# Find the equation of Line L that passes through the points (1,3) and (-3,4).

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First let us determine the standard line formula:

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

To calculate the slope (m) we have the points (1,3) and (-3,4)

m =(y2-y1)/(x2-x2)

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

= 1/-4= -1/4

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

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

y= -x/4 +13/4

We'll use the formula that describes the equation of the line which passes through 2 given points:

(y2-y1)/(y-y1) = (x2-x1)/(x-x1)

Let's note the given points: A1(1,3) and A2(-3,4)

Now, we'll substitute the coordinates of A1, A2, into the equation:

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

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

We'll cross multiply and we'll get:

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

x-1 = -4y+12

We'll move all terms to one side and we'll get the general form of the equation of the line that passes through A1, A2:

**x+4y-13=0**

We assume that the equation of the line L is y = mx+c. Since this pass through (1,3), the coordinates (1,3) should satisfy y = mx+c. So

3 = m*1 + c.......(1). The line L also passes through the point (-3,4). So the coordinates of the point(-3,4) should also satisfy the equation, y = mx+c

4=m*(-3)+ c......(2). Solve for m and c from the two equations (1) and (2) :

Eq(1)-eq(2) eliminates c : 3-4 = (m*1+c) - (m*(-3)+c) = 4m. Or

m = -1/4. Substituting m = -1/4 in (1) we get:

3 = (-1/4)*1+c. Or c = 3+1/4 = 13/4.

So substituting m=-1/4 and c =13/4 in the assumed line y =mx+c, we get:

So the required line is y = (-1/4)x+13/4 is the required line. Multiply by 4 if ypu wish to get away with the denominator.The same would change to a form like:

4y = -x+13 . Or

x+4y-13 = 0, which is in the form of ax+by+c = 0.