the points j(-8,9) and k(-2,-5) are endpoints of a diameter of circle C. What equation would represent circle C
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Hello!
<u>Number 22</u>: We'd plot the first point at 0 since there is no stated y-intercept. Next, we'd use our slope to determine where to plot the next point, and that would create our line. According to the problem, our slope is
, which automatically tells us that the slope would be going downwards because it's negative.
To plot our point, use the slope while going down and across from our y-intercept, which is 0. Go down 1, and over 2.
Your points should be at (0, 0) and (-1, 2)
<u /><u>Number 23:</u> This one will be a bit trickier since the equation is not in slope-intercept form. First, let's convert it to slope-intercept form.
Flip some of those numbers around to get our equation in slope-intercept form:
Now to graph this, we do the same as we did for the last problem. Plot our first point at (0, 2), since 2 is our y-intercept. Afterwards, go up 2 and over 3, then plot the other point.
Your points should be at (0, 2) and (4, 3)
<u>Number 22</u>: We'd plot the first point at 0 since there is no stated y-intercept. Next, we'd use our slope to determine where to plot the next point, and that would create our line. According to the problem, our slope is
To plot our point, use the slope while going down and across from our y-intercept, which is 0. Go down 1, and over 2.
Your points should be at (0, 0) and (-1, 2)
<u /><u>Number 23:</u> This one will be a bit trickier since the equation is not in slope-intercept form. First, let's convert it to slope-intercept form.
Flip some of those numbers around to get our equation in slope-intercept form:
Now to graph this, we do the same as we did for the last problem. Plot our first point at (0, 2), since 2 is our y-intercept. Afterwards, go up 2 and over 3, then plot the other point.
Your points should be at (0, 2) and (4, 3)
