Question

What is the equation for the line perpendicular to y= -5/3x + 11 1/3 and containing P(-2,3)?


A. y-2= -3/5(x-3)
B. y= -5/3x + 4 1/3
c. y= -3/5x + 4 1/5
D. y= 3/5x + 4 1/5

I think it is B..... Am I right?

Answer 1

I think it's D

Answer 2

How so? Can you explain? @Vincentli

Answer 3

@amistre64 @bibby @Callisto @jim_thompson5910

Answer 4

y = -5/3x + 11 1/3
in y = mx + b form, the slope is in the m position. In this problem, the slope is -5/3. However, we are looking for a perpendicular line, so we need the negative reciprocal of the slope we found in the given equation. All that means is " flip " the slope and change the sign. So the slope we need is the negative reciprocal of -5/3 and that is 3/5 (see how I flipped the slope and changed the sign).

Now we use y = mx + b
slope(m) = 3/5
(-2,3)...x = -2 and y = 3
now we sub
3 = 3/5(-2) + b
3 = - 6/5 + b
3 + 6/5 = b
15/5 + 6/5 = b
21/5 = b
4 1/5 = b

so your perpendicular equation is : y = 3/5x + 4 1/5...answer D

Answer 5

Sorry for replying late. Was busy... But yea, @texaschic101 explained it all