equation for line through origin and points  TOPIC_SOLVED

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equation for line through origin and points

Postby santaclaus on Thu Mar 12, 2009 1:25 am

A circle of radius 2 is centered at the origin and goes through the point (-1 sqare root of 3)

1. find an equation for the line through the origin and the point (-1, square root of 3) :confused:

2. find an equation for the tangent line to the circle at the radius at the point of tangency -- no idea what this means?????? :confused: :confused:
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Postby stapel_eliz on Thu Mar 12, 2009 2:21 am

santaclaus wrote:A circle of radius 2 is centered at the origin and goes through the point (-1 sqare root of 3)

1. find an equation for the line through the origin and the point (-1, square root of 3)

You are given two points. Plug them into slope equation to find the slope of the line.

Pick one of the points (it doesn't matter which), and plug it and the slope you found into the point-slope form to find the equation of the line.

santaclaus wrote:2. find an equation for the tangent line to the circle at the radius at the point of tangency -- no idea what this means?

To learn what "tangents" are, try here.

I would assume, from the way the exercise is arranged, that they mean "the point of tangency" to be the point they listed at the beginning.

You are given the center of the circle and a point on the circle. Plug these into the slope formula to find the slope of the radius line from the center to this point. Then use what you learned about tangent lines (in particular, how they are perpendicular to the curve to which they are tangents) and what you know about perpendicular slopes and the point-slope formula to find the necessary equation.

If you get stuck, please reply (after having studied the various lessons, of course) with a clear listing of your work and reasoning so far. Thank you! :D
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Re: equation for line through origin and points

Postby santaclaus on Tue Mar 24, 2009 2:11 pm

so, would I do, square root 3 = -1 + b? then b=square root 3 +1?
put it into the equation after finding slope like so? y=1/square root 3 + square root 3 +1?

btw, I found slope to be 0- -1/0-square root 3 = -1/square root 3
and perpendicular slope is just square root 3
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Postby stapel_eliz on Tue Mar 24, 2009 2:22 pm

santaclaus wrote:so, would I do, square root 3 = -1 + b? then b=square root 3 +1?
put it into the equation after finding slope like so? y=1/square root 3 + square root 3 +1?

I'm sorry, but I don't understand what you mean by the above...? (I'm assuming, though, that this is with respect to the first of the two exercises...?) :confused:

Assuming that the second point is meant to be (-1, sqrt[3]), you have two points; namely, (-1, sqrt[3]) and (0, 0). Plug these into the slope formula.

Since the line passes through the origin, the y-value at which the line crosses the y-axis has been given to you; it is not any of the values you have listed above. (Think: If the point where y = 0 is on the line, then where does the line cross the y-axis?) :oops:

Plug the y-intercept and the slope value into the slope-intercept form of the line equation.

santaclaus wrote:btw, I found slope to be 0- -1/0-square root 3 = -1/square root 3 and perpendicular slope is just square root 3

I think you mean [0 - (-1)] / (0 - sqrt[3]). (Division by zero, as written above, is not defined.)

However, the slope formula, as the lesson in the above link explains, is the subtraction of the y-values divided by the subtraction of the x-values. You appear to be subtracting the x-values in the numerator....
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