Three points are on the same line if the slope of the line through the first two points is the same as the slope of the line through the second two points. For what value of c are the three points (2,4), (6,3), and (-5,c) on the same line?
+1904 To solve this problem, first find the slope of the first two points.
\(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\)
\(m=slope\)
\({y}_{2}=\)y-point in second point
\({y}_{1}=\)y-point in first point
\({x}_{2}=\)x-point in second point
\({x}_{1}=\)x-point in first point
\({y}_{2}=3\)
\({y}_{1}=4\)
\({x}_{2}=6\)
\({x}_{1}=2\)
\(m=\frac{3-4}{6-2}\)
\(m=\frac{-1}{6-2}\)
\(m=\frac{-1}{4}\)
\(m=-\frac{1}{4}\)
Next, do the same for the second and third points and solve for c.
\(m=-\frac{1}{4}\)
\({y}_{2}=c\)
\({y}_{1}=3\)
\({x}_{2}=-5\)
\({x}_{1}=6\)
\(-\frac{1}{4}=\frac{c-3}{-5-6}\)
\(-\frac{1}{4}=\frac{c-3}{-11}\)
\(-\frac{1}{4}\times-11=\frac{c-3}{-11}\times-11\)
\(\frac{11}{4}=\frac{c-3}{-11}\times-11\)
\(\frac{11}{4}=c-3\)
\(\frac{11}{4}+3=c-3+3\)
\(\frac{11}{4}+\frac{12}{4}=c-3+3\)
\(\frac{11+12}{4}=c-3+3\)
\(\frac{23}{4}=c-3+3\)
\(5.75=c-3+3\)
\(5.75=c-0\)
\(5.75=c\)
\(c=5.75\)
+1904 To solve this problem, first find the slope of the first two points.
\(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\)
\(m=slope\)
\({y}_{2}=\)y-point in second point
\({y}_{1}=\)y-point in first point
\({x}_{2}=\)x-point in second point
\({x}_{1}=\)x-point in first point
\({y}_{2}=3\)
\({y}_{1}=4\)
\({x}_{2}=6\)
\({x}_{1}=2\)
\(m=\frac{3-4}{6-2}\)
\(m=\frac{-1}{6-2}\)
\(m=\frac{-1}{4}\)
\(m=-\frac{1}{4}\)
Next, do the same for the second and third points and solve for c.
\(m=-\frac{1}{4}\)
\({y}_{2}=c\)
\({y}_{1}=3\)
\({x}_{2}=-5\)
\({x}_{1}=6\)
\(-\frac{1}{4}=\frac{c-3}{-5-6}\)
\(-\frac{1}{4}=\frac{c-3}{-11}\)
\(-\frac{1}{4}\times-11=\frac{c-3}{-11}\times-11\)
\(\frac{11}{4}=\frac{c-3}{-11}\times-11\)
\(\frac{11}{4}=c-3\)
\(\frac{11}{4}+3=c-3+3\)
\(\frac{11}{4}+\frac{12}{4}=c-3+3\)
\(\frac{11+12}{4}=c-3+3\)
\(\frac{23}{4}=c-3+3\)
\(5.75=c-3+3\)
\(5.75=c-0\)
\(5.75=c\)
\(c=5.75\)
