# Find the equation of the line that passes through (5,-3) and perpendicular to 3x+2y=3

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The equation for the line is:

y-y1= m(x-x1)   where (x1,y1) is any point passes through the line and m is the slope>

==> y-(-3) = m(x-5)

==> y= m(x-5) -3

Since the line is perpendicular to 3x+2y=3, then the product of the slopes equals (-1)

The slope for 3x+2y =3 is  (-3/2)

Then the slope for the line is (2/3)

Then the equation is:

y= (2/3)(x-5) -3

y= (2/3)x -10/3 -3

y= (2/3)x -19/3

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Let's recall the fact that 2 line are perpendicular if the product of their slopes is -1.

For this reason, we'll put the given equation of the line into the standard form: y = mx + n.

3x+2y=3

We'll isolate the term in y to the left side:

2y = 3 - 3x

We'll divide by 2 both sides:

y = (-3/2)x + 3/2

The slope of this line is m1 = -3/2.

That means that the slope of the perpendicular line will be found from the relation:

m1*m2 = -1

m2 = -1/m1

m2 = -1/(-3/2)

m2 = 2/3

The equation of the line that passes through the point (5,-3) and is pepedicular to the line 3x+2y=3 is:

y - (-3) = m2*(x - 5)

y + 3 = (2/3)(x - 5)

y + 3 = (2/3)x - 10/3

y = (2/3)x - 10/3 - 3

y = (2/3)x - 19/3

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We find the slope of the given line in the form y = mx + c to find its slope:

3x + 2y = 3

2y = - 3x + 3

y = (-3/2)x = 3

Therefore the slope of the given line is -3/2.

Slope of the a line perpendicular to another is inverse of the slope of first line, multiplied by -1

Therefore slope of perpendicular to given line = -1/(-3/2) = 2/3

Therefore the equation of the perpendicular may be given as:

y = (2/3)x + c

This perpendicular line passes through the point (5, -3). Therefore to get the value of c in its equation we substitute the values of the coordinates of the point in the equation.

- 3 = (2/3)5 + c

- 3 = 10/3 + c

c = - 3 - 10/3 = - 19/3

Substituting this value in equation of perpendicular:

y = (2/3)x - 19/3

Multiplying this equation by -3, and taking all the terms on left hand side we get:

2x - 3y - 19 = 0

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An equation of a line perpendicular to a line ax+by+c= 0 is bx-ay+k = 0 where k is a constant to be determined by the condtion that it passes through aparticular point.

So 3x+2y = 3 or 3x+2y-3 = 0 has the perpendicular 2x-3y+k = 0.This passes through the point(5,-3).So (5,-3) should satisfy 32x-3y + k =0:

2(5)-3(-2)+k = 0.

10+6+k  = 0

k = -16.

Therefore , the line 2x-3y = -16 or 2x-3y+16 is the line passing through the point(5 ,-2) and perpendicular to 3x+2y=3.