# Find the equation of the line through (4, 3) to the line that is perpendicular to the line 3x+y=7

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The line equation is:

y-y1 = m(x-x1 ) where m is the slope and (x1,y1) is a point pass through the line:

We have the point (4,3)

Now for the slope:

Since the line is perpendicular to the line 3x+y=7

Then we could determine the slope.

First we will determine the slope for 3x+y=7

==> y = -3x+7

The slope = -3

The slope for the perpendicular line is 1/3 ( flip and change sign)

Then the equation for the line is:

y-3= 1/3(x-4)

y= (1/3)x -4/3 +3

y= (1/3)x + 5/3

The equation of a line perpendicular to a line ax+by+c is of the form bx-ay +k.

Therefore the line perpendicular to 3x+y = 7 or 3x+y-7 = 0 is of the form: x-3y+k = 0, where k could be determined by the fact that x-3y+k = 0should pass through the given point (4,3): So

4-3(3)+k = 0.

4-9+k = 0

-5+k = 0

k=5.

Therefore, the required line is x-3y+5 = 0 is the line perpendicular to 3x+y = 7.

If 2 lines are perpendicular then the product of their slopes is:

m1*m2=-1

Let's find the slope of the line 3x+y=7.

We'll put the equation into the standard form: y = mx + n, where m is the slope of the line.

For this reason, we'll isolate y to the left side and we'll subtract 3x both sides:

y = 7-3x

y = -3x + 7

The slope of the line y = -3x + 7 is m1 = -3.

The slope of the line that passes through the point (4,3) is:

m1*m2=-1

-3*m2=-1

We'll divide by -3, both sides:

m2 = 1/3

Now, we'll write the equation of a line which passes through a given point and tyhe valueo of it's slope is known.

y-3 = m2(x-4)

y-3 = (1/3)(x-4)

We'll add 3 both sides:

y = (1/3)x - 4/3 + 3

y = x/3 + 5/3