Question

The graph of $y - x$ against $y + x$ is a line passing through the origin in the first quadrant with inclination greater than $45^\circ$. Which of the following shows the graph of $y$ against $x$?

Accepted Answer

A line with negative slope steeper than -1. Let $Y = y - x$ and $X = y + x$. The graph shows $Y = m X$ where $m = an(45^\circ + heta) > 1$. $y - x = m(y + x) \implies y(1 - m) = x(1 + m) \implies y = \frac{1 + m}{1 - m} x$. Since $m > 1$, $1 - m < 0$ and $1 + m > 0$, so slope $k = \frac{1 + m}{1 - m} < 0$. Moreover, $|k| = \frac{m + 1}{m - 1} > 1$, so the line has a steep negative slope. This corresponds to Option (4).

Coordinate Transformation Graph

Explanation

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