+267 A line passing through the distinct points (2a+4,3a^2) and (3a+4,5a^2) has slope a+3. Find the value of a.
BubbleTanks you're on fire today :D
+577 hmm i dont think there is a solution cause the line just doesnt go through both points at once. sorry :P
+2446 I believe I have gotten the answer to this riddle! Using the slope formula, one can find the value for a, if a solution exists.
\(m=\frac{y_2-y_1}{x_2-x_1}\)
Just plug in the appropriate values for the x- and y-corrdinates and solve.
| \(\frac{5a^2-3a^2}{3a+4-(2a+4)}=a+3\) | Now, solve for a. First, distribute the negation to all the terms inside the parentheses. |
| \(\frac{5a^2-3a^2}{3a+4-2a-4}=a+3\) | Simplify the numerator and denominator in the fraction because both happen to have like terms. |
| \(\frac{2a^2}{a}=a+3\) | Both the numerator and denominator can cancel out an "a." |
| \(2a=a+3\) | Subtract a from both sides. |
| \(a=3\) | Wow! There is actually a solution! |
In case you are wondering, this is correct. Click here to view the corresponding graph.
