Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

Please Help!(: Find the point on the line 2 x + 5 y - 7 =0 which is closest to the point ( -7, -7 ).

OpenStudy (loser66):

|dw:1374754927626:dw|

OpenStudy (loser66):

so the line contains (-7,-7) has the slope of 5/2. you know why? because it's perpendicular to the given line. (We are asked to find the closest gap, that means the perpendicular line) you have slope m = 5/2, you have point (-7,-7), just plug into the formula y = m +b to find out the equation of the line. NOT finish yet. let those line equal, solve for the intersection point, and apply distance formula to find out the distance between the point (-7,-7) and the intersection point ha!! my solution is so bad, but I don't know any simpler way. Good luck

OpenStudy (anonymous):

\[y=\frac{ 5 }{ 2 }x+\frac{ 5 }{ 2 } ?\]

OpenStudy (loser66):

I don't know, just give you method, carefully calculate is yours

OpenStudy (amistre64):

expand a circle centered at -7,-7 until it hits the stated line :) \[f(x,y) =(x+7)^2+(y+7)^2~:~g(x,y)=2x+5y-7 \] \[fx=2(x+7)~:~k~gx=2k~:~x=k-7\] \[fy=2(y+7)~:~k~gy=5k~:~y=\frac52k-7\] plug x and y into g to solve for k \[0=2(k-7)+5(\frac52k-7)−7\] \[7=2k-14+\frac{25}2k-35\] \[56=\frac{29}2k\] \[\frac{112}{29}=k\] therefore (if i dint mess up) \[x = \frac{112}{29}-7\] \[y=\frac52\frac{112}{29}-7\]

OpenStudy (anonymous):

Ohhhh! okay. that made more sense! the whole g(x,y) and k got me a bit confused though... @amistre64

OpenStudy (amistre64):

it was over kill to say the least ... but i needed some lagrange multiplier practice. we could have just as well made an oversized circle; found the points of intersection and midpointed them

OpenStudy (anonymous):

Well. thank you! (: Over-all it was easy to follow along. @amistre64

OpenStudy (amistre64):

youre welcome

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!