Question

Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, −1).

y = one thirdx − 3

y = one thirdx + 17

y = −3x − 3

y = −3x + 17

Answer 1

find the slope of the line first\[3x + y = 7\]\[y = -3x + 7\]\[m_1 = -3\]because the line is perpendicular\[m_1m_2 = -1\]\[\left( -3 \right)m_2 = -1\]\[m_2 = \frac{1}{3}\]after that use the formula to find the equation of a line that passes through a point (x1,y1) and has slope m\[y - y_1 = m(x - x_1)\]\[y - \left( -1 \right) = \frac{1}{3}\left( x - 6 \right)\]\[y + 1 = \frac{1}{3}x - 2\]\[y = \frac{1}{3}x - 3\]