Line JK passes through points J(–4, –5) and K(–6, 3). If the equation of the line is written in slope-intercept form, y = mx + b
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Answer:
Step-by-step explanation:
slopes and equations:
find the equation thur ( 6,1 ) and (-2,-3)
find the slope m
m = (y2-y1) / (x2-x1 )
m = (-3 - 1) / (-2 -6)
m = -4 / -8
m = 1/2
now use the point-slope formula with our known slope
y-y1 = m(x-x1)
y-1 = 1/2(x-6)
y - 1 = 1/2x -3
y = 1/2x -3 +1
y = x -2
Find the equation parallel to y = 3x + 6 and thur (0,1)
Parallel means the same slope, the slope is 3 for the equation above.
use the slope-intercept formula again with the point given and the slope 3
y-1 = 3(x -0)
y - 1 = 3x
y = 3x +1
Find the equations perpendicular to 2x + y = 8 and the same y intercept as 4y = x + 3.
put both equations into proper form
y = -2x +6
y = x +
perpendicular means reciprocal slope and change the sign, the 2nd equation has an intercept of , so
y = 1/2x +
there you go Amanda :)