Perimeter & Area on the Coordinate Plane

Q: When do I need the distance formula?

Only for slanted segments. If two vertices share a $y$-coordinate, the segment is horizontal — subtract the $x$-coordinates. If they share an $x$-coordinate, it's vertical — subtract the $y$-coordinates. Save $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for everything else.

When a figure sits on a coordinate grid, nobody hands you the side lengths — you find them from the vertices. Horizontal and vertical sides can be counted or subtracted directly. Diagonal sides need the distance formula. Once every side length is known, perimeter and area work exactly the way they always do.

This is really two skills glued together: reading lengths off a grid, and the area and perimeter formulas you already know. The grid part is what's new, so that's where to focus.

Horizontal and vertical lengths: subtract coordinates

For a horizontal segment, the $y$ -coordinates match, so the length is the difference of the $x$ -coordinates: from $(-2, 3)$ to $(4, 3)$ is $4 - (-2) = 6$ units. For a vertical segment, subtract the $y$ -coordinates instead.

You can also just count grid squares, but subtracting is faster and safer once the numbers get big or negative. Length is always positive — if the subtraction comes out negative, take the absolute value.

Diagonal lengths: the distance formula

A slanted side can't be counted square by square. Use the distance formula, which is the Pythagorean theorem in coordinate form: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

From $(1, 2)$ to $(4, 6)$ : the run is $4 - 1 = 3$ and the rise is $6 - 2 = 4$ , so $d = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ . Every diagonal side of a polygon gets this treatment before you can add up the perimeter.

Putting it together

Perimeter: find every side length, then add. Area: for rectangles and other familiar shapes, find the base and height from the coordinates and use the usual formula. The base and height of a figure on the grid are almost always a horizontal length and a vertical length.

For a triangle with a horizontal base, the height is the vertical distance from the base up to the third vertex — subtract $y$ -coordinates, even though the slanted sides are what you see.

Worked examples

Example 1: rectangle from vertices

A rectangle has vertices $(0, 0)$ , $(5, 0)$ , $(5, 3)$ , and $(0, 3)$ . Find its perimeter and area.

Length along the bottom: subtract

x

-coordinates

\ell = 5 - 0 = 5

Width along the side: subtract

y

-coordinates

w = 3 - 0 = 3

Perimeter

P = 2(5) + 2(3) = 16

Area

A = (5)(3) = 15

Answer: $P = 16$ units, $A = 15$ square units

Example 2: a diagonal side

Find the length of the segment from $(-1, 2)$ to $(5, 10)$ .

Start with the distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Substitute the coordinates

d = \sqrt{(5 - (-1))^2 + (10 - 2)^2}

Simplify inside

d = \sqrt{6^2 + 8^2} = \sqrt{100}

Take the square root

d = 10

Answer: $d = 10$ units

Example 3: triangle area from coordinates

A triangle has vertices $(0, 0)$ , $(8, 0)$ , and $(3, 5)$ . Find its area.

The base is horizontal

b = 8 - 0 = 8

The height is the vertical distance to the third vertex

h = 5 - 0 = 5

Apply the triangle area formula

A = \dfrac{1}{2}(8)(5) = 20

Answer: $A = 20$ square units

Try one yourself

Common questions

When do I need the distance formula?

Only for slanted segments. If two vertices share a $y$ -coordinate, the segment is horizontal — subtract the $x$ -coordinates. If they share an $x$ -coordinate, it's vertical — subtract the $y$ -coordinates. Save $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for everything else.

What if the subtraction gives a negative number?

Take the absolute value — lengths are always positive. Inside the distance formula it takes care of itself, because squaring makes the differences positive anyway.

How do I find the height of a triangle on the grid?

Pick a horizontal or vertical side as the base if you can. The height is then the perpendicular distance from that base to the opposite vertex, which is just a difference of $y$ -coordinates (or $x$ -coordinates). The slanted sides are not the height.

Can I just count squares instead?

For short horizontal and vertical sides, yes — counting works. But subtracting coordinates is faster with larger numbers and is the only method that generalizes to diagonal sides, so it's worth practicing from the start.

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