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Sagot :
[tex]1.) \\\\(-2, 1)\ \ and \ \ (6,-4) \\\\ First \ find \ the \ slope \ of \ the \ line \ thru \ the \ points \: \\ \\ m= \frac{y_{2}-y_{1}}{x_{2}-x_{1} } \\ \\m=\frac{ -4-1}{6+2} = -\frac{5}{8}[/tex]
[tex]The\ slope \ intercept \ form:y=mx+b\\\\ y= -\frac{5}{8}x+b\\\\the\ point\ (-2, 1).\\ Put\ x=-2\ and\ y=1\ to\ equation\ of\ line:\\\\1= -\frac{5}{8}\cdot (-2)+b \\1= \frac{5}{4}+b \\b=1-\frac{5}{4}=\frac{4}{4 }-\frac{5}{4}= \frac{1}{4} \\ \\ y=-\frac{5}{8}x -\frac{1}{4}[/tex]
[tex]2.) \\\\(2, 5) \ \ and \ \ m = \frac{2}{3} \\\\ The\ slope \ intercept \ form:y=mx+b\\\\ y= \frac{2}{3}x+b\\\\the\ point\ ( 2, 5).\\ Put\ x= 2\ and\ y=5\ to\ equation\ of\ line:\\\\5= \frac{2}{3}\cdot 2+b \\5= \frac{4}{3}+b \\b=5-\frac{4}{3}=\frac{15}{3 }-\frac{4}{3}= \frac{11}{3} \\ \\ y= \frac{2}{3}x +\frac{11}{3} [/tex]
[tex]The\ slope \ intercept \ form:y=mx+b\\\\ y= -\frac{5}{8}x+b\\\\the\ point\ (-2, 1).\\ Put\ x=-2\ and\ y=1\ to\ equation\ of\ line:\\\\1= -\frac{5}{8}\cdot (-2)+b \\1= \frac{5}{4}+b \\b=1-\frac{5}{4}=\frac{4}{4 }-\frac{5}{4}= \frac{1}{4} \\ \\ y=-\frac{5}{8}x -\frac{1}{4}[/tex]
[tex]2.) \\\\(2, 5) \ \ and \ \ m = \frac{2}{3} \\\\ The\ slope \ intercept \ form:y=mx+b\\\\ y= \frac{2}{3}x+b\\\\the\ point\ ( 2, 5).\\ Put\ x= 2\ and\ y=5\ to\ equation\ of\ line:\\\\5= \frac{2}{3}\cdot 2+b \\5= \frac{4}{3}+b \\b=5-\frac{4}{3}=\frac{15}{3 }-\frac{4}{3}= \frac{11}{3} \\ \\ y= \frac{2}{3}x +\frac{11}{3} [/tex]
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