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AMC 10 Math Olympiad USA Math Olympiad

Sum of Co-ordinates | AMC-10A, 2014 | Problem 21

Try this beautiful sum of Co-ordinates based on co-ordinate Geometry from AMC-10A, 2014. You may use sequential hints to solve the problem.

Try this beautiful problem from Algebra based on Sum of Co-ordinates

Sum of Co-ordinates – AMC-10A, 2014- Problem 21


Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$-axis at the same point. What is the sum of all possible $x$-coordinates of these points of intersection?

  • \(16\)
  • \(12\)
  • \(-8\)
  • \(-4\)

Key Concepts


Geometry

Co-ordinate

Check the Answer


But try the problem first…

Answer: \(-8\)

Source
Suggested Reading

AMC-10A (2014) Problem 21

Pre College Mathematics

Try with Hints


First hint

The given equations are $y=ax+5$\(\Rightarrow x=\frac{-5}{a}\)…..(1)

$y=3x+b$\(\Rightarrow x=\frac{-b}{3}\)…………(2)

Since two lines intersect the $x$-axis at the same point,then at first we have to find out the common point on x -axis………

Now the intercept form of the given two equations will be

\(\frac{x}{(-5/a)} +\frac{y}{5}=1\) ,Therefore the straight line intersect x-axis at the point (\(\frac{-5}{a},0\))

\(\frac{x}{(-b/3)}+\frac{y}{b}=1\) Therefore the straight line intersect x-axis at the point (\(\frac{-b}{3},0)\).

Can you now finish the problem ……….

Second Hint

Since two lines intersect the $x$-axis at the same point so we may say that

(\(\frac{-5}{a},0\))=(\(\frac{-b}{3},0)\)

\(\Rightarrow ab=15\)

The only possible pair (a,b) will be \((1,15),(3,5),(5,3),(15,1)\)

can you finish the problem……..

Final Step

Now if we put the values \((1,15),(3,5),(5,3),(15,1)\) in (1) & (2) we will get \(-5\),\(\frac{-5}{3}\),\(-1\),\(\frac{1}{3}\)

Therefore the sun will be \(-8\)

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