Geometry Problem 2

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Here’s and example of a SMART MATH problem for GEOMETRY.

Geometry Problems

Geometry Problem 2

The coordinates of P, Q and R are \left( \frac{2}{3},\frac{3}{2} \right), (1, –3) and (xy) respectively. If R is the midpoint of PQ, find the values of and y.

  1. \frac{-5}{6},\frac{-3}{4}
  2. \frac{-3}{4},\frac{5}{3}
  3. \frac{5}{6},\frac{-3}{4}
  4. \frac{3}{4},\frac{5}{3}
  5. \frac{3}{4},\frac{-5}{3}

The Usual Way

Using the midpoint formula:




Hence, answer is \frac{5}{6},\frac{-3}{4}

(Ans: 3)

Estimated Time to arrive at the answer = 45 seconds.

The Smart Way

Since the y coordinate of point Q is negative with a value greater than that of point P (i.e. 3 > \frac{3}{2}, the sum of these ( -3+\frac{3}{2}) will be negative. Thus the y coordinate of the answer should be negative and the x coordinate positive. This eliminates options ‘1’, ‘2’ and ‘4’.

Now looking at the coordinates of P and Q, we observe that the ‘x’ and ‘y’ coordinates of P are fractions with denominators 3 and 2 respectively, whereas the coordinates of Q are integers.

Hence, half of the sum of ‘x’ coordinates of P and Q would yield a fraction with denominator 3 x 2 = 6.

Similarly, half of the sum of y coordinates of P and Q would yield a fraction with denominator 2 x 2 = 4.

The only option that satisfies this condition is option ‘3’.

(Ans: 3)

Estimated Time to arrive at the answer = 10 seconds.

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