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
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.
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...
solve similarly for second one...
the choices in the first one looks wrong by the way.. answer for first one should be y = x/3 -3
ya.. i messed up something.. i was just checking over it again.. i guess m wrong..
@nphuongsun93 : However my answer fits in option.. 4th one.. ??
yea idk that's why it looks wrong =.=, perpendicular to -3 slope should be 1/3 lol
@Hero , @waterineyes , Guys.. any comments ??
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...
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