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lbvjy [14]
3 years ago
8

What is the standard form equation of the line shown below? Graph of a line going through negative 1, 5 and 2, 4

Mathematics
1 answer:
just olya [345]3 years ago
3 0

Answer:

x+3y=14

Step-by-step explanation:

Standard form for an equation of a line is ax+by=c . Some books have restrictions on a,b, and c.

So first thing I'm going to do is state the equation I'm going to use to get there.

I'm going to use the point-slope form because we are given a point (2 really) and we can find the slope using two points (we have).

Line up points and subtract. You will then put second difference over the first difference.  This will give you the slope. You could just use (y2-y1)/(x2-x1).

( -1 , 5)

-(  2, 4)

-----------

-3     1

So the slope is 1/-3 or -1/3 .

So the equation in point-slope form, y-y1=m(x-x1), is y-4=-1/3 (x-2) .

y-4=-1/3 (x-2)

First step: I'm going to get rid of the fraction by multiplying both sides by 3.

3y-12=-1(x-2)

Second step: Distribute

3y-12=-x+2

Third step: add x on both sidfes

x+3y-12=2

Fourth step: add 12 on both sides

x+3y=14.

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b) { (a,a)(b,b)(c,c)(c,d)(d,c)(d,d)(e,e)(e,f)(f,e)(f,f)(g,g)

c)  {(a,a)(a,b)(b,b)(b,a)(a,c)(c,a)(b,c)(c,b)(c,c)(d,a)(a,d)(b,d)(d,b)(d,c)(c,d)(d,d)(c,c)(c,f)(f,c)(f,f)(c,g)(g,c)(f,g)(g,f)(g,g)

Step-by-step explanation:

a) {a,b} , {c,d} , {e,f,g}

  {a , b } = {(a,a) (b,b) (a,b) (b,a)}

  {c , d } = {(c,c) (d,d) (c,d) (d , c) }

  {e,f,g} = {(e,e) (e,f)  (f,e)  (e,g) (f,f) (f,g) ( g,e) (g,f) (g,g)}  

b) {a} ,{b},{c,d} ,{e,f},{g}

     { (a,a)(b,b)(c,c)(c,d)(d,c)(d,d)(e,e)(e,f)(f,e)(f,f)(g,g)}

c)  {a,b,c,d},{e,f,g}

   {(a,a)(a,b)(b,b)(b,a)(a,c)(c,a)(b,c)(c,b)(c,c)(d,a)(a,d)(b,d)(d,b)(d,c)(c,d)(d,d)(c,c)(c,f)(f,c)(f,f)(c,g)(g,c)(f,g)(g,f)(g,g)

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