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

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Homework Help Overview

The discussion revolves around finding the coordinates of the foot of the perpendicular from a given point A (-5, 5) to a specified line represented by the equation 10x + 2y - 3 = 0. The subject area pertains to coordinate geometry and the properties of lines.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the gradients of the original line and the perpendicular line, along with the equations derived for the perpendicular line. There are attempts to solve the equations simultaneously, but some participants express uncertainty about the correctness of their results.

Discussion Status

Participants are actively engaging with the problem, with some providing corrections to earlier attempts and others questioning specific steps in the calculations. There is a recognition of mistakes made in the algebraic manipulation, and guidance is offered on how to properly handle the equations.

Contextual Notes

Some participants note that they are returning to basic mathematics and may be struggling with foundational concepts. There is an acknowledgment of potential errors in the setup and calculations, which are being discussed without reaching a definitive conclusion.

MMCS
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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:
 

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