though (6,9), perpendicular to x=5
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the line x=5 is a vertical line going through the point (5,0)
to be perpendicular means to go through at a 90 degree angle, so our other line must be horizontal.
all horizontal lines are in the form y= a number
since our equation has to pass through (6,9), and the y value for this point is 9, then our equation is
y=9
to be perpendicular means to go through at a 90 degree angle, so our other line must be horizontal.
all horizontal lines are in the form y= a number
since our equation has to pass through (6,9), and the y value for this point is 9, then our equation is
y=9
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If it is perpendicular to x=5, which is a horizontal line, then the gradient of the line passing through (6,9), which will then be vertical, must be 0, as any 2 points on this vertical line will have exactly the same y co-ordinate, and using the formula m=(y - y(1))/(x - x(1)), for where y=y(1), then
m=0/(x - x(1))=0, and so to find the equation going through the point (6,9), rearrange the above formula of the gradient to get y - y(1)=m(x - x(1)), replace the m with 0, the x(1) with 6, and the y(1) with 9, and so you get
y - 9=0(x - 6)=0
y=9
And to the guy who answered that you should do your own homework is wrong, as everyone needs help sometimes, especially with mathematics as most people nowadays don't get it, I mean I only have like 5 students in my calculus class, but I'm the only one who can understand what's going most of the time.
m=0/(x - x(1))=0, and so to find the equation going through the point (6,9), rearrange the above formula of the gradient to get y - y(1)=m(x - x(1)), replace the m with 0, the x(1) with 6, and the y(1) with 9, and so you get
y - 9=0(x - 6)=0
y=9
And to the guy who answered that you should do your own homework is wrong, as everyone needs help sometimes, especially with mathematics as most people nowadays don't get it, I mean I only have like 5 students in my calculus class, but I'm the only one who can understand what's going most of the time.
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Y=9