Question

y3 + y2-y-1 please answer me
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Answer 1

Answer:

, y1), and pt2, given by (x2, y2), how can I locate a third point

pt3, given by (x3, y3), which is perpendicular to the line joining pt1

and pt2 and is of certain distance d from either pt1 or pt2?

I've established two equations with two unknowns as following:

m = (y2-y1)/(x2-x1), which implies (y3-y1)/(x3-x1) = -1/m (eqn. 1)

(y3-y1)^2 + (x3-x1)^2 = d^2 (eqn. 2)

Since x1, y1, x2, y2, and d are known, I can solve for x3 and y3

variables using either the substitution or the elimination method.

There will also be two values for both x3 and y3, since the equation

becomes a quadratic equation.

To simplify the whole thing, is there a standard equation I can use to

solve x3 and y3?

Your response will be greatly

Answer 2

$\huge\boxed{\underline{\underline{\green{\tt Solution}}}}$

${y}^{3} + {y}^{2} - y - 1$

$= {y}^{2} (y + 1) - 1(y + 1)$

$= ( {y}^{2} - 1)(y + 1)$

$= (y - 1)(y + 1)(y + 1)$

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