# What is the equation of the line (in y=mx+b form) that passes through (-3,1) and is perpendicular to the line 2x-5y=-17?

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We need to find the line in the form (y= mx + b).

Given the point (-3, 1) passes through the line.

The, we will write the line equation in the slope form.

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

Then, we will substitute the point.

==> ( y -1) = m (x+3)

But, given the perpendicular line 2x-5y = -17.

We know that the product of the slopes = -1.

Let us determine the slope of the perpendicular line.

==> 2x -5y = -17

==> -5y = -2x - 17

==> y= (2/5)x +17/5.

Then, the perpendicular slope is 2/5.

==> (2/5)* m = -1

==> m = -5/2

==> (y-1) = (-5/2)( x+ 3)

==> y-1 = (-5/2)x - 15/2

==> y= (-5/2)x - 15/2 + 1

**==> y= (-5/2)x - 13/2**

We'll start from the fact that 2 lines are perpendicular if and only if the product of their slopes is -1.

For this reason, we'll put both equation in the point slope form:

y = mx + n

We'll start by the given equation:

2x-5y=-17

We'll isolate -5y to the left side:

-5y = -2x - 17

We'll divide by -5:

y = 2x/5 + 17/5

m1 = 2/5

The slope of the perpendicular line is:

m2 = -1/m1

m2 = -5/2

We'll write the equation of the line that passes through a given point and it has the slope m2.

y - 1 = m2(x + 3)

y - 1 = -5x/2 - 15/2

We'll add 1 both sides:

y = -5x/2 - 15/2 + 1

y = -5x/2 - 13/2

The equation of the perpendicular line to the given line 2x-5y=-17, whose slope is m2 = -5/2 and it is written in the point slope form. is:

**y = -5x/2 - 13/2**