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Beauty of Coordinate geometry

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Beauty of Coordinate geometry [#permalink]

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Coordinate geometry



Mathematics is a wonderful subject where you can solve one problem in numerous ways.
Coordinate geometry simplifies the Geometry much further and makes it easy.

Here is a problem from Veritas question bank simple but wonderful to test various concepts.

Attachment:
Geometry_Img48.png
Geometry_Img48.png [ 7.7 KiB | Viewed 3717 times ]

If the x-coordinate of point E is 4, what is its y-coordinate?

    A. \(\frac{-1}{2}\)
    B. \(1\)
    C. \(\frac{3}{2}\)
    D. \(2\)
    E. \(\frac{7}{2}\)



Different Methods to solve the above problem

Let me consider

\(A (0,3)\)
\(B (6,0)\)
\(E (4,y)\)
Origin \(O(0,0)\)
Lets drop a perpendicular from E to X axis at \((4,0)\) and say it \(F.\)




1. Traditional distance method



Distance between any two points \((x_1,y_1)\) and \((x_2,y_2)\) in coordination plane is

\(\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\)

\(AE+EB=AB\)

\(\sqrt{4^2+(y-3)^2}\)+\(\sqrt{y^2+2^2}\)=\(\sqrt{6^2+3^2}\)

subtract \(\sqrt{y^2+2^2}\) on both sides

\(\sqrt{y^2-6y+25}\)=\(\sqrt{45}\)-\(\sqrt{y^2+4}\)

squaring on both sides

\(y^2-6y+25=45+y^2+4-2\sqrt{45}\sqrt{y^2+4}\)

After cancelling similar terms like y^2 and simplifying the terms, Equation looks like

\(3y+12=\sqrt{45*(y^2+4)}\)

squaring on both sides again, we get

\(9(y^2+16+8y)=45(y^2+4)\)

on bringing all terms to one side and simplification, we get \((y-1)^2=0\)

i.e., \(y=1\)

This method is quite time consuming and cumbersome.




2. Form an equation and Substitute the value



Two Intercept Form for the Equation of a Line

\(\frac{x}{a}+\frac{y}{b}=1\), where a is the x-intercept and b is the y-intercept.

with points A and B, make an equation for line AB

\(\frac{x}{6}+\frac{y}{3}=1\)

Since point E lies on line AB, substitute \(E(4,y)\) in above equation

\(\frac{4}{6}+\frac{y}{3}=1\)

solving for y, we get \(y=1\)


Equation of line can also be formed with two points A(x1,y1), B(x2,y2) and slope of line AB m.
It can be represented as

\(m=\frac{(y_2-y_1)}{(x_2-x_1)}\)

equation of line is \(y-y_1=m(x-x_1)\) or \(y-y_2=m(x-x_2)\)

now we get \(y=1\) as we substitute \(E(4,y)\) as above.




3. Midpoint method



Midpoint between (0,3) and (6,0) is (3, (3/2) = (3, 1.5).

Because point E is below the midpoint with an x-coordinate of 4 and the y-coordinate had to be less than 1.5.
The only answer choice below 1.5 yet above 0 is 1 as in option B.




4. Slope method



Equate the slope of Line AE and EB since both the lines Coincide.

\(\frac{(y-3)}{(4-0)}=\frac{(0-y)}{(6-4)}\)

we get solve for y now.




5. Trigonometry



GMAT doesnt require you to know trigonometric functions and values of 0, 30,45,60,90 degree angles but it is always good to know
Those who are not aware of trigonometry, please refer to overview section of below link.
https://en.wikipedia.org/wiki/Trigonometry

Now Lets drop a perpendicular from E to X axis at (4,0) and say it F. It appears as in fig.
Attachment:
New Doc 5_1.jpg
New Doc 5_1.jpg [ 18.68 KiB | Viewed 3679 times ]


\(tan∠ABO=tan∠EBF\)

\(tan(θ)\)=opposite side/adjacent side

In OAB, \(tan∠ABO=\frac{OA}{OB}\)
In EBF, \(tan∠EBF=\frac{EF}{FB}\)

\(\frac{OA}{OB}\)=\(\frac{EF}{FB}\)

i.e., \(\frac{3}{6}\)=\(\frac{y}{(6-4)}\)

now we can solve for y.




6. Similar triangles



OAB and EFB are similar triangles for EF is parallel to OA.

Triangle Proportionality Theorem.
If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

\(\frac{BE}{BA}=\frac{BF}{BO}\)=\(\frac{EF}{OA}\)

to solve in simple terms, we take

\(\frac{BF}{BO}\)=\(\frac{EF}{OA}\)

\(\frac{2}{6}=\frac{y}{3}\) and hence \(y=1\).


:) Mathematics may not teach us how to add love or how to minus hate. But it gives every reason to hope that every problem has not one but multiple solutions. :)

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Re: Beauty of Coordinate geometry [#permalink]

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Re: Beauty of Coordinate geometry [#permalink]

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New post 15 Jul 2017, 02:48
I have a question, how useful each method is in solving a gmat question?
It is because these methods seem to prove your creativity rather than gmat skills.

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Re: Beauty of Coordinate geometry   [#permalink] 15 Jul 2017, 02:48
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