color(blue)("General introduction")General introduction
consider the equation of a straight line in the standard form of:
y=mx+cy=mx+c
In this case mm is the slope (gradient) and cc is some constant value
A straight line that is perpendicular to this would have the gradient of [-1xx 1/m][−1×1m] so its equation is:
color(white)(.).
y=[(-1)xx1/m]x+k" "->" "y=-1/mx+ky=[(−1)×1m]x+k → y=−1mx+k
Where kk is some constant value that is different to that of cc
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color(blue)("Determine the given line equation")Determine the given line equation
Given " "color(green)(6x+y=1) 6x+y=1
Subtract color(red)(6x)6x from both sides
color(green)(6xcolor(red)(-6x)+y" "=" "1color(red)(-6x)6x−6x+y = 1−6x
But 6x-6x=06x−6x=0
0+y=-6x+10+y=−6x+1
color(blue)(y=-6x+1)" "->" "y=mx+c" "color(blue)(larr" Given line")y=−6x+1 → y=mx+c ← Given line
So m=-6m=−6
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color(blue)("Determine perpendicular line equation")Determine perpendicular line equation
y=-1/m x+k" "->" "y=-(1/(-6))x+ky=−1mx+k → y=−(1−6)x+k
y=+1/6 x+k" " larr" Perpendicular line"y=+16x+k ← Perpendicular line
We are told that this passes through the known point
(x,y)->(6,-3)(x,y)→(6,−3)
Substitute these values in the equation to find kk
y=1/6 x+k" "->" "-3=1/(cancel(6))(cancel(6))+k
-3=1+k
Subtract 1 from both sides
-4=k
So the equation is
y=-1/mx+k" "->" "color(blue)(ul(bar(|color(white)(2/2)y=1/6x-4" "|)))
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