Write an equation in slope-intercept form for a line perpendicular to the graph of the equation that passes through the x-intercept of that line. y=2/3-6
is the slope 2/3 and the y-intercept-6
-3/2
so would it be y=-3/2+6
If it is not clear we can start from the beginning.
can we im so confused
Sure. No problem
y = 2/3x - 6 is the equation of the line. The general equation of a line is y = mx + b where m is the slope of the line and b is the Y intercept. y intercept just means that is where the graph will cut the y axis. But here they are interested in the x intercept. x intercept means where does the line cut the axis? To find that all we have to do is set y = 0. Why y = 0? Because all points on x axis has y = 0.
So let us find the x intercept of y = 2/3x - 6 by setting y = 0 and solving for x. 0 = 2/3x - 6 Add 6 to both sides 6 = 2/3x multiply both sides by 3 18 = 2x divide both sides by 2 9 = x. So the x intercept is 9. That is where the line cuts the x axis. And at that point the y value will be 0. So the point will be (9, 0)
so would it be y=9x+0
No. What we found was the point where the original line cuts the x axis. I will show it in a drawing:|dw:1382672980547:dw|
We did the calculations so far to find the x intercept of the line y = 2/3x - 6 The point where it cuts the x axis is (9,0) Now we need to find the equation of a line that is perpendicular to the original line and also passes through (9,0)
The slope of the line y = 2/3x - 6 is: 2/3 The slope of the perpendicular line will be the negative reciprocal of the slope of the original line. So it is -3/2 Put it in y = mx + b y = -3/2x + b
The perpendicular line passes through (9, 0). So put x = 9 and y = 0 in y = -3/2x + b and solve for b.
y = -3/2x + b x = 9, y = 0 0 = (-3/2)(9) + b 0 = -27/2 + b b = 27/2 Put it in y = -3/2x + b and you get: y = -3/2x + 27/2 This is the equation of the line that is perpendicular to the original line and passes through the x intercept of the original line.
This is how the two lines will look when graphed: |dw:1382673964114:dw|
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