Question

line ps is perpendicular to line dn. line dn is represented by the equation y=-x - 5. line ps passes through the point p(-6,-17). determine the equation of line ps in slope - intercept form.

Explanation:

Step1: Identify slope of line DN

The equation of line DN is \(y=-x - 23\), which is in slope - intercept form \(y = mx + b\) where \(m\) is the slope. So, the slope of line DN, \(m_1=-1\).

Step2: Find slope of line PS

If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of line PS be \(m_2\). Since \(m_1\times m_2=-1\) and \(m_1 = - 1\), then \((-1)\times m_2=-1\), so \(m_2 = 1\).

Step3: Use point - slope form to find equation of line PS

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)\) is a point on the line and \(m\) is the slope. The line PS passes through the point \(P(-6,-17)\) and has a slope \(m = 1\). Substitute \(x_1=-6\), \(y_1=-17\) and \(m = 1\) into the point - slope form:

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Answer:

\(y=x - 11\)