Asked by Gray
wirte an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given points
(2,-1) y = (3/2)x +6
(2,-1) y = (3/2)x +6
Answers
Answered by
Jen
Slope intercept form is y = mx + b where m = gradient and b = y intercept.
equation (1) is y = -4x - 1 is already in this form, m=-4 and b = -1
perpendicular lines m1 x m2 = -1 (i.e. each slope is negative inverse of the other)
-4 x m2 = -1 => m2 = 1/4
equation of perpendicular line is y = 1/4 x + c
substitute point values (1, -3) to find c:
-3 = 1/4 (1) + c => -3 - 1/4 = c => c = -13/4
hence required equation is y = 1/4 x - 13/4 which is equivalent to 4y = x - 13
now do the same with equation (2) y = x + 3 where m = 1 and b = ?
m1 x m2 = -1
1 x m2 = -1 => m2 = ?
required equation of perpendicular line is therefore y =
you have figure out what y equals
equation (1) is y = -4x - 1 is already in this form, m=-4 and b = -1
perpendicular lines m1 x m2 = -1 (i.e. each slope is negative inverse of the other)
-4 x m2 = -1 => m2 = 1/4
equation of perpendicular line is y = 1/4 x + c
substitute point values (1, -3) to find c:
-3 = 1/4 (1) + c => -3 - 1/4 = c => c = -13/4
hence required equation is y = 1/4 x - 13/4 which is equivalent to 4y = x - 13
now do the same with equation (2) y = x + 3 where m = 1 and b = ?
m1 x m2 = -1
1 x m2 = -1 => m2 = ?
required equation of perpendicular line is therefore y =
you have figure out what y equals
Answered by
mathhelper
since the new line is parallel to y = (3/2)x + 6, it will differ only in the constant
so let new line be
y = (3/2)x + b, but (2,-1) lies on it, so
-1 = (3/2)(2) + b
-1 = 3 + b
b = -4
y = (3/2)x - 4
so let new line be
y = (3/2)x + b, but (2,-1) lies on it, so
-1 = (3/2)(2) + b
-1 = 3 + b
b = -4
y = (3/2)x - 4
Answered by
Jen
i did to much
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