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.
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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 problemLet 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/TrigonometryNow Lets drop a perpendicular from E to X axis at (4,0) and say it F. It appears as in fig.
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\(tan∠ABO=tan∠EBF\)
\(tan(θ)\)=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.