Question

which line is perpendicular to a line that has a slope of (\frac{1}{2})? (\bigcirc) line ab (\bigcirc) line cd (\bigcirc) line fg (\bigcirc) line hj

Explanation:

Step1: Recall perpendicular slope rule

For two lines to be perpendicular, the product of their slopes is -1. If a line has slope \( m_1=\frac{1}{2} \), the slope \( m_2 \) of the perpendicular line satisfies \( m_1\times m_2=-1 \). So \( m_2 = - \frac{1}{m_1}=-2 \)? Wait, no: \( \frac{1}{2}\times m_2=-1 \implies m_2=-2 \)? Wait, no, wait: the correct formula is that if two lines are perpendicular, their slopes are negative reciprocals. So slope of perpendicular line to \( \frac{1}{2} \) is \( -2 \)? Wait, no: negative reciprocal of \( \frac{1}{2} \) is \( -2 \)? Wait, no: reciprocal of \( \frac{1}{2} \) is 2, negative of that is -2. Wait, but let's check the lines.

Wait, maybe I made a mistake. Wait, the negative reciprocal of \( \frac{1}{2} \) is \( -2 \)? Wait, no: \( \frac{1}{2} \times (-2) = -1 \), yes. But let's check the lines. Let's find the slope of each line.

First, line HJ: points H and J. From the graph, H is at (-2, 3)? Wait, no, looking at the grid: H is at (-1, 3)? Wait, the grid: x-axis from -5 to 5, y-axis from -5 to 5. Let's find coordinates:

Line HJ: H is (-1, 3), J is (1, -1)? Wait, no, J is at (1, -1)? Wait, no, looking at the graph: H is at (-1, 3), J is at (1, -1)? Wait, let's calculate slope of HJ: \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{1 - (-1)} = \frac{-4}{2} = -2 \).

Line AB: points A and B. A is (-5, -2), B is (4, 2)? Wait, no, A is (-5, -2), B is (4, 2)? Wait, slope of AB: \( \frac{2 - (-2)}{4 - (-5)} = \frac{4}{9} \)? No, maybe I misread. Wait, the red line AB: let's take two points. A is (-5, -2), B is (4, 2)? No, maybe A is (-5, -2) and B is (4, 2)? Wait, no, the slope of AB: let's see, from A (-5, -2) to B (4, 2): \( \frac{2 - (-2)}{4 - (-5)} = \frac{4}{9} \), not -2.

Line CD: points C (-4, 0) and D (4, -4). Slope: \( \frac{-4 - 0}{4 - (-4)} = \frac{-4}{8} = -\frac{1}{2} \).

Line FG: points F (-3, -3) and G (0, 3). Slope: \( \frac{3 - (-3)}{0 - (-3)} = \frac{6}{3} = 2 \). Wait, no, G is at (0, 3)? Wait, the blue line FG: F is (-3, -3), G is (0, 3). So slope is \( \frac{3 - (-3)}{0 - (-3)} = \frac{6}{3} = 2 \).

Wait, the original line has slope \( \frac{1}{2} \). The negative reciprocal of \( \frac{1}{2} \) is \( -2 \), because \( \frac{1}{2} \times (-2) = -1 \). So line HJ has slope -2, which is the negative reciprocal of \( \frac{1}{2} \), so they are perpendicular.

Step2: Confirm with slope formula

For two lines to be perpendicular, \( m_1 \times m_2 = -1 \). Given \( m_1 = \frac{1}{2} \), we need \( m_2 = -2 \) (since \( \frac{1}{2} \times (-2) = -1 \)). Calculating slope of HJ: using points H (-1, 3) and J (1, -1), slope \( m = \frac{-1 - 3}{1 - (-1)} = \frac{-4}{2} = -2 \). Thus, line HJ has slope -2, which is perpendicular to the line with slope \( \frac{1}{2} \).

Answer:

line HJ