+0  
 
0
443
7
avatar+272 

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

WhichWitchIsWhich  Nov 10, 2017
 #1
avatar+578 
+1

a=6

 

the fire will never go out

 

8)

OfficialBubbleTanks  Nov 10, 2017
 #2
avatar+272 
+1

wait it wasnt 6 tho :I

WhichWitchIsWhich  Nov 10, 2017
 #3
avatar+272 
+1

o its 6 or -3 nvm

WhichWitchIsWhich  Nov 10, 2017
 #4
avatar+578 
+1

hmm i dont think there is a solution cause the line just doesnt go through both points at once. sorry :P

OfficialBubbleTanks  Nov 10, 2017
 #6
avatar+7025 
0

For any 2 points, there is a line passing through it.

MaxWong  Nov 12, 2017
 #5
avatar+2266 
+2

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.

TheXSquaredFactor  Nov 10, 2017
 #7
avatar+7025 
0

Using the slope formula:

Slope = \(\dfrac{5a^2-3a^2}{(3a+4)-(2a+4)}\) = 2a

while it equals a+3.

Therefore 

a + 3 = 2a

a = 3

MaxWong  Nov 12, 2017

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