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The line with equation x+3y=12 meets the x and y axes at the points A and B respectively. Find the equation of the perpendicular bisector of AB
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This requires a couple of steps. The words look all googly and weird but its pretty basic:
You have the equation for a line. You are told to find the bisector (the midpoint) of the line formed by the points A and B, which are your X and Y intercepts. Therefore:
To find the X-int, we set Y=0 and solve:
Alternatively, we could have used out original equation.
To find the Y=int, we set X=0 and solve. This one is straight forward. Y=4.
Now we have the co-ordinates of point A and B: (12,0) and (0,4) respectively. Our question asks us to find the equation of the perp-bisector. Now for a line to be perpendicular to another line, it must have the a slope equal to the negative reciprocal of the line it is to be perpendicular to. The slope of our original line is simple to see:. Therefore the negative reciprocal is 3. Thus the slope of our perpendicular line is 3.
But we aren't done. We have a very specific point we need to use in order to make sure this line bisects the X and Y intercepts of our original equation. Thus, we need to use the midpoint formula, to find the midpoint between (12,0) and (0,4):
Now, we have a point (6,2), a slope of 3 and a form called the point-slope form of an equation:
 \Longrightarrow y=3x-36+2 \Longrightarrow y=3x-34)
I am concerned that so many people are posting questions here without showing any attempt at the problems themselves. "Math-DumbassX" (and that name shows very low self-esteem) were you able, at least to determine the points A and B? Do you know what "meets the x and y axes" means?Originally Posted by Math-DumbassX
Next Question .... Sorry :P
The line with equation x+3y=12 meets the x and y axes at the points A and B respectively. Find the equation of the perpendicular bisector of AB
Help Please
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