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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