Draw the graphs of the following sets of lines. In each case, reflect on the role of $a$ and $b$.

(A) $y = 4x$, $y = 2x$, $y = x$

(B) $y = -6x$, $y = -3x$, $y = -x$

(C) $y = 5x$, $y = -5x$

(D) $y = 3x - 1$, $y = 3x$, $y = 3x + 1$

$(E) \ y = -2x - 3, \ y = -2x, \ y = 2x + 3$ [Page No. 36]

(A) y = 4x, y = 2x, y = x

All the lines have b = 0 and positive slopes; as the value of a increases, the line becomes steeper.

(B) $y = -6x, y = -3x, y = -x$

Here, $b = 0$ and the slopes are negative, so each line decreases from left to right. As $|a|$ increases, the line becomes steeper in the downward direction.

(C) $y = 5x, y = -5x$

Both pass through origin. $y = 5x$ rises steeply; $y = -5x$ falls steeply. They are mirror images about the $x$ -axis (and also about the $y$ -axis).

(D) $y = 3x - 1, y = 3x, y = 3x + 1$

Same slope $3 \Rightarrow$ all three lines are parallel. $y$ -intercepts are $-1, 0, 1$ respectively.

Plot points:

$$y = 3x - 1: (0, -1), (1, 2)$$

$$y = 3x: (0, 0), (1, 3)$$

$$y = 3x + 1: (0, 1), (1, 4)$$

(E) $y = -2x - 3, y = -2x, y = 2x + 3$

$y = -2x - 3$ and $y = -2x$ are parallel (both slope $-2$). $y = 2x + 3$ has slope 2 (opposite sign), so it is NOT parallel to the others.

Plot points:

$$y = -2x - 3: (0, -3), (1, -5)$$

$$y = -2x: (0, 0), (1, -2)$$

$$y = 2x + 3: (0, 3), (1, 5)$$Concept Applied
In y = ax + b, ‘a’ is the slope (which decides steepness and direction) and ‘b’ is the y- intercept (where the line crosses the y-axis).