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QUESTION IMAGE

reflect the triangle across the dashed line and enter the new coordinat…

Question

reflect the triangle across the dashed line and enter the new coordinates.
c(-2,4)
b(1,3)
a(-1,1)
a(-1,-1)
enter the number that belongs in the green box.
b(1,?)
c( , )

Explanation:

Step1: Identify the reflection line

The dashed line is the x - axis (since point A(-1,1) reflects to A'(-1,-1), the y - coordinate changes sign, so the reflection is over the x - axis). The rule for reflecting a point \((x,y)\) over the x - axis is \((x,-y)\).

Step2: Apply the reflection rule to point B

For point B(1,3), using the reflection rule over the x - axis \((x,y)\to(x,-y)\), we substitute \(x = 1\) and \(y=3\). So the new y - coordinate is \(- 3\)? Wait, no, wait. Wait, looking at point A: A(-1,1) and A'(-1,-1). Wait, the distance from A to the x - axis is 1 unit (since y = 1, distance from y = 0 is 1). The reflection A' is 1 unit below the x - axis, so y - coordinate is - 1. For point B(1,3), the distance from B to the x - axis is 3 units (y = 3). So the reflection B' should be 3 units below the x - axis. So the y - coordinate of B' is \(0 - 3=-3\)? Wait, no, wait the x - axis is y = 0. The formula for reflection over x - axis is \((x,y)\to(x, - y)\). So for B(1,3), it should be (1, - 3)? But wait, maybe the dashed line is not the x - axis. Wait, looking at point A(-1,1) and A'(-1,-1). The midpoint between A and A' is \((\frac{-1 + (-1)}{2},\frac{1+(-1)}{2})=(-1,0)\), so the reflection line is the x - axis (y = 0). Wait, but maybe I made a mistake. Wait, let's check the vertical distance. The y - coordinate of A is 1, A' is - 1. The difference is \(1-(-1) = 2\), so the distance from A to the reflection line is 1, and from A' to the reflection line is 1. So the reflection line is y = 0. So for point B(1,3), the reflection over y = 0 (x - axis) is (1, - 3)? But wait, maybe the dashed line is a horizontal line. Wait, no, the grid: let's count the units. From A(-1,1) to A'(-1,-1): the y - coordinate changes by - 2. So the reflection line is the line halfway between y = 1 and y=-1, which is y = 0. So the rule is (x,y)→(x, - y). So for B(1,3), applying the rule, we get (1, - 3)? Wait, but maybe the dashed line is y = 0. Wait, but let's re - examine. Wait, the y - coordinate of A is 1, A' is - 1. So the reflection is over the line y = 0. So the formula is correct. So B(1,3) reflected over y = 0 is (1, - 3)? Wait, but that seems odd. Wait, maybe the dashed line is not the x - axis. Wait, maybe it's a horizontal line, but let's check the vertical distance. The y - coordinate of A is 1, A' is - 1. So the distance from A to the line is 1, from A' to the line is 1. So the line is y = 0. So the reflection of (x,y) over y = 0 is (x, - y). So for B(1,3), it's (1, - 3). But wait, maybe the dashed line is y = 0. So the answer should be - 3? Wait, but let's check again. Wait, the problem says "reflect the triangle across the dashed line". From A(-1,1) to A'(-1,-1), the transformation is (x,y)→(x, - y), so the reflection is over the x - axis. So for B(1,3), the reflection is (1, - 3). But wait, maybe I misread the graph. Wait, maybe the dashed line is y = 0. So the y - coordinate of B' is - 3.

Answer:

-3