## Calculus 10th Edition

The two points are $(6, -8)$ and $(-6, 8).$
$\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2)=\dfrac{d}{dx}(100)\rightarrow$ $2x+\dfrac{dy}{dx}(2y)=0\rightarrow\dfrac{dy}{dx}=-\dfrac{x}{y}.$ Slope $= \dfrac{3}{4}\rightarrow \dfrac{dy}{dx}=\dfrac{3}{4}\rightarrow-\dfrac{x}{y}=\dfrac{3}{4}\rightarrow y=-\frac{4}{3}x$ Substituting $y=-\frac{4}{3}x$ in the original equation gives: $x^2+(-\frac{4}{3}x)^2=100\rightarrow\frac{25}{9}x^2=100\rightarrow x^2=36\rightarrow x=6$ or $x=-6$ When $x=6\rightarrow y=-\frac{4}{3}(6)=-8.$ When $x=-6\rightarrow y=-\frac{4}{3}(-6)=8.$ The two points are $(6, -8)$ and $(-6, 8).$