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secondary 4 | E Maths
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secondary 4 chevron_right E Maths chevron_right Singapore

Someone please help me for part D thanks in advance

Date Posted: 11 months ago
Views: 49

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Reflected along y-axis. So y axis is like a mirror plane, and your reflected line is the mirror image of line l1. Point D (where x = 0) is on y axis so y intercept remains the same. Other points on the line are now on the negative portion of the x-axis so the x coordinates are negative . This causes your gradient to become the negative of the original. And negative of -2 is just 2. Since it was y = -2x + 10, the reflected line would be y = 2x + 10
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