Question

What is the equation of the line that passes through the points (4/5,1/5) and (1/2,3/2)?​

Answer 1

Answer:

$\large\boxed{y=-\dfrac{13}{3}x+\dfrac{11}{3}}-\bold{slope\ intercept\ form}\\\boxed{13x+3y=11}-\bold{standard\ form}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have two points

$\left(\dfrac{4}{5},\ \dfrac{1}{5}\right),\ \left(\dfrac{1}{2},\ \dfrac{3}{2}\right)$

Convert fractions to the decimals

(divide the numerator by the denominator) :

$\dfrac{4}{5}=0.8,\ \dfrac{1}{5}=0.2,\ \dfrac{1}{2}=0.5,\ \dfrac{3}{2}=1.5$

$\left(\dfrac{4}{5},\ \dfrac{1}{5}\right)=(0.8,\ 0.2)\\\\\left(\dfrac{1}{2},\ \dfrac{3}{2}\right)=(0.5,\ 1.5)$

Calculate the slope:

$m=\dfrac{1.5-0.2}{0.5-0.8}=\dfrac{1.3}{-0.3}=-\dfrac{13}{3}$

Put the value of slope and the coordinates of the first point to the equation of a line:

$0.2=-\dfrac{13}{3}(0.8)+b$      multiply both sides by 3

$0.6=(-13)(0.8)+3b$

$0.6=-10.4+3b$           add 10.4 to both sides

$11=3b$           divide both sides by 3

$\dfrac{11}{3}=b\to b=\dfrac{11}{3}$

Finally:

$y=-\dfrac{13}{3}x+\dfrac{11}{3}$ - slope-intercept form

Convert to the standard form (Ax + By = C):

$y=-\dfrac{13}{3}x+\dfrac{11}{3}$      multiply both sides by 3

$3y=-13x+11$          add 13x to both sides

$13x+3y=11$ - standard form

Answer 2

Answer:

$y-\frac{3}{2}=\frac{-13}{3}(x-\frac{1}{2})$ point-slope form

$13x+3y=11$ (standard form)

Let me know if you prefer another form.

Step-by-step explanation:

The slope of a line can be found using $\frac{y_2-y_1}{x_2-x_1}$ provided you are given two points on the line.

We are.

Now you can use that formula.  But I really love to just line up the points vertically then subtract them vertically then put 2nd difference over 1st difference.

 (4/5  ,  1/5)

-( 1/2  ,  3/2)

-----------------

3/10          -13/10

2nd/1st = $\frac{\frac{-13}{10}}{\frac{3}{10}}=\frac{-13}{3}$ is our slope.

So the following is point-slope form for a linear equaiton:

$y-y_1=m(x-x_1) \text{ where } m \text{ is slope and } (x_1,y_1) \text{ is a point on the line }$    

Plug in a point $(x_1,y_1)=(\frac{1}{2},\frac{3}{2}) \text{ and } m=\frac{-13}{3}$.

This gives:

$y-\frac{3}{2}=\frac{-13}{3}(x-\frac{1}{2})$

I'm going to distribute:

$y-\frac{3}{2}=\frac{-13}{3}x-\frac{-13}{6}$

Now I don't like these fractions so I'm going to multiply both sides by the least common multiply of 2,3, and 6 which is 6:

$6y-9=-26x+13$

Add 26x on both sides:

$26x+6y-9=13$

Add 9 on both sides:

$26x+6y=22$ This is actually standard form of a line.

It can be simplified though.

Divide both sides by 2:

$13x+3y=11$ (standard form)