Question #07602

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Answer 1 · 2017-03-31T22:10:27

The point is $(- 3, - 4)$ $(-3,-4)$

Given: $sec (θ) = - \frac{5}{3}$ $sec(theta) = -5/3$

We know that $sec (θ) = \frac{r}{x}$ $sec(theta) = r/x$

We know that r cannot be negative, therefore, $x = - 3$ $x = -3$ and $r = 5$ $r = 5$

$r = \sqrt{x^{2} + y^{2}}$ $r = sqrt(x^2+y^2)$

$5 = \sqrt{- 3^{2} + y^{2}}$ $5 = sqrt(-3^2+ y^2)$

$25 = 9 + y^{2}$ $25 = 9+ y^2$

$y^{2} = 16$ $y^2 = 16$

$y = \pm 4$ $y = +-4$

I quadrant 3, both x and y are negative so we choose the negative:

$y = - 4$ $y = -4$

The point is $(- 3, - 4)$ $(-3,-4)$