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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
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