Question

which point is on the line that passes through point r and is perpendicular to line pq? (-6,10) (-4,-8) (0,-1) (2,4)

Explanation:

Step1: Find the slope of line \(PQ\)

Let \(P(-6,4)\) and \(Q(4, - 4)\). The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).

$$m_{PQ}=\frac{-4 - 4}{4-(-6)}=\frac{-8}{10}=-\frac{4}{5}$$

Step2: Find the slope of the line perpendicular to \(PQ\)

If two lines are perpendicular, \(m_1\times m_2=-1\). Let \(m_1 =-\frac{4}{5}\), then \(m_2=\frac{5}{4}\) (since \(-\frac{4}{5}\times m_2=-1\Rightarrow m_2 = \frac{5}{4}\)). Point \(R(4,2)\)
The equation of the line using the point - slope form \(y - y_1=m(x - x_1)\) (where \(x_1 = 4,y_1=2,m=\frac{5}{4}\)) is \(y - 2=\frac{5}{4}(x - 4)\), which simplifies to \(y=\frac{5}{4}x-3\)

Step3: Check each point

• For \((-6,10)\): \(y=\frac{5}{4}x-3\), substitute \(x=-6\), \(y=\frac{5}{4}\times(-6)-3=-\frac{15}{2}-3=-\frac{15 + 6}{2}=-\frac{21}{2}

eq10\)

• For \((-4,-8)\): substitute \(x =-4\), \(y=\frac{5}{4}\times(-4)-3=-5 - 3=-8\)
• For \((0,-1)\): substitute \(x = 0\), \(y=\frac{5}{4}\times0-3=-3

eq-1\)

• For \((2,4)\): substitute \(x = 2\), \(y=\frac{5}{4}\times2-3=\frac{5}{2}-3=-\frac{1}{2}

eq4\)

Answer:

\((-4,-8)\)