Intergration: Algebra, Writing Equations of Lines

[Redfrog2]

I have two problems that deal with writing an equation in slope-intercept form of a line that satifises certain conditions. Two of such problems my teacher has given me, I have no idea where to start in how to put the conditions in an equation. Can anyone help me in finding out on how to do this?

1. Perpendicular to the y-axis, passes through (-6, 4)

2. Parallel to the y-axis, passes through (-7,3)

 

[Janitor]

Number 1 is easy if you remember that one possibility for y=mx+b is that m=0.

Number 2 is tough if you have to express the equation with a naked y on the left. On the other hand, if you are allowed to consider x as the dependent variable, then number 2 becomes just as easy as number 1.

 

[M98Ranger]

To write an equation in slope-intercept form, we need to use the formula y = mx + b, where m is the slope and b is the y-intercept. In these problems, we are given a point that the line passes through, so we can use that to find the y-intercept.

1. Perpendicular to the y-axis means that the line is vertical, and therefore has an undefined slope. To find the y-intercept, we can plug in the given point (-6, 4) into the formula. This gives us 4 = m(-6) + b. Since the slope is undefined, we can set m = 0, which gives us 4 = 0 + b. Therefore, the y-intercept is b = 4. The equation of the line is y = 0x + 4, or simply y = 4.

2. Parallel to the y-axis means that the line has a slope of 0. Using the same formula, we can plug in the given point (-7, 3) and set the slope m = 0. This gives us 3 = 0(-7) + b. So, the y-intercept is b = 3. The equation of the line is y = 0x + 3, or simply y = 3.

In summary, to write an equation of a line in slope-intercept form, we need to find the slope and y-intercept. The conditions given in these problems help us determine these values, and then we can plug them into the formula to write the equation. I hope this helps in solving your problems!