I need to find an equation perpendicular to the line y=8, containing the point (3,4)?

Question

Also, I'm looking for the general form of the equation of a circle with a center @ the point (-3,1) tangent to the y-axis.

Answer 1

A line to perpendicular to a horizontal line (y=k, where k is a constant) would be a vertical line (one where x=k). The k in this case would be the x value of the point (3,4) which is 3. So your equation is x=3.

If you ever had an equation where y=mx+b, then the perpendicular line would have a slope of -1/m (called the negative reciprocal). In this case, however, ytour line has a slope of zero and -1/0 does not exist unless you consider a vertical line having that slope. If you do, your answer would be an equation of x.

For the circle, the center helps you write the equation (which is always of the form (x-a)2+(y-b)2=r2 (where you see the 2's, that means squared). The a's and b's from the equation come from the circle's center and the r is its radius. The radius of this circle, since it is tangent to the y-axis, must be the shortest distance from the center to the y-axis. This is 1, since the y-coordinate is 1. Your a and b are obviously -3 and 1. Thus your equation is (x+3)2+(y-1)2=1 You can expand the terms if you want.

Answer 2

a)
A circle has an equation of the form:

(x - h)^2 + (y - k)^2 = r^2

r is the circle's radius and (h,k) is the center

So with your point:

(x - 3)^2 + (y - 1)^2 = r^2

To find radius, remember the y-axis is where x=0. Since the the x coordinate of the center is x = -1, the radius is from -1 to 0 or just "1"

(x - 3)^2 + (y - 1)^2 = 1

b) Perpendicular lines have slopes that are negative reciprocals of each other. The line y=8, has a zero slope, and the negative reciprocal, -1/0, is undefined, so the new line has no slop. It is just a vertical line. Since it goes through point (3,4) The line is just x = 3.

Answer 3

The slope of the given line is -5, so the slope of a line perpendicular to it really is (a million/5). So now we seem on the equation y=(-a million/5)x+b. Plug contained in the given aspect, because it may fulfill the equation no matter if it really is on the line. 8=(a million/5)(6)+b sparkling up this equation for b and also you get b=34/5. hence, the line is y=(a million/5)x+(34/5)

Answer 4

If you want an x=3 containing Y=8, your line must contain that point. Then, it would also contain x=3, y=4. so your line is x=3.

For your circle. x=3, which is the radius. Y=1 is just the point of the centre and does not affect the cirle.

Answer 5

If a line is perpendicular to another line(y=8) then the slope will be the negitive recipricol of the given line. You want to plug the numbers into point-slope form... I am sorry, I don't remember what the formula is, I haven't been in school for 2 months... sorry....

Answer 6

x=3

For the circle:

(x+3)²+(y-1)²=9