Q. 36 - Algebra 2

Question

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

passes through $\left(\mathbf{2},-\mathbf{5}\right)$, perpendicular to the graph of $\mathbf{y}=\frac{\mathbf{1}}{\mathbf{4}}\mathbf{x}+\mathbf{7}$

Step-by-Step Solution

Answer

The equation of the required straight line in slope-intercept form is $y=-4x+3$.

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 $\left(2,-5\right)$ and is perpendicular to $y=\frac{1}{4}x+7$.

Then, $\left(h,k\right)=\left(2,-5\right)$.

Step 3. Determine the slope

Comparing $y=\frac{1}{4}x+7$ to $y=mx+c$, $m=\frac{1}{4}$.

So, the slope of $y=\frac{1}{4}x+7$ is $\frac{1}{4}$.

Thus, the slope of the required line, as it is perpendicular to $y=\frac{1}{4}x+7$, is $-4$.

Then, $m=-4$.

Step 4. Write the equation

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

 

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

 

$y+5=-4\left(x-2\right)$  (Simplify)

 

$y+5=-4x+8$  (Distributive property)

 

$y+5-5=-4x+8-5$  (Subtract 5 from both sides)

 

$y=-4x+3$   (Simplify)

 

So, the required equation of the straight line in slope-intercept form is $y=-4x+3$.