# Find the eqation of the line throught the point (5,-2) perpendicular to the line with the equation 3x-2y=5

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### 1 Answer

We'll recall the fact that the product of the slopes of two perpendicular lines is -1.

Therefore, we'll re-write the given equation in the slope intercept form:

y = mx + n, where m is the slope and n is the y intercept.

3x - 2y = 5

We'll isolate -2y to the left side:

-2y = -3x + 5

We'll divide by -2:

y = 3x/2 - 5/2

Comapring the equations, we'll get the slope m1 = 3/2

The slope of the perpendicular line is determined from relation:

m1*m2 = -1

m2 = -1/m1

m2 = -1/(3/2)

m2 = -2/3

We'll write the point slope form of the equation of the line:

y - y2 = m2*(x - x2), where x2 = 5, y2 = -2 and m2 = -2/3

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

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

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

y = -2x/3 + 4/3

**The equation of the line, which is passing through the point (5,-2) and it is perpendicular to the line 3x - 2y = 5, is: y = -2x/3 + 4/3.**