Question 42777
#1 Give the center of the circle with the equation x^2+2x+y^2-10y+22=0
{{{x^2 + 2x + y^2 - 10y + 22 = 0}}}
{{{(x^2 + 2x) + (y^2 - 10y) = -22}}}
{{{(x + 1)^2 + (y - 5)^2 = -22 + 25 + 1}}}
{{{(x + 1)^2 + (y - 5)^2 = 4}}}
Center: (-1,5)
#2 Find the equation for the line with slope 1.2 and the y-intercept 3.
{{{y = mx + b}}}
{{{y = 1.2x + 3}}}
#3 Find the slope of the line through the points A(-1,6) and B(5,2)
M = (y2 - y1)/(x2 - x1) = (6 - 2)/(-1 - 5) = 4/-6 = -2/3
#4 Find the point on the positive y-axis that is a distance 5 from the point P(3,4)
{{{d = sqrt((x - x)^2 + (y - y)^2)}}}
{{{5 = sqrt((3 - 0)^2 + (4 - y)^2)}}}
{{{5 = sqrt(9 + (4 - y)^2)}}}
{{{5 = sqrt(9 + y^2 - 8y + 16)}}}
{{{5 = sqrt(y^2 - 8y + 25)}}}
{{{25 = y^2 - 6y + 25}}}
{{{0 = y^2 - 8y}}}
{{{0 = y(y - 8)}}}
{{{y = 0}}} or {{{y = 8}}}
(0,0) and (0,8)