Find the value of x so that the line passing through the points will have the given slope: (x,6) and (10,-3) with a slope of -3/2.

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[tex](x,6) \\ x_1=x \\ y_1=6 \\ \\ (10,-3) \\ x_2=10 \\ y_2=-3 \\ \\ \hbox{the slope: } m=\frac{y_2-y_1}{x_2-x_1}=\frac{-3-6}{10-x}=\frac{-9}{10-x} \\ \\ m=-\frac{3}{2} \\ \Downarrow \\ -\frac{3}{2}=\frac{-9}{10-x} \\ -\frac{3}{2}(10-x)=-9 \\ 10-x=-9 \times (-\frac{2}{3}) \\ 10-x=3 \times 2 \\ 10-x=6 \\ -x=6-10 \\ -x=-4 \\ \boxed{x=4}[/tex]