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Question
- find the equation of a line that passes through the point (7, -4) and is perpendicular to the equation y = -7x + 12. y = \square x \square \square
Step1: Find the slope of the perpendicular line
The given line is \( y = -7x + 12 \), so its slope \( m_1 = -7 \). The slope of a line perpendicular to it, \( m_2 \), satisfies \( m_1 \times m_2 = -1 \). So \( -7 \times m_2 = -1 \), which gives \( m_2=\frac{1}{7} \).
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 \( (x_1,y_1)=(7, - 4) \) and \( m=\frac{1}{7} \). Substituting these values, we get \( y - (-4)=\frac{1}{7}(x - 7) \).
Step3: Simplify the equation
Simplify \( y + 4=\frac{1}{7}x - 1 \). Subtract 4 from both sides: \( y=\frac{1}{7}x-1 - 4 \), so \( y=\frac{1}{7}x - 5 \).
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\( y=\frac{1}{7}x - 5 \)