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Question 397360: If the line
y=A+B(x-3)
is perpendicular to the line
y=2x
and
it contains the point
(5,15)
, what are the values of
A and B?
Found 2 solutions by ewatrrr, lwsshak3:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
Using the standard slope-intercept form for an equation of a line y = mx + b
where m is the slope and b the y-intercept.
y = A+B(x-3)
y = Bx - 3B + A | ⊥ to y = 2x, therefore it's slope will be m = -1/2
B = -1/2
y = (-1/2)x +3/2 + A |using ordered pair pt(5,15) to solve for A
15 = -5/2 + 3/2 + A
16 = A
CHECKING our Answer
y = 16 -(1/2)(x-3)
15 = 16 -(1/2)(5-3)
15 = 16 - 1
Answer by lwsshak3(11628)  (Show Source):
You can put this solution on YOUR website! If the line
y=A+B(x-3)
is perpendicular to the line
y=2x
and
it contains the point
(5,15)
, what are the values of
A and B?
y=A+B(x-3)
This equation is a straight line of the standard form, y=mx+b, with m being the slope and b the y-intercept.
In the given equation, B is the slope and A, the y-intercept.
This line is perpendicular to another line, y=2x, whose slope=2.
The negative reciprocal of this slope is the slope of the given line which would be -1/2 which is also equal to B
The equation now reads y=-1/2(x-3)+A
To find A, use the (x,y) coordinates given, and solve for A.
15=-1/2(5-3)+A
15=-1+A
A=16
The equation now reads, y=(-1/2)(x-3)+16
Ans:
B=-1/2
A=16
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