Chapter 1.3, Problem 40AYU

In Problems 39-46, find an equation of the line $L$.

To determine

To find: The equation of the line $\text{L}$ with slope $\frac{1}{3}$ and containing the point $\left(-1,1\right)$ and $\left(2,2\right)$.

Answer to Problem 40AYU

Solution:

$\text{<mi>x</mi> <mo>−</mo> 3<mi>y</mi> + 4 = 0}$

Explanation of Solution

Given:

The following diagram

Slope of $\text{L}$ is $\frac{1}{3}$ and the points on the line are $\left(-1,1\right)$ and $\left(2,2\right)$

Formula used:

Point slope form of equation of line $\text{(<mi>y</mi> <mo>−</mo> <mi>y</mi>}₁\text{) = <mi>m</mi>(<mi>x</mi> <mo>−</mo> <mi>x</mi>}₁\text{)}$

Calculation:

The equation of the line with the slope $\frac{1}{3}$ and containing the point $\left(-1,1\right)$ can be found by using the point slope form with $\text{<mi>m</mi> = }\frac{1}{3}$, $x₁ =-1$ and $y₁ =1$.

$\text{(<mi>y</mi> <mo>−</mo> <mi>y</mi>}₁\text{) = <mi>m</mi>(<mi>x</mi> <mo>−</mo> <mi>x</mi>}₁\text{)}$

$\left(y-1\right)= \frac{1}{3}\left(x-\left(-1\right)\right)$

$\text{3}\left(\text{<mi>y</mi> <mo>−</mo> 1}\right)\text{ = }\left(\text{<mi>x</mi> + 1}\right)$Multiply 3 to both sides

$\text{3<mi>y</mi> <mo>−</mo> 3 = <mi>x</mi> + 1}$

$\text{3<mi>y</mi> <mo>−</mo> 3 + 3 = <mi>x</mi> + 1 + 3}$Adding 3 to both sides

$\text{3<mi>y</mi> = <mi>x</mi> + 4}$

$3\text{<mi>y</mi> }-3\text{<mi>y</mi> }= \text{<mi>x</mi> }+4-3y$

$0=\text{ <mi>x</mi> }-3\text{<mi>y</mi> }+4$

$\text{<mi>x</mi> }-3\text{<mi>y</mi> }+4=0$