Question 595376: Not sure how to go about this other than turning the equation into y=mx+b, plug the points in and solve, then re write in standard form, not sure.
1) Write an equation of the line, In standard form, parallel to Y+4x=8 thru (-6,3)
2) Write an equation of a circle, centered at (4, -1) passing thru (8,3) In standard form.
3) Write an equation of a line perpendicular to 2y=6x-4 passing thru (5,-2) In standard form.
Really appreciate the help and Guidance if any.
Thank You
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! 1) parallel lines have the same ratio of the x and y coefficients (slope), so you can use the same equation
___EXCEPT, the constant term is different because the y-intercept is different
___ plug in the new point to find the new constant term ___ (3) + 4(-6) = ? ___ -21 = ?
___ the new equation is ___ y + 4x = -21
2) a general circle equation is ___ (x - h)^2 + (y - k)^2 = r^2 ___ the center is (h,k) and the radius is r
___ in this case, the distance formula (from the center to the point on the circumference) will give r^2
___ (x - 4)^2 + (y + 1)^2 = 32 (not sure about "standard" form)
3) perpendicular lines have negative-reciprocal slopes, so swap the x and y coefficients and change one of the signs
___ like the 1st question, plug in the new point to find the new constant term
___ 6(-2) = -2(5) + ? ___ -2 = ?
___ the new equation is 6y = -2x - 2
|
|
|
