What is the equation of the line perpendicular to y=2x-3 and that passes through (0;0)
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Two lines are perpendicular when the product of their slopes is -1. Comparing the given equation to the point slope form of an equation of a line, we'll get the slope m1 = 2.
y = mx + n
y = 2x - 3
We also know that:
m1*m2 = -1, where m2 is the slope of the perpendicular line.
m2 = -1/m1
m2 = -1/2
The equation of the perpendicular line, that passes through the origin and has the slope m2 is:
y - 0 = m2(x - 0)
y = -x/2
y = -0.5x
We have to find the equation of the line perpendicular to y=2x-3 and that passes through (0,0).
Now y = 2x - 3, has a slope 2
The slope of the line perpendicular to this is the neagtive inverse of -1/2.
Therefore the equation of the line passing through (0 , 0) and with the slope -1/2 is
y = (-1/2)*x
=> 2y = -x
=> x + 2y = 0
The required equation of the line perpendicular to y = 2x - 3 and that passes through (0, 0) is x + 2y = 0.
Given the point (0,0) passes through the line.
Then we will write the equation of the line:
(y-y1) = m(x-x1) where (x1,y1) is any point passes through the line, and m is the slope.
==> (y-0) = m( x-0)
==> y= mx.
Now we will find the slope (m).
We are given that the equation of the perpendicular line is y=2x-3.
Then we know that the slope of the perpendicular line is 2.
Also, we know that the product of the slopes of two perpendicular lines is -1.
==> 2*m = -1
==> m = -1/2,
Now we will substituteinto the equation.
==> y= (-1/2)x
=> y= -(1/2)x
==> 2y + x = 0
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