Math worksheet for plotting coordinates on a grid, with exercises on identifying and creating shapes.

Let’s solve this step by step.

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Part 1: Write down the coordinates of the points on the first grid.

We read coordinates as (x, y) — x is how far right from 0, y is how far up from 0.

- Point A: Go right 2, up 5 → (2, 5)
- Point B: Go right 4, up 5 → (4, 5)
- Point C: Go right 4, up 3 → (4, 3)
- Point D: Go right 2, up 1 → (2, 1)
- Point E: Go right 1, up 2 → (1, 2)
- Point F: Go right 2, up 3 → (2, 3)
- Point G: Go right 1, up 4 → (1, 4)

✔ Double-check each one against the graph — all match.

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Part 2: Plot these coordinates on this grid: (1,2), (2,4), (4,3)

You don’t need to plot them here since we’re just solving, but note:
These are three corners of a shape — likely part of a square or rectangle.

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Part 3: Plot the 4th point so that the four coordinates make a square. What is the coordinate of the 4th point?

Given points:
P1 = (1,2)
P2 = (2,4)
P3 = (4,3)

We need to find P4 such that all four form a square.

Let’s think geometrically.

Plotting mentally:

- From (1,2) to (2,4): move right 1, up 2 → vector <1,2>
- From (2,4) to (4,3): move right 2, down 1 → vector <2,-1>

Wait — those vectors are perpendicular? Let’s check dot product:
<1,2> • <2,-1> = (1)(2) + (2)(-1) = 2 - 2 = 0 → YES, perpendicular!

And lengths:
|<1,2>| = √(1²+2²) = √5
|<2,-1>| = √(4+1) = √5 → same length!

So if we go from (1,2) with vector <2,-1>, we get to (1+2, 2-1) = (3,1)
Or from (4,3) with vector <-1,-2>? Wait — better way:

In a square, opposite sides are equal and parallel.

If we have points A=(1,2), B=(2,4), C=(4,3), then to complete the square, we can use vector addition.

From A to B: <1,2>
From B to C: <2,-1>

Then from C to D should be same as from B to A reversed? Or from A to D same as from B to C?

Actually, in order: A→B→C→D→A

Vector AB = <1,2>
Vector BC = <2,-1>
Then vector CD should be <-1,-2> (opposite of AB)
So D = C + <-1,-2> = (4-1, 3-2) = (3,1)

Check vector DA: from D(3,1) to A(1,2) = <-2,1> — which should be opposite of BC <2,-1> → yes, <-2,1> = -<2,-1>

Perfect.

Also, check distance from D to A: √[(3-1)² + (1-2)²] = √[4+1] = √5 — matches others.

So 4th point is (3,1)

Alternative method: midpoint of diagonals must be same.

Diagonal AC: from (1,2) to (4,3) → midpoint = ((1+4)/2, (2+3)/2) = (2.5, 2.5)

Diagonal BD: from (2,4) to (x,y) → midpoint = ((2+x)/2, (4+y)/2)

Set equal:
(2+x)/2 = 2.5 → 2+x = 5 → x=3
(4+y)/2 = 2.5 → 4+y = 5 → y=1

Same answer: (3,1)

✔ Confirmed.

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Part 4: Plot another square that has no horizontal or vertical lines. What are the 4 coordinates?

This means a rotated square — not aligned with axes.

We can pick any starting point and build a square using perpendicular vectors of equal length.

Simple example: Start at (1,1). Use vector <1,2> for one side, then perpendicular vector <-2,1> (since dot product 1*(-2)+2*1 = 0, and same length √5).

So:

Point 1: (1,1)
Point 2: (1+1, 1+2) = (2,3)
Point 3: (2-2, 3+1) = (0,4)
Point 4: (0-1, 4-2) = (-1,2) ← wait, negative? Maybe avoid negatives.

Try starting at (2,1):

Use vector <1,1> and perpendicular <-1,1> (dot product: 1*(-1)+1*1=0, both length √2)

Points:

P1: (2,1)
P2: (2+1,1+1) = (3,2)
P3: (3-1,2+1) = (2,3)
P4: (2-1,3-1) = (1,2)

Check: All points within grid? Yes: (2,1), (3,2), (2,3), (1,2)

Is it a square? Sides:

(2,1)-(3,2): dist √2
(3,2)-(2,3): dist √2
(2,3)-(1,2): dist √2
(1,2)-(2,1): dist √2

Angles: vectors between sides are perpendicular — yes.

No horizontal/vertical lines — correct, all slanted.

Perfect.

We could also choose other sets, but this is simple and fits in the grid.

Another option without repeating previous points: try (1,3), (2,5), (4,4), (3,2) — let’s verify:

(1,3) to (2,5): <1,2>
(2,5) to (4,4): <2,-1> → perp and same length → then next: (4-1,4-2)=(3,2); then (3-2,2+1)=(1,3)? Wait:

From (4,4) add <-1,-2> → (3,2)
From (3,2) add <-2,1> → (1,3) — yes.

Points: (1,3), (2,5), (4,4), (3,2)

All within grid? y=5 is okay (grid goes to 6). x=4 ok.

But maybe simpler to stick with smaller numbers.

I’ll go with: (1,2), (2,3), (3,2), (2,1) — wait, that’s diamond-shaped, centered at (2,2). But (1,2) to (2,3) is diagonal? Actually, that’s a square rotated 45°, sides not axis-aligned.

Distances: each side √[(1)^2 + (1)^2] = √2 — good.

Vectors: from (1,2) to (2,3): <1,1>; from (2,3) to (3,2): <1,-1> — dot product 1*1 + 1*(-1)=0 → perpendicular. Good.

But (1,2) was already used in part 2 — doesn’t matter, question says “another square”, doesn’t say new points.

To be safe, let’s pick entirely new points.

How about: (1,1), (2,3), (4,2), (3,0) — but y=0 might be edge.

Better: (2,2), (3,4), (5,3), (4,1) — check:

(2,2) to (3,4): <1,2>
(3,4) to (5,3): <2,-1> → perp, same length → then (5-1,3-2)=(4,1); then (4-2,1+1)=(2,2)? From (4,1) to (2,2): <-2,1> — which is -<2,-1>? No, <-2,1> vs <2,-1> — actually, <-2,1> = -<2,-1>? -<2,-1> = <-2,1> — yes!

So points: (2,2), (3,4), (5,3), (4,1)

All within grid? x=5 is allowed (grid goes to 6), y=4,3,2,1 — all fine.

And no horizontal/vertical lines — perfect.

But perhaps too big? The grid only goes to x=6, y=6 — so (5,3) is ok.

Alternatively, keep it small: (1,3), (2,5), (4,4), (3,2) — as before.

Let me confirm distances:

(1,3) to (2,5): dx=1, dy=2 → dist 5
(2,5) to (4,4): dx=2, dy=-1 → dist √5
(4,4) to (3,2): dx=-1, dy=-2 → dist √5
(3,2) to (1,3): dx=-2, dy=1 → dist √5

Vectors:
AB = <1,2>, BC=<2,-1> → dot product 0 → perp
BC=<2,-1>, CD=<-1,-2> → dot product 2*(-1) + (-1)*(-2) = -2 + 2 = 0 → perp
etc.

Yes, square.

And no horizontal/vertical lines — all slopes are non-zero and non-infinite.

So I’ll use: (1,3), (2,5), (4,4), (3,2)

But to make it even simpler and fit nicely, how about:

(0,2), (1,4), (3,3), (2,1) — but x=0 might be edge.

Grid starts at x=0, so ok.

But let's pick one that’s clearly inside.

Final choice: (1,1), (2,3), (4,2), (3,0) — y=0 is bottom line, acceptable.

But to avoid y=0, let’s shift up: (1,2), (2,4), (4,3), (3,1) — wait, that’s exactly the points from parts 2 and 3! So we can’t reuse that.

Ah, important: Part 4 says “another square” — meaning different from the one in part 3.

So we must choose a different set.

Let’s do: (0,3), (1,5), (3,4), (2,2)

Check:

(0,3) to (1,5): <1,2>
(1,5) to (3,4): <2,-1> → perp, same length
(3,4) to (2,2): <-1,-2>
(2,2) to (0,3): <-2,1>

All good. Points: (0,3), (1,5), (3,4), (2,2)

Within grid? x=0 to 3, y=2 to 5 — yes.

No horizontal/vertical lines — yes.

Perfect.

But to make it very clear and simple, I'll use a standard rotated square like:

(1,2), (2,4), (4,3), (3,1) — but that’s the same as part 3.

No — part 3 used (1,2),(2,4),(4,3) and found (3,1) — so that’s one square.

For part 4, we need a different square.

Let’s create one with vector <1,1> and <-1,1>:

Start at (1,1):
P1: (1,1)
P2: (1+1,1+1) = (2,2)
P3: (2-1,2+1) = (1,3)
P4: (1-1,3-1) = (0,2)

Points: (1,1), (2,2), (1,3), (0,2)

Check distances: each side √[(1)^2 + (1)^2] = √2

Vectors:
(1,1)->(2,2): <1,1>
(2,2)->(1,3): <-1,1> → dot product 1*(-1)+1*1=0 → perp
Good.

No horizontal/vertical lines — yes.

All points within grid: x=0 to 2, y=1 to 3 — yes.

And different from part 3.

Perfect.

So final answer for part 4: (1,1), (2,2), (1,3), (0,2)

But (0,2) is on the edge — acceptable.

To avoid x=0, start at (2,1):

P1: (2,1)
P2: (3,2)
P3: (2,3)
P4: (1,2)

Points: (2,1), (3,2), (2,3), (1,2)

This is a diamond centered at (2,2). All sides √2, angles 90 degrees.

No horizontal/vertical lines — correct.

And all within grid, no zero coordinates.

Best choice.

So: (2,1), (3,2), (2,3), (1,2)

Note: (1,2) was used in part 2, but that’s ok — the square is different.

The square in part 3 was (1,2),(2,4),(4,3),(3,1) — different points.

So this is fine.

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Now compile all answers.

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Final Answer:
1)
A (2, 5)
B (4, 5)
C (4, 3)
D (2, 1)
E (1, 2)
F (2, 3)
G (1, 4)

2) [Plotting not required for answer]

3) (3, 1)

4) (2, 1), (3, 2), (2, 3), (1, 2)