Question

Avani wrote the linear equation y = 5 x + 4. Then, she wrote the equation of the line that is perpendicular to y = 5 x + 4 and that passes through (15,–2). If her new equation is in the form y = negative one-fifth x + b, what is the value of b?

Answer 1

Avani wrote the linear equation y = 5x + 4. Then, she wrote the equation of the line that is perpendicular to y = 5x+ 4 and that passes through (15,–2). If her new equation is in the form  $y-\frac{1}{5}x+b$, what is the value of b?

$\bf y=mx+b$

• m is the slope
• b is the y-intercept

The equation of a line that is perpendicular to another line always has the opposite reciprocal slope. In the original equation, the slope is 5. The opposite of 5, which refers to the sign, is -5. The reciprocal of -5 is -1/5. That means the slope of the new line is -1/5.

The perpendicular line:

• Has a slope of $-\frac{1}{5}$

• Passes through the point (15, -2)

To find the equation of a line that passes through a specific point, you need to use slope-intercept form.

$\bf y-y_{1} =m(x-x_{1})$

• y₁ is the y-value of the point
• m is the slope
• x₁ is the x-value of the point

y₁ is -2, m is -1/5,  and x₁ is 15.

Substitute the known values into the equation:

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

Open the parentheses and distribute:

$-\frac{1}{5}(x-15) \rightarrow -\frac{1}{5} \cdot x\ + -\frac{1}{5} \cdot -15 \rightarrow -\frac{1}{5}x +3$

The new equation will be:

$y+2=-\frac{1}{5}x+3$

Lastly, you need to move the positive 2 to the other side to keep it in y = mx + b form.

$y+2-2=-\frac{1}{5}x+3-2 \rightarrow y=-\frac{1}{5}x+1$

The equation of the perpendicular line is:

$\bf y=-\frac{1}{5}x+1$

The value of b is 1.