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If you know the slope of the line that your equation is perpendicular too, you find the negative reciprocal of it and use it as the slope for the line. (negative reciprocal = flip the slope over and change its sign. Ex: a slope of 2 has a negative reciprocal of -1/2. ) Then you use the given point, and put your equation in point-slope form.
The general equation for point slope form is
Y-y1=m(x-x1)
The y1 is the y coordinate of the given point.
X1 is the x coordinate of the given point.
M is the slope that you found earlier.
You now have your equation.
If you are asked to put it in slope intercept form, simply distribute the numbers and solve the equation for y.
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What is the equation of a line that is perpendicular to and passes through the point (8 3)?
That depends on the equation that it is perpendicular too which has not been given but both equations will meet each other at right angles.
What is the equation of a line that is perpendicular to and passes through the point (4 6)?
That would depend on its slope which has not been given.
What is the equation of the line that contains the point (21) and is perpendicular to the line y3x-4?
If you mean: y=3x-4 and the point (2, 1) then the perpendicular equation is 3y=-x+5
How many planes can be perpendicular to a given line at a given point?
Only one
How do you write an equation that is perpendicular to a given line and passes through the given point?
you make an equation of the line: standard form: (y-y1)= m(x-x1) so if the point is (2, -2) and you want to make it perpendicular to the line with a slope (m) of 1/2, the perpendicular slope is the negative recipricle which is -2 so the equation would be: (y--2)= -2(x-2) (y+2)= -2(x-2) y+2= -2x +4 -2 -2 Slope Intercept Form: y= -2x +2
