Q. 37 - Algebra 2

Question

Write an equation in slope-intercept form for the line that satisfies each set of conditions.

passes through $\left(\mathbf{6},-\mathbf{5}\right)$, perpendicular to the line whose equation is $\mathbf{3}\mathbf{x}-\frac{\mathbf{1}}{\mathbf{5}}\mathbf{y}=\mathbf{3}$

Step-by-Step Solution

Answer

The equation of the required straight line in slope-intercept form is $y=-\frac{1}{15}x-\frac{23}{5}$.

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 product of slopes of perpendicular lines is $-1$.

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

It is given that the line passes through $\left(6,-5\right)$ and is perpendicular to $3x-\frac{1}{5}y=3$.

Then, $\left(h,k\right)=\left(6,-5\right)$.

Step 3. Determine the slope

$3x-\frac{1}{5}y=3$  (Given equation)

 

$3x-\frac{1}{5}y-3x=3-3x$  (Subtract $3x$ from both sides)

 

$-\frac{1}{5}y=3-3x$  (Simplify)

 

$-\frac{1}{5}y\left(-5\right)=-5\left(3-3x\right)$  (Multiply both sides by -5)

 

$y=15x-15$  (Simplify)

 

Comparing $y=15x-15$ to $y=mx+c$, $m=15$.

 

So, the slope of $y=15x-15$ is $15$.

Thus, the slope of the required line, as it is perpendicular to $y=15x-15$, is $-\frac{1}{15}$.

 Then, $m=-\frac{1}{15}$.

Step 4. Write the equation

Put $m=-\frac{1}{15}$ and $\left(h,k\right)=\left(6,-5\right)$ in $y-k=m\left(x-h\right)$ to get,

 

$y-\left(-5\right)=-\frac{1}{15}\left(x-6\right)$

 

$y+5=-\frac{1}{15}\left(x-6\right)$  (Simplify)

 

$y+5=-\frac{1}{15}x+\frac{2}{5}$  (Distributive property)

 

$y+5-5=-\frac{1}{15}x+\frac{2}{5}-5$  (Subtract 5 from both sides)

 

$y=-\frac{1}{15}x-\frac{23}{5}$  (Simplify)

 

So, the required equation of the straight line in slope-intercept form is $y=-\frac{1}{15}x-\frac{23}{5}$