Question

what is an equation of the line that passes through the point (6, 5) and is perpendicular to the line 3x + 5y = 40?

Explanation:

Step1: Find the slope of the given line

First, rewrite the equation \(3x + 5y = 40\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept).
Subtract \(3x\) from both sides: \(5y=-3x + 40\).
Divide each term by 5: \(y=-\frac{3}{5}x + 8\).
The slope of the line \(3x + 5y = 40\) is \(m_1=-\frac{3}{5}\).

Step2: Find the slope of the perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of the perpendicular line be \(m_2\).
We know that \(m_1\times m_2=-1\). Substituting \(m_1 = -\frac{3}{5}\) into the equation: \(-\frac{3}{5}\times m_2=-1\).
Solve for \(m_2\): \(m_2=\frac{5}{3}\) (by multiplying both sides by \(-\frac{5}{3}\)).

Step3: Use the point - slope form to find the equation of the line

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(6,5)\) and \(m = \frac{5}{3}\).
Substitute the values into the formula: \(y - 5=\frac{5}{3}(x - 6)\).

Step4: Simplify the equation to slope - intercept form (or standard form)

Expand the right - hand side: \(y - 5=\frac{5}{3}x-10\).
Add 5 to both sides: \(y=\frac{5}{3}x-10 + 5\).
Simplify: \(y=\frac{5}{3}x-5\).
If we want to write it in standard form (\(Ax+By = C\), where \(A\), \(B\), and \(C\) are integers and \(A\geq0\)):
Multiply through by 3 to get rid of the fraction: \(3y = 5x-15\).
Rearrange to get \(5x-3y=15\) (or we can also have \(y=\frac{5}{3}x - 5\) as the slope - intercept form).

Answer:

The equation of the line is \(y=\frac{5}{3}x - 5\) (or \(5x-3y = 15\))