Question

the triangle nop is a dilation of the triangle nop. what is the scale factor of the dilation? simplify your answer and write it as a proper fraction, an improper fraction, or a whole number.

Explanation:

Step1: Find coordinates of original and dilated points

First, identify coordinates of points in triangle \( NOP \) and \( N'O'P' \). Let's take point \( N \) and \( N' \), \( O \) and \( O' \), \( P \) and \( P' \).
- For \( O \): From graph, \( O \) is at \( (-5, 1) \)? Wait, no, looking at the grid: \( O \) seems at \( (-5, 1) \)? Wait, maybe better to take \( P \) and \( P' \). \( P \) is at \( (-5, -2) \), \( P' \) is at \( (-10, -4) \). \( O \): \( O \) is at \( (-5, 1) \), \( O' \) is at \( (-10, 2) \). \( N \): \( N \) is at \( (5, -2) \)? Wait, no, \( N \) is at \( (5, -2) \)? Wait, \( N' \) is at \( (8, -4) \)? Wait, maybe I misread. Let's re-examine:

Looking at the grid, \( P \) is at \( (-5, -2) \), \( P' \) is at \( (-10, -4) \). \( O \) is at \( (-5, 1) \), \( O' \) is at \( (-10, 2) \). \( N \): Let's see, \( N \) is at \( (5, -2) \)? Wait, \( N' \) is at \( (10, -4) \)? Wait, the x-axis: from -10 to 10, y from -10 to 10.

Wait, \( P \) is at \( (-5, -2) \), \( P' \) is at \( (-10, -4) \). So the vector from \( P \) to \( P' \): \( x \)-coordinate change: \( -10 - (-5) = -5 \), \( y \)-coordinate change: \( -4 - (-2) = -2 \). But dilation scale factor is the ratio of corresponding side lengths or distance from center. Wait, dilation center: since all lines from \( N \) to \( N' \), \( O \) to \( O' \), \( P \) to \( P' \) should pass through the center. Let's find the center of dilation. The center is the intersection of lines \( NN' \), \( OO' \), \( PP' \).

Let's take \( O \) and \( O' \): \( O(-5, 1) \), \( O'(-10, 2) \). The line through \( O \) and \( O' \): slope is \( (2 - 1)/(-10 - (-5)) = 1/(-5) = -1/5 \). Equation: \( y - 1 = -1/5(x + 5) \), so \( y = -x/5 - 1 + 1 = -x/5 \).

Take \( P(-5, -2) \) and \( P'(-10, -4) \): slope is \( (-4 - (-2))/(-10 - (-5)) = (-2)/(-5) = 2/5 \). Wait, no, that can't be. Wait, maybe I got the coordinates wrong. Let's look again:

Wait, \( P \) is at \( (-5, -2) \) (x=-5, y=-2), \( P' \) is at \( (-10, -4) \) (x=-10, y=-4). So the distance from \( P \) to origin? No, dilation scale factor is \( \frac{\text{length of } N'O'P'}{\text{length of } NOP} \). Let's take the length of \( PP' \) and \( PP \)? Wait, no, dilation scale factor is \( \frac{\text{coordinate of } P'}{\text{coordinate of } P} \) if center is origin? Wait, no, center of dilation: let's check the ratio of \( x \)-coordinates of \( P \) and \( P' \). \( P \) has \( x = -5 \), \( P' \) has \( x = -10 \). So \( -10 / -5 = 2 \)? Wait, \( y \)-coordinate: \( P \) has \( y = -2 \), \( P' \) has \( y = -4 \). \( -4 / -2 = 2 \). Similarly, \( O \): \( O \) has \( x = -5 \), \( O' \) has \( x = -10 \), \( -10 / -5 = 2 \); \( y \) of \( O \) is 1, \( y \) of \( O' \) is 2, \( 2 / 1 = 2 \). \( N \): Let's find \( N \)'s coordinates. \( N \) is at \( (5, -2) \)? Wait, \( N' \) is at \( (10, -4) \). So \( x \) of \( N \) is 5, \( x \) of \( N' \) is 10: \( 10 / 5 = 2 \); \( y \) of \( N \) is -2, \( y \) of \( N' \) is -4: \( -4 / -2 = 2 \). So the scale factor is \( 2 \)? Wait, but wait, maybe I mixed up original and dilated. Wait, the problem says \( N'O'P' \) is a dilation of \( NOP \), so \( NOP \) is original, \( N'O'P' \) is image. So scale factor \( k = \frac{\text{length of image}}{\text{length of original}} \). So for \( P \) to \( P' \): original \( P(-5, -2) \), image \( P'(-10, -4) \). So the vector from center? Wait, if we take the ratio of coordinates (assuming center is origin, but is it? Wait, the lines from \( N \) to \( N' \), \( O \) to \( O' \), \( P \) to \( P' \) should pass through the center. Let's check the line through \( O(-5,1) \) and \( O'(-10,2) \): the slope is \( (2-1)/(-10+5) = 1/(-5) = -1/5 \). Equation: \( y - 1 = -1/5(x + 5) \) → \( y = -x/5 -1 +1 = -x/5 \). So when \( x=0 \), \( y=0 \). So center is at origin (0,0). So the scale factor is the ratio of distance from center to image point over distance from center to original point.

Distance from center (0,0) to \( P(-5, -2) \): \( \sqrt{(-5)^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29} \). Distance from center to \( P'(-10, -4) \): \( \sqrt{(-10)^2 + (-4)^2} = \sqrt{100 + 16} = \sqrt{116} = 2\sqrt{29} \). So ratio is \( 2\sqrt{29}/\sqrt{29} = 2 \). Alternatively, since coordinates of \( P' \) are \( (-10, -4) = 2*(-5, -2) = 2*P \), so scale factor \( k = 2 \). Wait, but let's check \( O \): \( O(-5,1) \), \( O'(-10,2) = 2*(-5,1) = 2*O \). \( N \): Let's find \( N \)'s coordinates. From the graph, \( N \) is at \( (5, -2) \), \( N' \) is at \( (10, -4) = 2*(5, -2) = 2*N \). So yes, each coordinate of \( N'O'P' \) is 2 times that of \( NOP \). So scale factor is 2.

Wait, but maybe I made a mistake. Let's recheck. Original triangle \( NOP \): points \( N(5, -2) \), \( O(-5, 1) \), \( P(-5, -2) \). Dilated triangle \( N'O'P' \): \( N'(10, -4) \), \( O'(-10, 2) \), \( P'(-10, -4) \). So the length of \( OP \): \( O(-5,1) \) to \( P(-5,-2) \): vertical distance, \( |1 - (-2)| = 3 \). Length of \( O'P' \): \( O'(-10,2) \) to \( P'(-10,-4) \): vertical distance, \( |2 - (-4)| = 6 \). So ratio \( 6/3 = 2 \). Similarly, length of \( PN \): \( P(-5,-2) \) to \( N(5,-2) \): horizontal distance, \( |5 - (-5)| = 10 \). Length of \( P'N' \): \( P'(-10,-4) \) to \( N'(10,-4) \): horizontal distance, \( |10 - (-10)| = 20 \). Ratio \( 20/10 = 2 \). So scale factor is 2.

Step2: Confirm scale factor

Scale factor \( k = \frac{\text{length of image side}}{\text{length of original side}} \). As seen from \( OP \) and \( O'P' \), \( 6/3 = 2 \), or from coordinates (since \( P' = 2*P \), \( O' = 2*O \), \( N' = 2*N \) when center is origin), so scale factor is 2.

Answer:

2