+0

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.

0
5
542
7
+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

Nov 10, 2017

#1
+578
+1

a=6

the fire will never go out

8)

Nov 10, 2017
#2
+272
+1

wait it wasnt 6 tho :I

WhichWitchIsWhich  Nov 10, 2017
#3
+272
+1

o its 6 or -3 nvm

WhichWitchIsWhich  Nov 10, 2017
#4
+578
+1

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

Nov 10, 2017
#6
+7220
0

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

MaxWong  Nov 12, 2017
#5
+2339
+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.

Nov 10, 2017
#7
+7220
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

Nov 12, 2017