Q5. - Algebra 2

Question

Graph the line that goes through $\left(\mathbf{4},-\mathbf{3}\right)$ and is parallel to the line whose equation is $\mathbf{2}\mathbf{x}+\mathbf{5}\mathbf{y}=\mathbf{10}$.

Step-by-Step Solution

Answer

The graph of the required line is given as follows:

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 slopes of parallel lines are equal.

 

The equation of a straight-line having slope m and passing through the point $\left(h,k\right)$ is given as $y-k=m\left(x-h\right)$.

Step 2 – List the given data

The given line is parallel to $2x+5y=10$ and goes through the point $\left(4,-3\right)$.

Step 3 – Find the slope

The required line is parallel to the line $2x+5y=10$.

 

Converting this straight-line equation to slope-intercept form,

 

$2x+5y=10$  (Given equation)

 

$2x+5y-2x=10-2x$  (Subtract $2x$ from both sides)

 

$5y=-2x+10$  (Simplify and rearrange)

 

$\frac{5y}{5}=\frac{-2x+10}{5}$  (Divide both sides by 5)

 

$y=-\frac{2}{5}x+2$ (Simplify)

 

Comparing with $y=mx+c$, $m=-\frac{2}{5}$ and $c=2$.

 

So, slope of the line $2x+5y=10$ and thus the required line is $m=-\frac{2}{5}$.

Step 4 – Find the equation

Put $m=-\frac{2}{5}$ and $\left(h,k\right)=\left(4,-3\right)$ in $y-k=m\left(x-h\right)$ to get,

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

 

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

 

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

 

$5y+15=-2\left(x-4\right)$  (Simplify)

 

$5y+15=-2x+8$  (Simplify)

 

$2x+5y+15=-2x+8+2x$  (Add $2x$ to both sides)

 

$2x+5y+15=8$  (Simplify)

 

$2x+5y+15-15=8-15$  (Subtract 15 from both sides)

 

$2x+5y=-7$  (Simplify)

 

So, $2x+5y=-7$ is the equation of the required line.

Step 5 – Graph the equation

Graphing the obtained equation of the straight line,