Question

write the equation of a line that is perpendicular to $y = \frac{7}{5}x + 6$ and that passes through the point $(2, -6)$.

Explanation:

Step1: Find the slope of the perpendicular line

The slope of the given line \( y = \frac{7}{5}x + 6 \) is \( m_1=\frac{7}{5} \). The slope of a line perpendicular to it, \( m_2 \), satisfies \( m_1\times m_2=- 1 \). So \( m_2=-\frac{5}{7} \).

Step2: Use point - slope form to find the equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(2,-6) \) and \( m =-\frac{5}{7} \).
Substitute the values into the formula: \( y-(-6)=-\frac{5}{7}(x - 2) \).
Simplify the left - hand side: \( y + 6=-\frac{5}{7}(x - 2) \).
Expand the right - hand side: \( y+6=-\frac{5}{7}x+\frac{10}{7} \).
Subtract 6 from both sides: \( y=-\frac{5}{7}x+\frac{10}{7}-6 \).
Convert 6 to a fraction with denominator 7: \( 6=\frac{42}{7} \).
Then \( y=-\frac{5}{7}x+\frac{10 - 42}{7}=-\frac{5}{7}x-\frac{32}{7} \).

Answer:

The equation of the line is \( y =-\frac{5}{7}x-\frac{32}{7} \)