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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