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line v has an equation of $y = \\frac{10}{9}x + 1$. line w includes the…

Question

line v has an equation of $y = \frac{10}{9}x + 1$. line w includes the point $(-3, 3)$ and is perpendicular to line v. what is the equation of line w? write the equation in slope - intercept form. write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

Explanation:

Step1: Find the slope of line w

The slope of line \( v \) is \( \frac{10}{9} \). For two perpendicular lines, the product of their slopes is -1. Let the slope of line \( w \) be \( m \). So, \( \frac{10}{9} \times m=-1 \), then \( m = -\frac{9}{10} \).

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 \( (x_1,y_1)=(-3,3) \) and \( m = -\frac{9}{10} \). Substitute these values: \( y - 3=-\frac{9}{10}(x + 3) \).

Step3: Convert to slope - intercept form

Expand the right - hand side: \( y - 3=-\frac{9}{10}x-\frac{27}{10} \). Then add 3 to both sides. Since \( 3=\frac{30}{10} \), we have \( y=-\frac{9}{10}x-\frac{27}{10}+\frac{30}{10} \). Simplify the right - hand side: \( y = -\frac{9}{10}x+\frac{3}{10} \).

Answer:

\( y=-\frac{9}{10}x+\frac{3}{10} \)