpendicular to y=-3 and passes through the point (5,2).

How would you do that one? I could make a graph and do it but i think the teacher wants us to do it algebraiicly(is that a word?).
 
And so, the answer would be y=1/3x+2/3 but when i graph it, the lines dont intersect at right angles!
Click to expand...

Good work! I'll bet you have your x-axis and y-axis set to different scales. that distorts the graph, so perpendicular lines would not appear perpendicular.

Find the equation of a line that is perpendicular to y=-3 and passes through the point (5,2).
Click to expand...

Y = -3 is a horizontal line, which has a slope of zero. This presents a special case, because the reciprocal of zero, 1/0, is undefined. However, since we know the first line is horizontal, we also know that's a line perpendicular to it is a vertical line. Vertical lines have the eqn x = c, where c is some constant. So what x-value (c value) must your line pass through in the problem you have stated?
 
well, since the line has to pass through 5,2 and be perpendicular(A straight line at right angles to another line) to a horizontal line, what id need is a vertical line according to the definition. And since x of (5,2) is 5, then id say that x=5. If thats the answer....
 
since the line has to pass through 5,2 and be perpendicular(A straight line at right angles to another line) to a horizontal line, what id need is a vertical line according to the definition. And since x of (5,2) is 5, then id say that x=5.
Click to expand...

That's it. Good work.
 
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