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# How do you solve?

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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?

Guest Nov 12, 2017

#1
+7096
+1

the slope between two points   $$=\,\frac{\text{difference in y values}}{\text{difference in x values}}$$

the slope between  (2, 4)  and  (6, 3)   $$=\,\frac{3-4}{6-2}$$

the slope between  (2, 4)  and  (6, 3)   $$=\,-\frac{1}{4}$$

And we know that the slope between  (6, 3)  and  (-5, c)  also equals  -$$\frac14$$ .  So we know

$$-\frac14\,=\,\frac{c-3}{-5-6}\\~\\ -\frac14\,=\,\frac{c-3}{-11}$$

Multiply both sides of the equation by  -11 .

$$\frac{11}4\,=\,c-3$$

$$\frac{11}4+3\,=\,c \\~\\ c\,=\,\frac{23}{4}\,=\,5.75$$

Here is a graph to show that these points lie on the same line.

hectictar  Nov 12, 2017
#1
+7096
+1

the slope between two points   $$=\,\frac{\text{difference in y values}}{\text{difference in x values}}$$

the slope between  (2, 4)  and  (6, 3)   $$=\,\frac{3-4}{6-2}$$

the slope between  (2, 4)  and  (6, 3)   $$=\,-\frac{1}{4}$$

And we know that the slope between  (6, 3)  and  (-5, c)  also equals  -$$\frac14$$ .  So we know

$$-\frac14\,=\,\frac{c-3}{-5-6}\\~\\ -\frac14\,=\,\frac{c-3}{-11}$$

Multiply both sides of the equation by  -11 .

$$\frac{11}4\,=\,c-3$$

$$\frac{11}4+3\,=\,c \\~\\ c\,=\,\frac{23}{4}\,=\,5.75$$

Here is a graph to show that these points lie on the same line.

hectictar  Nov 12, 2017