Question

1. what is the equation of a line that is perpendicular to y=x-5 and passes through (12,-12)?

options:

• ( y = (3/4)x - 21 )
• ( y = (-4/5)x + 5 )
• ( y = (-3/4)x - 3 )
• ( y = (3/4)x + 3 )

Explanation:

Step1: Find slope of given line

The line \( y = x - 5 \) is in slope - intercept form \( y=mx + b \), where \( m \) is the slope. So the slope of this line, \( m_1=1 \).
For two perpendicular lines, the product of their slopes \( m_1\times m_2=- 1 \). Let the slope of the perpendicular line be \( m_2 \). Then \( 1\times m_2=-1\), so \( m_2=-1 \)? Wait, no, wait, maybe there is a typo in the original line? Wait, the original line is \( y = x - 5 \), slope \( m = 1 \). Wait, but the options have slopes like \( \frac{3}{4},-\frac{4}{5},-\frac{3}{4},\frac{3}{4} \). Wait, maybe the original line is \( y=\frac{4}{3}x - 5 \)? Wait, the user's question: "What is the equation of a line that is perpendicular to \( y = x-5 \) and passes through \( (12,-12) \)?" Wait, maybe there is a mistake in the problem statement, or maybe I misread. Wait, if the original line is \( y=\frac{4}{3}x - 5 \), then the slope \( m_1=\frac{4}{3} \), and the slope of the perpendicular line \( m_2=-\frac{3}{4} \) (since \( m_1\times m_2=-1\)). Let's check the options. The third option has slope \( -\frac{3}{4} \). Let's use point - slope form \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(12,-12) \) and \( m = -\frac{3}{4} \).

Step2: Use point - slope form

Point - slope form: \( y-(-12)=-\frac{3}{4}(x - 12) \)
Simplify: \( y + 12=-\frac{3}{4}x+9 \)
Subtract 12 from both sides: \( y=-\frac{3}{4}x+9 - 12 \)
\( y=-\frac{3}{4}x-3 \)
Which matches the third option.

Answer:

\( y = (-\frac{3}{4})x - 3 \) (the third option, i.e., the option with \( y = (-3/4)x-3 \))