Problema Solution
determine the points of intersection of curve y^2-10x+10y-55=0 and a line 2x-y-10=0
Answer provided by our tutors
click here to see the graph and the intersection of the two graphs:
from 2x-y-10=0 follows y = 2x - 10
plug y = 2x - 10 into y^2-10x+10y-55=0
(2x - 10)^2 - 10x + 10(2x - 10) - 55 = 0
by solving we find:
x1 = 9.02
x2 = -1.52
click here to see the step by step solution of the quadratic equation:
for x1 = 9.02 we have y1 = 2*9.02 - 10 = 8.04
for x2 = -1.52 we have y2 = 2*(-1.52) - 10 = -13.04
the points of intersection are: (9.02, 8.04) and (-1.52, -13.04).