Algebra line prob: gen. form of line through (-2, 5) with...

xtrmk

New member
Joined
Aug 30, 2006
Messages
22
I forgot how to do these problems, help is really appreciated

Write the general form of the equation of the line passing through the point (-2,5) with slope -3/4.

I started out by y = (-3/4)x + b

I plugged in y and x and got b to equal 3.5

Is the answer y = (-3/4)x+3.5 or do I have to do something else?
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
In the equation "y = mx + b", the slope is "m", so you've plugged "-3/4" into the correct spot. There are various ways to proceed from there, but I think yours is the most direct: plug-n-chug:

. . . . .y = (-3/4)x + b

. . . . .x = -2, y = 5

. . . . .[5] = (-3/4)[-2] + b
. . . . .5 = 6/4 + b
. . . . .5 = 3/2 + b
. . . . .10/2 = 3/2 + b
. . . . .10/2 - 3/2 = 3/2 - 3/2 + b
. . . . .7/2 = b
. . . . .b = 3.5

Looks good to me! :D

Note: Different books mean different things by "general form". You'll have to check your text and/or your notes for this.

Eliz.

P.S. Thank you for showing your work!!
 

xtrmk

New member
Joined
Aug 30, 2006
Messages
22
stapel said:
In the equation "y = mx + b", the slope is "m", so you've plugged "-3/4" into the correct spot. There are various ways to proceed from there, but I think yours is the most direct: plug-n-chug:

. . . . .y = (-3/4)x + b

. . . . .x = -2, y = 5

. . . . .[5] = (-3/4)[-2] + b
. . . . .5 = 6/4 + b
. . . . .5 = 3/2 + b
. . . . .10/2 = 3/2 + b
. . . . .10/2 - 3/2 = 3/2 - 3/2 + b
. . . . .7/2 = b
. . . . .b = 3.5

Looks good to me! :D

Note: Different books mean different things by "general form". You'll have to check your text and/or your notes for this.

Eliz.

P.S. Thank you for showing your work!!
What is the Ax+Bx=0 form called? Is there such thing? I think i have to write it that way
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
xtrmk said:
What is the Ax+Bx=0 form called?
I'm not familiar with this. I've only heard of "y = mx + b", "y - y<sub>1</sub> = m(x - x<sub>1</sub>)", "Ax + By + C = 0", and "Ax + By = C". I don't know what "(A + B)x = 0" would be (other than some oddly-defined vertical line). Sorry.

Eliz.
 

xtrmk

New member
Joined
Aug 30, 2006
Messages
22
stapel said:
xtrmk said:
What is the Ax+Bx=0 form called?
I'm not familiar with this. I've only heard of "y = mx + b", "y - y<sub>1</sub> = m(x - x<sub>1</sub>)", "Ax + By + C = 0", and "Ax + By = C". I don't know what "(A + B)x = 0" would be (other than some oddly-defined vertical line). Sorry.

Eliz.
sorry I meant the Ax+By = C way. If i was to write it this way, how would i continue with the work I already did ?
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
xtrmk said:
I meant the Ax+By = C way. If i was to write it this way, how would i continue with the work I already did ?
Since the coefficients in this form are required to be integers, you would start by multiplying through to get rid of any fractions or decimals.

Then you would move the x-term over with the y-term, leaving "C" by itself.

If necessary, you would multiply everything by -1, so that "A" is positive.

Eliz.
 
Top