Question

The equation for line g is y = 14x – 44. Suppose lines g and h are perpendicular, and the point (–2, 1) lies on line h. Write an equation in standard form for line h.

Answer 1

Answer: x+14y=12

Step-by-step explanation: got this off an assignment on mcgrawhill still dont know how to get that tho good luck

Answer 2

Line h is y = -(1/14)x + (8/7)

Step-by-step explanation: A perpendicular line will have a slope that is the negative inverse of the reference line. The reference line in this problem is y=14x-44. The slope of 14 is inverted to 1/14 and multiplied by -1 to yield a new slope of -(1/14) for the perpendicular line. The perpendicular line will have the form of y = -(1/14)x + b.

Any value of b will still result in a line perpendicular to y=14x-44. But we want a line for h that goes through point (-2,1). To find a value of b that moves the line so that it intersects (-2,1), substitute the value (-2,1) for (x,y) in the equation y = -(1/14)x + b, and then solve for b.

y= -(1/14)*(x) + b

1 = -(1/14)*(-2) + b for (-2,1)

1 = (2/14) + b

1- (1/7) = b

b = (6/7)

The equation for line h becomes y = -(1/14)x + (6/7)

See the attached worksheet.