Q12. - Algebra 2

Question

What is an equation of the line through $\left(\mathbf{2},-\mathbf{4}\right)$ and $\left(-\mathbf{3},-\mathbf{1}\right)$?

  (A).$\mathbf{y}=-\frac{\mathbf{3}}{\mathbf{5}}\mathbf{x}+\frac{\mathbf{26}}{\mathbf{5}}$             (B) $\mathbf{y}=-\frac{\mathbf{3}}{\mathbf{5}}\mathbf{x}-\frac{\mathbf{14}}{\mathbf{5}}$   

   (C)$\mathbf{y}=\frac{\mathbf{3}}{\mathbf{5}}\mathbf{x}-\frac{\mathbf{26}}{\mathbf{5}}$                 (D) $\mathbf{y}=\frac{\mathbf{3}}{\mathbf{5}}\mathbf{x}+\frac{\mathbf{14}}{\mathbf{5}}$

Step-by-Step Solution

Answer

The equation of the required straight line in slope-intercept form is $y=-\frac{3}{5}x-\frac{14}{5}$. So, (B) 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 $\left(2,-4\right)$ and $\left(-3,-1\right)$.

Then, $\left(h,k\right)=\left(2,-4\right)$ and $\left(a,b\right)=\left(-3,-1\right)$.

Step 3 – Write the equation

Put $\left(h,k\right)=\left(2,-4\right)$ and $\left(a,b\right)=\left(-3,-1\right)$ in $\frac{y-k}{x-h}=\frac{b-k}{a-h}$ to get,

 

$\frac{y-\left(-4\right)}{x-2}=\frac{-1-\left(-4\right)}{-3-2}$

 

$\frac{y+4}{x-2}=\frac{-1+4}{-5}$  (Simplify)

 

$\frac{y+4}{x-2}=-\frac{3}{5}$  (Simplify)

 

$\frac{y+4}{x-2}\left(x-2\right)=-\frac{3}{5}\left(x-2\right)$  (Multiply both sides by (x-2)

 

$y+4=-\frac{3}{5}\left(x-2\right)$  (Simplify)

 

$y+4=-\frac{3}{5}x+\frac{6}{5}$  (Distributive property)

 

$y+4-4=-\frac{3}{5}x+\frac{6}{5}-4$  (Subtract 4 from both sides)

 

$y=-\frac{3}{5}x-\frac{14}{5}$  (Simplify)

 

So, the required equation of the straight line in slope-intercept form is $y=-\frac{3}{5}x-\frac{14}{5}$.