Which of the following is an equation of the line perpendicular to 3x + 6y = 12 and passing through (4, 0)? *help please*

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Which of the following is an equation of the line perpendicular to 3x + 6y = 12 and passing through (4, 0)? *help please*

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Rearrange equation into standard form by dividing through by 6 so 3x + 6y = 12 is now written 1/2x + y = 2 Now make y the subject y = 2 -1/2x A perpendicular to this line will have a gradient -2 (inverse negative of original gradient) so the line will be y = -2x + c (we now have to find the constant) Put your coordinate values into this equation. 0 = -2(4) + c 0 = -8 + c Therefore c = 8 (adding 8 to both sides) Answer y = 8 -2x or 2x +y -8 = 0
thanks darling your AMAZING (:
Bailey I noticed an error on my previous post. All is ok up to here... Please take care to look at the changes from -2 to 2 A perpendicular to this line will have a gradient 2 (inverse negative of original gradient) so the line will be y = 2x + c (we now have to find the constant) Put your coordinate values into this equation. 0 = 2(4) + c 0 = 8 + c Therefore c = -8 (subtracting 8 to both sides) Answer y = 2x - 8 Sorry for this silly mistake, I hope it is not to late for you.

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