Points A and B are in the first quadrant; poi | vedclass

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(D) Let $OA = OB = r$. Since the slope of $OA$ is $1$,the angle $\alpha$ that $OA$ makes with the $x$-axis is $45^\circ$. Thus,the coordinates of $A$

Vedclass · Accepted Answer

$-1/2$

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(D) Let $OA = OB = r$. Since the slope of $OA$ is $1$,the angle $\alpha$ that $OA$ makes with the $x$-axis is $45^\circ$. Thus,the coordinates of $A$ are $(r \cos 45^\circ, r \sin 45^\circ) = (r/\sqrt{2}, r/\sqrt{2})$.Since the slope of $OB$ is $7$,let $\beta$ be the angle $OB$ makes with the $x$-axis. Then $	an \beta = 7$. This implies $\sin \beta = 7/\sqrt{1^2 + 7^2} = 7/\sqrt{50} = 7/(5\sqrt{2})$ and $\cos \beta = 1/\sqrt{50} = 1/(5\sqrt{2})$.Thus,the coordinates of $B$ are $(r \cos \beta, r \sin \beta) = (r/(5\sqrt{2}), 7r/(5\sqrt{2}))$.The slope of $AB$ is given by $m = \frac{y_B - y_A}{x_B - x_A} = \frac{\frac{7r}{5\sqrt{2}} - \frac{r}{\sqrt{2}}}{\frac{r}{5\sqrt{2}} - \frac{r}{\sqrt{2}}} = \frac{\frac{7r - 5r}{5\sqrt{2}}}{\frac{r - 5r}{5\sqrt{2}}} = \frac{2r}{-4r} = -\frac{1}{2}$.

Points $A$ and $B$ are in the first quadrant; point $O$ is the origin. If the slope of $OA$ is $1$,the slope of $OB$ is $7$,and $OA = OB = r$,then the slope of $AB$ is:

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