Finding Distance after Reflection

Transcript

Finding Distance after Reflection We flipped ∆ ADM across the y-axis, so our figure looks like Now, here x-coordinate of all points become negative y-coordinate remains the same Let’s find the distance Let’s find the distance AD and A’D’ Distance AD Since A (3, 4) & D (7, 1) AD = √((𝟕−𝟑)^𝟐+(𝟏−𝟒)^𝟐 ) = √(4^2+(−3)^2 ) = √(𝟒^𝟐+𝟑^𝟐 ) = √(16+9) = √25 = √(5^2 ) = 5 Distance A’D’ Since A’ (–3, 4) & D’ (–7, 1) A’D’ = √((−𝟕−(−𝟑))^𝟐+(𝟏−𝟒)^𝟐 ) = √(〖(−7+3) 〗^2+(−3)^2 ) = √(〖(−4)〗^2+(−3)^2 ) = √(𝟒^𝟐+𝟑^𝟐 ) = √(16+9) = √25 = √(5^2 ) = 5 Distance A’D’ Since A’ (–3, 4) & D’ (–7, 1) A’D’ = √((−𝟕−(−𝟑))^𝟐+(𝟏−𝟒)^𝟐 ) = √(〖(−7+3) 〗^2+(−3)^2 ) = √(〖(−4)〗^2+(−3)^2 ) = √(𝟒^𝟐+𝟑^𝟐 ) = √(16+9) = √25 = √(5^2 ) = 5 We see that on reflection, length of the side is the same This is because any negative number becomes positive when you square it (e.g., (-3)2 = 9), the final distances remain completely unchanged. Thus, distance A'D' is still 5 units, just like AD. Similarly, let’s find MD, M’D’ and AM, A’M’ Distance MD Since M (9, 6) & D (7, 1) MD = √((𝟕−𝟗)^𝟐+(𝟏−𝟔)^𝟐 ) = √(〖(−2) 〗^2+(−5)^2 ) = √(𝟐^𝟐+𝟓^𝟐 ) = √(4+25) = √𝟐𝟗 Distance M’D’ Since M’ (–9, 6) & D’ (–7, 1) A’D’ = √((−𝟕−(−𝟗))^𝟐+(𝟏−𝟔)^𝟐 ) = √(〖(−7+9) 〗^2+(−5)^2 ) = √(2^2+(−5)^2 ) = √(𝟐^𝟐+𝟓^𝟐 ) = √(4+25) = √𝟐𝟗 Thus, they are the same Now, finding AM and A’M’ Distance AM Since A (3, 4) & M (9, 6) AM = √((𝟗−𝟑)^𝟐+(𝟔−𝟒)^𝟐 ) = √(𝟔^𝟐+𝟐^𝟐 ) = √(36+4) = √𝟒𝟎 Distance A’M’ Since A’ (–3, 4) & M’ (–9, 6) A’M’ = √((−𝟗−(−𝟑))^𝟐+(𝟔−𝟒)^𝟐 ) = √(〖(−9+3) 〗^2 + 2^2 ) = √(〖(−6)〗^2+ 2^2 ) = √(𝟔^𝟐+𝟐^𝟐 ) = √(36+4) = √𝟒𝟎 Thus, all 3 sides of the triangle is the same on reflection