Question

Please Help ME!!!!!!!!!!!!!!!!!!!!!!

Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1).

y = x − 3

y = x + 17

y = −3x − 3

y = −3x + 17

Write the equation of the line that is perpendicular to the line y = x + 4 and passes through the point (−6, 3).

y = x − 1

y = x + 3

y = −x − 1

y = −x + 3

Answer 1

for the first one, change it to standard from first.
y = -3x + 7

perpendicular = negative reciprocal (-1/x)

so the new line with a slope that is perpendicular to the ^^^^ is

y = x/3 + b

now substitute the point into the equation above to find b and u r done.

Answer 2

Equation of 1st line,
3x + y = 7
y = -3x + 7

comparing with y = mx + c.., so m = -3

Now equation of perpendicular line and passing through (6, -1)

y - y0 = m (x - x0)
y - (-1) = -3 (x - 6)
y + 2 = -3x + 18
y = -3x + 16...

Answer 3

solve similarly for second one...

Answer 4

the choices in the first one looks wrong by the way..

answer for first one should be y = x/3 -3

Answer 5

ya.. i messed up something.. i was just checking over it again..

i guess m wrong..

Answer 6

@nphuongsun93 : However my answer fits in option.. 4th one.. ??

Answer 7

yea idk that's why it looks wrong =.=, perpendicular to -3 slope should be 1/3 lol

Answer 8

@Hero , @waterineyes , Guys.. any comments ??

Answer 9

Typo error !!! it shd be 17, not 18..

y - y0 = m (x - x0)
y - (-1) = -3 (x - 6)
y + 1 = -3x + 18
y = -3x + 17...