Question

Write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (1,7) and perpendicular to 3 x + 7y = 1.a) The equation of the line in slope-intercept form is   enter your response here.write the equation in standard form also.

Answer 1

Answer:

a.)

$y=-\frac{3}{7}x+\frac{1}{7}$

b.)

$7x-3y=-14$

Given:

$3x+7y=1$

To write the given equation to its slope-intercept form, we'll just have to write it in terms of y.

$\begin{gathered} 3x+7y=1 \\ 7y=-3x+1 \\ y=-\frac{3}{7}x+\frac{1}{7} \end{gathered}$

Now, perpendicular lines have the negative reciprocal of the other line.

$m_{\perp}=-\frac{1}{m}$

Since the slope of the given equation is -3/7,

$\begin{gathered} m_{\perp}=-\frac{1}{m} \\ =-\frac{1}{-\frac{3}{7}} \\ m=\frac{7}{3} \end{gathered}$

Now that we got the slope, we will substitute this to the following equation with the point (1,7)

$\begin{gathered} y=mx+b \\ 7=\frac{7}{3}(1)+b \\ 7=\frac{7}{3}+b \\ b=7-\frac{7}{3} \\ b=\frac{14}{3} \end{gathered}$

With this, we now know that the y-intercept of the equation that we are looking for is 14/3. Substituting the slope and y-intercept to the equation and we will get:

$\begin{gathered} y=mx+b \\ y=\frac{7}{3}x+\frac{14}{3} \end{gathered}$

Then, to write this in its standard form,

$\begin{gathered} y=\frac{7}{3}x+\frac{14}{3} \\ 3y=7x+14 \\ -7x+3y=14 \end{gathered}$

Since the first term cannot be negative, we will multiply the entire equation by -1. Therefore the answer for b is 7x - 3y = -14