Question

1. are lines h and j perpendicular? check answer

Explanation:

Step1: Find slope of line h

Line h passes through (0, 4) and (-2, 0). Slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$. So $m_h = \frac{0 - 4}{-2 - 0} = \frac{-4}{-2} = 2$.

Step2: Find slope of line j

Line j passes through (0, 4) and (4, 0). Using slope formula: $m_j = \frac{0 - 4}{4 - 0} = \frac{-4}{4} = -1$? Wait, no, wait. Wait, line j: let's check points. Wait, maybe I misread. Wait, line j: looking at the graph, line j goes from, say, (0,4) to (4, something? Wait, no, the grid: let's take two points on line j. Let's see, line j is a red line. Let's pick (0,4) and (4, 0)? No, wait, maybe ( -4, 0) and (0,4)? Wait, no, the line h: let's re-examine. Wait, line h: from (0,4) to (-2, 0)? Wait, no, maybe (0,4) and (-1, 2)? Wait, no, the grid is with x from -4 to 4, y from -4 to 4. Let's correctly identify points.

Wait, line h: let's take two points. Let's say line h passes through (0, 4) and (-2, 0). So change in y: 0 - 4 = -4, change in x: -2 - 0 = -2. So slope $m_h = \frac{-4}{-2} = 2$.

Line j: let's take two points. Let's say line j passes through (0, 4) and (4, 0). Change in y: 0 - 4 = -4, change in x: 4 - 0 = 4. So slope $m_j = \frac{-4}{4} = -1$? Wait, no, that can't be. Wait, maybe I got the lines wrong. Wait, the line j: looking at the graph, line j is a red line going from left to right, passing through (0,4) and (4, 0)? No, wait, maybe ( -4, 0) and (0,4)? Wait, no, the slope of perpendicular lines should be negative reciprocals. Wait, maybe I made a mistake. Let's try again.

Wait, line h: let's take points (0, 4) and (-1, 2). Then change in y: 2 - 4 = -2, change in x: -1 - 0 = -1. So slope $m_h = \frac{-2}{-1} = 2$.

Line j: take points (0, 4) and (2, 2). Change in y: 2 - 4 = -2, change in x: 2 - 0 = 2. So slope $m_j = \frac{-2}{2} = -1$? No, that's not right. Wait, maybe the correct points: let's look at the graph again. The line h: from (0,4) to (-2, 0) – so when x decreases by 2, y decreases by 4, so slope 2. Line j: from (0,4) to (4, 0) – x increases by 4, y decreases by 4, slope -1. Wait, but 2 and -1 are not negative reciprocals. Wait, maybe I messed up the lines. Wait, maybe line h is the steeper one. Wait, maybe line h passes through (0, -4) and (-2, 0)? No, the y-intercept of h is 4? Wait, the graph: line h has a y-intercept at 4, going down to the left. Line j has a y-intercept at 4, going down to the right. Wait, maybe the correct slopes: let's calculate slope of h: points (0,4) and (-2, 0). Slope: (0-4)/(-2-0) = (-4)/(-2) = 2. Slope of j: points (0,4) and (4, 0). Slope: (0-4)/(4-0) = (-4)/4 = -1. Wait, but 2 and -1 are not negative reciprocals. Wait, maybe I got the lines wrong. Wait, maybe line h is the one with y-intercept -4? No, the graph shows line h crossing y-axis at 4. Wait, maybe the other line: line k? No, the question is about h and j. Wait, maybe I made a mistake in points. Let's try again. Let's take line h: two points: (-2, 0) and (0, 4). So slope is (4 - 0)/(0 - (-2)) = 4/2 = 2. Line j: two points: (0, 4) and (4, 0). Slope is (0 - 4)/(4 - 0) = -4/4 = -1. Wait, but 2 and -1: the product is -2, not -1. So they are not perpendicular. Wait, but maybe I picked the wrong points. Let's check another pair for line j. Let's take (2, 2) and (4, 0). Slope: (0 - 2)/(4 - 2) = -2/2 = -1. Same as before. For line h: (-1, 2) and (0, 4). Slope: (4 - 2)/(0 - (-1)) = 2/1 = 2. So slope of h is 2, slope of j is -1. Product is -2 ≠ -1. So they are not perpendicular. Wait, but maybe I misread the lines. Wait, maybe line h is the one with negative slope? Wait, no, line h is going up to the left? No, going from (0,4) to (-2, 0) – as x decreases, y decreases, so slope is positive. Line j is going from (0,4) to (4, 0) – as x increases, y decreases, slope negative. Wait, maybe the correct answer is no.

Answer:

No, lines h and j are not perpendicular because the product of their slopes (2 and -1) is -2, not -1 (which is required for perpendicular lines).