Question

determine the equation, in slope - intercept form, of the line sq that is perpendicular to line rj and passes through the point s (7, 3).
slope of line bj: ( m_1 )
slope of line sq: ( m_2 )
point - slope form of line sq: ( y - y_1 = m(x - x_1) )

Explanation:

Step1: Find slope of Line B (let's assume Line B is \( y = x + 10 \))

The equation \( y = x + 10 \) is in slope - intercept form \( y=mx + b \), where \( m \) is the slope. So, the slope of Line B (\( m_1 \)) is \( 1 \).

Step2: Find slope of perpendicular line SQ

If two lines are perpendicular, the product of their slopes is \( - 1 \). Let the slope of Line SQ be \( m_2 \). We know that \( m_1\times m_2=-1 \). Since \( m_1 = 1 \), then \( 1\times m_2=-1 \), so \( m_2=-1 \).

Step3: Write point - slope form of Line SQ

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(7,3) \) and \( m = m_2=-1 \). Substituting these values, we get \( y - 3=-1(x - 7) \).

Answer:

• Slope of Line B (\( m_1 \)): \( 1 \)
• Slope of Line SQ (\( m_2 \)): \( - 1 \)
• Point - Slope Form of Line SQ: \( y - 3=-1(x - 7) \)