Question

which line is perpendicular to a line that has a slope of $-\frac{5}{6}$?

line jk
line lm
line no
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 using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

• For line \(LM\): Let \(L(- 4,-3)\) and \(M(0,3)\). Then \(m_{LM}=\frac{3-(-3)}{0 - (-4)}=\frac{3 + 3}{0+4}=\frac{6}{4}=\frac{3}{2}\).
• For line \(NO\): Let \(N(-6,-5)\) and \(O(0,0)\). Then \(m_{NO}=\frac{0-(-5)}{0-(-6)}=\frac{0 + 5}{0+6}=\frac{5}{6}\).
• For line \(PQ\): Let \(P(-5,4)\) and \(Q(0,-2)\). Then \(m_{PQ}=\frac{-2 - 4}{0-(-5)}=\frac{-6}{5}\).
• For line \(JK\): Let \(J(-6,1)\) and \(K(0,-4)\). Then \(m_{JK}=\frac{-4 - 1}{0-(-6)}=\frac{-5}{6}\).

Answer:

line \(LM\)