The Coordinate Plane: Additional Practice (Set 1)

Drop from $T$ to the x-axis: it sits at $4$. Slide across to the y-axis: it sits at $-1$.

$T = (4, -1).$

Find a horizontal side from $(-1, -1)$ to $(5, -1)$ and a vertical side from $(-1, -1)$ to $(-1, 6)$.

$l = |5 - (-1)| = 6, \qquad w = |6 - (-1)| = 7.$

Then the area is

$A = 6 \times 7 = 42.$

Both coordinates are negative, the lower-left region.

$(-, -) \;\Rightarrow\; \text{Quadrant III}.$

Both points have $x = -7$, so the segment is vertical. Subtract the y-coordinates and take the absolute value.

$|2 - (-6)| = |2 + 6| = 8.$

They are $8$ units apart.

The fourth corner shares its x-coordinate with $(4, 5)$ and its y-coordinate with $(-2, 1)$.

$(\,x = 4, \; y = 1\,) \;\Rightarrow\; (4, 1).$

Moving up changes only the y-coordinate, adding $3$.

$y = -5 + 3 = -2 \;\Rightarrow\; (2, -2).$

Its y-coordinate is $0$, so it lies on the horizontal axis.

$y = 0 \;\Rightarrow\; \text{on the x-axis}.$

A point on an axis is in no quadrant.

The given side has length

$|5 - (-1)| = 6.$

A square has four equal sides, so its perimeter is

$P = 4 \times 6 = 24.$

The side lengths come from the gaps in the x-coordinates and the y-coordinates.

$l = |2 - (-3)| = 5, \qquad w = |4 - (-2)| = 6.$

Then the perimeter is

$P = 2(5 + 6) = 22.$

Distance from the y-axis is $|x|$ and distance from the x-axis is $|y|$. Quadrant I has both coordinates positive.

$|x| = 6, \; |y| = 3, \; (+, +) \;\Rightarrow\; (6, 3).$

The horizontal side has length

$|3 - (-6)| = 9,$

and the vertical side is $4$. Add a length and a width, then double:

$P = 2(9 + 4) = 2 \times 13 = 26.$

The distance from the y-axis is the horizontal distance, which is the size of the x-coordinate.

$|-8| = 8.$

The point is $8$ units from the y-axis.