Question

Find the equation of the line that is perpendicular to the given line and passes through the given point. Enter the right side of the equation as a single fraction. y = 5x − 1 2 ; (3, 3)

Answer 1

y = 5x - 12....the slope here is 5. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal, u flip the slope and change the sign. So we have 5 or 5/1.....flip the slope making it 1/5.....change the sign making it -1/5. So the perpendicular line will have a slope of -1/5.

y = mx + b
slope(m) = -1/5
(3,3)....x = 3 and y = 3
now we sub into the formula and find b, the y int
3 = -1/5(3) + b
3 = -3/5 + b
3 + 3/5 = b
15/5 + 3/5 = b
18/5 = b

so ur equation is : y = -1/5x + 18/5

Answer 2

To answer this question we need to know two things, 1) a linear equation is written as y=mx+b. and 2) a perpendicular line is the negative reciprocal of the given slope. So, with this in mind, we now know that the new slope is -1/5x. In order for the equation to pass through the given points (3,3), we need to plug it in with the new equation. We plug both x and y with 3 while leaving the y-intercept to be variable b. 3=-1/5(3)+b. Now, we solve this equation by isolating the variable. -1/5*3 is -3/5, and we add this to 3, so it will be 3 3/5. This will be our new y-intercept. The equation is now y=-1/5x+3 3/5.