Question

Find an equation of a line perpendicular to the given line that contains the given point. Write the equation in slope-intercept form Line 5x-4y=6, point(-4,3)

Answer 1

Explanation

From the statement, we must find an equation:

0. of a line perpendicular to the line 5x - 4y = 6,

,

1. that contains the point (x, y) = (-4, 3).

We write the equation of the given line as:

$\begin{gathered} 4y+6=5x, \\ 4y=5x-6, \\ y=\frac{5}{4}x-\frac{6}{4}=\frac{5}{4}x-\frac{3}{2}. \end{gathered}$

We see that the slope of the given line is m₁ = 5/4.

The slope-intercept equation of the line that we must find is:

$y=m_2\cdot x+b_2.$

Where:

• m₂ is the slope,

,

• b₂ is the y-intercept.

1) From condition 1, we know that the line must be perpendicular to the given line. So their slopes must satisfy:

$m_1\cdot m_2=-1\Rightarrow\frac{5}{4}\cdot m_2=-1.$

Solving for m₂ we get:

$m_2=-\frac{4}{5}.$

2) Replacing the coordinates of the point from condition 2, and the slope m₂ = -4/5 in the equation of the line, we have:

$\begin{gathered} 3=(-\frac{4}{5})\cdot(-4)+b_2, \\ 3=\frac{16}{5}+b_2. \end{gathered}$

Solving for b₂ we get:

$b_2=3-\frac{16}{5}=-\frac{1}{5}.$

So the slope-intercept equation of the perpendicular line is:

$y=-\frac{4}{5}x-\frac{1}{5}.$Answer$y=-\frac{4}{5}x-\frac{1}{5}$