Question

what is the equation of the line that is perpendicular to the line $y = \frac{3}{5}x + 10$ and passes through the point $(15, -5)$? $y = -\frac{3}{5}x + 20$ $y = \frac{3}{5}x - 20$ $y = \frac{5}{3}x - 20$ $y = -\frac{5}{3}x + 20$

Explanation:

Step1: Find the slope of the perpendicular line

The slope of the given line \( y = \frac{3}{5}x + 10 \) is \( m_1=\frac{3}{5} \). For two perpendicular lines, the product of their slopes is \( - 1 \), i.e., \( m_1\times m_2=-1 \). So, \( \frac{3}{5}\times m_2=-1 \), solving for \( m_2 \) we get \( m_2 =-\frac{5}{3} \)? Wait, no, wait. Wait, no, the slope of a line perpendicular to a line with slope \( m \) is the negative reciprocal. So if \( m=\frac{3}{5} \), the negative reciprocal is \( -\frac{5}{3} \)? Wait, no, wait, let's recalculate. The negative reciprocal of \( \frac{3}{5} \) is \( -\frac{5}{3} \)? Wait, no, reciprocal of \( \frac{3}{5} \) is \( \frac{5}{3} \), then negative of that is \( -\frac{5}{3} \)? Wait, no, wait, the product of slopes of perpendicular lines is -1. So \( m_1\times m_2=-1 \), so \( m_2=\frac{-1}{m_1} \). So if \( m_1 = \frac{3}{5} \), then \( m_2=\frac{-1}{\frac{3}{5}}=-\frac{5}{3} \). Wait, but let's check the options. Wait, the first option has slope \( -\frac{3}{5} \), second \( \frac{3}{5} \), third \( \frac{5}{3} \), fourth \( -\frac{5}{3} \). Wait, maybe I made a mistake. Wait, no, the given line is \( y=\frac{3}{5}x + 10 \), so slope \( m_1=\frac{3}{5} \). The slope of the perpendicular line \( m_2 \) should satisfy \( m_1\times m_2=-1 \), so \( m_2=\frac{-1}{m_1}=\frac{-1}{\frac{3}{5}}=-\frac{5}{3} \). Now, we use the point - slope form \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(15,-5) \) and \( m =-\frac{5}{3} \). So \( y-(-5)=-\frac{5}{3}(x - 15) \), which simplifies to \( y + 5=-\frac{5}{3}x+25 \), then \( y=-\frac{5}{3}x + 20 \). Let's check the options. The fourth option is \( y =-\frac{5}{3}x + 20 \). Wait, let's verify. Plug \( x = 15 \) into the fourth option: \( y=-\frac{5}{3}\times15 + 20=-25 + 20=-5 \), which matches the point \( (15,-5) \). And the slope is \( -\frac{5}{3} \), which is the negative reciprocal of \( \frac{3}{5} \) (since \( \frac{3}{5}\times(-\frac{5}{3})=-1 \)). So the correct equation is \( y =-\frac{5}{3}x + 20 \).

Step2: Verify the equation with the point

We have the equation \( y =-\frac{5}{3}x + 20 \). Substitute \( x = 15 \) into the equation: \( y=-\frac{5}{3}\times15+20=-25 + 20=-5 \), which is the y - coordinate of the given point \( (15,-5) \). So the equation is correct.

Answer:

\( y =-\frac{5}{3}x + 20 \) (the fourth option: \( y =-\frac{5}{3}x + 20 \))