Write an equation, in standard form, of the line that passes through the point (-2, 8) and is perpendicular to the line y= 14x-16

Answer 1

You're given a line in slope-intercept format (Y=mX+b) and asked for a perpendicular line in standard form (aX+bY=c). Since the initial line could have a zillion lines perpendicular to it, we need a point on one of them to begin pinpointing a specific line. We're told that point is X=-2 and Y=8. We know that the slope of this perpendicular is the opposite-signed reciprocal (flipped fraction) of the slope (14/1) of the initial. We can plug in 8 for Y, -1/14 for m(slope), and -2 for X and solve for b (the point on the Y-axis where the perpendicular crosses).

8 = (-1/14)(-2) + b

8 = 2/14 (or 1/7) + b

Clearing the fraction by multiplying each item by 7 gives us

56 = 1 + 7b, 7b = 55, b = 55/7 or 110/14

Now back to the slope-intercept format.

Y = -1/14X + 110/14

Since the standard form requires whole numbers, we multiply each item by 14.

14Y = -X + 110

A final rearrangement gives us the aX + bY = c we need.

X + 14Y = 110

Answer 2

The slope of the line will be -1/14 ( the negative reciprocal of 14)

Now use the point (-2,8) to calculate b ( y intercept)

8 = -1/14(-2) +b

8 = 1/7 + b

b= 7 7/8 or 63/64

Final equation is y = (-1/14)x + 63/64

You can graph these 2 equations to confirm