Find Coordinates of Foot of Perpendicular from Point A (-5,5) to Line 10x+2y-3=0

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The discussion focuses on finding the coordinates of the foot of the perpendicular from point A (-5, 5) to the line represented by the equation 10x + 2y - 3 = 0. The user initially struggles with the calculations, particularly in forming the correct equation for the intercepting line. A mistake in rearranging the equation leads to incorrect results, prompting a request for assistance. Other participants point out the error in multiplying one side of the equation without applying the same operation to the other side. The user acknowledges the mistake and expresses gratitude for the clarification, indicating a need for more practice in basic math concepts.
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I apologise in advance for an extremely trivial question, i am returning to brush up on some basic maths and can't seem to arrive at the answer

Find the coordinates of the foot of the perpendicular from the point A with coordinates (-5,5) to the line 10x+2y-3= 0

1.) gradient of original line -10/2, gradient of intercepting line 1/5
2.) equation of intercepting line y-5 = 1/5(x+1) *5
3.)5y-25 = x+1 re-arranging gives 5y-x = 26
4.) solve simultaneously -10x-50y=-260 ( *-10 )
10x+2y=3

-48y=-257

y=5.354
Substituting y into either equation
x= -0.7708

From the program i know these answers arnt correct

Any pointers would be appreciated,

Thanks
 
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MMCS said:
I apologise in advance for an extremely trivial question, i am returning to brush up on some basic maths and can't seem to arrive at the answer

Find the coordinates of the foot of the perpendicular from the point A with coordinates (-5,5) to the line 10x+2y-3= 0

1.) gradient of original line -10/2, gradient of intercepting line 1/5
2.) equation of intercepting line y-5 = 1/5(x+1) *5
Mistake above. The known point on the intersecting line is (-5, 5), so the equation of this line would be y - 5 = (1/5)(x - (-5))

Why do you have "*5" on the right side of your equation?
MMCS said:
3.)5y-25 = x+1 re-arranging gives 5y-x = 26
4.) solve simultaneously -10x-50y=-260 ( *-10 )
10x+2y=3

-48y=-257

y=5.354
Substituting y into either equation
x= -0.7708

From the program i know these answers arnt correct

Any pointers would be appreciated,

Thanks
 
Hi,

I had the *5 to remove the fraction, don't know how i ended up with the 1 on the right hand side. That sorted it. Thanks for spotting that, such a silly mistake. Wouldnt think id been staring at it for half an hour. Probably suggests i need some sleep :bugeye:
 
MMCS said:
I apologise in advance for an extremely trivial question, i am returning to brush up on some basic maths and can't seem to arrive at the answer

Find the coordinates of the foot of the perpendicular from the point A with coordinates (-5,5) to the line 10x+2y-3= 0

1.) gradient of original line -10/2, gradient of intercepting line 1/5
2.) equation of intercepting line y-5 = 1/5(x+1) *5

Check the red part.

ehild
 
MMCS said:
Hi,

I had the *5 to remove the fraction
If you multiply one side by 5, you need to also multiply the other side by the same number. Maybe you did this, but it doesn't show in your work.

It's best to start a new line when you do this kind of operation.
MMCS said:
, don't know how i ended up with the 1 on the right hand side. That sorted it. Thanks for spotting that, such a silly mistake. Wouldnt think id been staring at it for half an hour. Probably suggests i need some sleep :bugeye:
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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