Question

which line is perpendicular to a line that has a slope of $-\frac{5}{6}$? line no line jk line lm line pq

Explanation:

Step1: Recall the slope - perpendicular relationship

If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of the given line be \(m_1=-\frac{5}{6}\), and the slope of the perpendicular line be \(m_2\). Then \(m_1\times m_2=-1\). Solving for \(m_2\), we get \(m_2=\frac{6}{5}\).

Step2: Calculate the slopes of the lines

• For line \(NO\): Using the slope formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\), if \(N(-6,-5)\) and \(O(0,0)\), then \(m_{NO}=\frac{0 - (-5)}{0-(-6)}=\frac{5}{6}\).
• For line \(JK\): If \(J(-7,1)\) and \(K(-4,-4)\), then \(m_{JK}=\frac{-4 - 1}{-4-(-7)}=\frac{-5}{3}\).
• For line \(LM\): If \(L(-5,-4)\) and \(M(0,3)\), then \(m_{LM}=\frac{3-(-4)}{0 - (-5)}=\frac{7}{5}\).
• For line \(PQ\): If \(P(-4,4)\) and \(Q(0,-2)\), then \(m_{PQ}=\frac{-2 - 4}{0-(-4)}=\frac{-6}{4}=-\frac{3}{2}\).

Answer:

line \(LM\)