Q53. - Algebra 2

Question

Find an equation of the line through $\left(\mathbf{0},-\mathbf{3}\right)$ and $\left(\mathbf{4},\mathbf{1}\right)$

A. $\mathbf{y}=-\mathbf{x}+\mathbf{3}$     (B) $\mathbf{y}=-\mathbf{x}-\mathbf{3}$        (C) $\mathbf{y}=\mathbf{x}-\mathbf{3}$            (D) $\mathbf{y}=\mathbf{x}+\mathbf{3}$

Step-by-Step Solution

Answer

The equation of the required straight line is $y=x-3$. So, (C) is the correct option.

Step 1 – State the concept

The slope intercept form of a straight-line equation is $y=mx+c$ where $m$ is the slope and $c$ is the y-intercept.

The equation of a straight-line passing through the points $\left(h,k\right)$ and $\left(a,b\right)$ is given as $\frac{y-k}{x-h}=\frac{b-k}{a-h}$.

Step 2 – List the given data

It is given that the line passes through $\left(0,-3\right)$ and $\left(4,1\right)$.

Then, $\left(h,k\right)=\left(0,-3\right)$ and $\left(a,b\right)=\left(4,1\right)$.

Step 3 – Write the equation

Put $\left(h,k\right)=\left(0,-3\right)$ and $\left(a,b\right)=\left(4,1\right)$ in $\frac{y-k}{x-h}=\frac{b-k}{a-h}$ to get,

 

$\frac{y-\left(-3\right)}{x-0}=\frac{1-\left(-3\right)}{4-0}$

 

$\frac{y+3}{x}=\frac{1+3}{4}$  (Simplify)

 

$\frac{y+3}{x}=\frac{4}{4}$  (Simplify)

 

$\frac{y+3}{x}=1$  (Simplify)

 

$\frac{y+3}{x}\left(x\right)=1\left(x\right)$  (Multiply both sides by $x$)

 

$y+3=x$  (Simplify)

 

$y+3-3=x-3$  (Subtract 3 from both sides)

 

$y=x-3$  (Simplify)

 

So, the required equation of the straight line is $y=x-3$.