The table below gives the equations of six lines - Junior Cycle Mathematics - Question Question 1 - 2012

Lines are parallel if they have the same slope. Here, Line 1 and Line 2 both have a slope of 3. Therefore, we can conclude that:

• Line 1: $y = 3x - 6$ (slope = 3)
• Line 2: $y = 3x + 12$ (slope = 3)

Thus, Line 1 and Line 2 are parallel.

To sketch Line 1 ($y = 3x - 6$), we can plot two points:

1. When $x = 0$, $y = -6$ (point $(0,-6)$)
2. When $x = 2$, $y = 0$ (point $(2,0)$)

Then connect these points to form a straight line.

Examining the slope in the diagram:

• The slope is negative and the y-intercept is 4. Comparing these properties, we find that it represents Line 5, which is given by $y = -2x + 4$.

To find which equation satisfies the values:

1. For $x = 7$, check: $y = 4x - 16 ightarrow y = 4(7) - 16 = 28 - 16 = 12$ (satisfies Line 6)
2. For $x = 9$, check: $y = 4(9) - 16 = 36 - 16 = 20$ (satisfies Line 6)
3. For $x = 10$, check: $y = 4(10) - 16 = 40 - 16 = 24$ (satisfies Line 6)

Thus, all points satisfy Line 6: $y = 4x - 16$.

Set the equations equal to each other:

For Line 4: $y = x - 7$ For Line 6: $y = 4x - 16$

Setting them equal:

$x - 7 = 4x - 16$

Rearranging gives:

$16 - 7 = 4x - x$ $9 = 3x$ $x = 3$

Now substituting $x = 3$ to find $y$:

$y = 3 - 7 = -4$

So, the value of $x$ is 3 and the corresponding value of $y$ is -4.

To verify: For Line 4: when $x = 3$, $y = 3 - 7 = -4$. For Line 6: when $x = 3$, $y = 4(3) - 16 = 12 - 16 = -4$. Both equations give the same value of $y = -4$ when $x = 3$, confirming the solution.