Coordinate Transformation Graph

CAT 2006 Slot 1 · Quantitative Ability · Medium · Quantitative Aptitude

This is a medium Quantitative Ability question from the CAT 2006 Slot 1 paper. It tests Quantitative Aptitude. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

The graph of yxy - x against y+xy + x is a line passing through the origin in the first quadrant with inclination greater than 4545^\circ. Which of the following shows the graph of yy against xx?

  1. A.

    A horizontal line below x-axis

  2. B.

    A line passing through origin in quadrant I & III

  3. C.

    A line with small positive slope

  4. D.

    A line with negative slope steeper than -1

  5. E.

    A line with small negative slope

Answer

A line with negative slope steeper than -1

Explanation

Let Y=yxY = y - x and X=y+xX = y + x. The graph shows Y=mXY = m X where m=tan(45+θ)>1m = \tan(45^\circ + \theta) > 1. yx=m(y+x)    y(1m)=x(1+m)    y=1+m1mxy - x = m(y + x) \implies y(1 - m) = x(1 + m) \implies y = \frac{1 + m}{1 - m} x. Since m>1m > 1, 1m<01 - m < 0 and 1+m>01 + m > 0, so slope k=1+m1m<0k = \frac{1 + m}{1 - m} < 0. Moreover, k=m+1m1>1|k| = \frac{m + 1}{m - 1} > 1, so the line has a steep negative slope. This corresponds to Option (4).

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