Question

use the polygon tool to draw a rhombus that includes the given side. start and end at the same point (the inside of the rhombus will be blue). polygon undo redo reset

Explanation:

Step1: Identify the given side's vector

The given side is from \((1,7)\) to \((5,9)\). The vector is \((5 - 1, 9 - 7)=(4,2)\).

Step2: Find the next vertex (perpendicular vector)

A rhombus has adjacent sides perpendicular (or same - direction but for rhombus, we can use the property that the other side should have a vector that is a 90 - degree rotation of the given vector. Rotating \((4,2)\) 90 degrees counter - clockwise gives \((- 2,4)\) (using the rotation formula \((x,y)\to(-y,x)\)). So from \((5,9)\), adding \((-2,4)\) gives \((5-2,9 + 4)=(3,13)\) (but this is outside the grid, so we can rotate clockwise: \((2,-4)\). Adding \((2,-4)\) to \((5,9)\) gives \((5 + 2,9-4)=(7,5)\).

Step3: Find the fourth vertex

Now, from \((7,5)\), we need to add the negative of the first vector \((-4,-2)\) to get back to the start. \((7-4,5-2)=(3,3)\)? Wait, no. Wait, the first vector is \((4,2)\), so from \((7,5)\), adding \((-4,-2)\) gives \((7 - 4,5-2)=(3,3)\)? No, let's correct. The first side is from \(A(1,7)\) to \(B(5,9)\). The vector \(\overrightarrow{AB}=(4,2)\). The vector \(\overrightarrow{BC}\) should be perpendicular to \(\overrightarrow{AB}\) and of the same length. The dot product of \(\overrightarrow{AB}=(4,2)\) and \(\overrightarrow{BC}=(x,y)\) should be zero: \(4x + 2y=0\), and \(x^{2}+y^{2}=4^{2}+2^{2}=20\). Solving, we can take \(x=-2,y = 4\) (since \((-2)^{2}+4^{2}=4 + 16 = 20\)). So \(C=B+\overrightarrow{BC}=(5-2,9 + 4)=(3,13)\) (invalid). Or \(x = 2,y=-4\) (since \(2^{2}+(-4)^{2}=4 + 16 = 20\)). So \(C=(5 + 2,9-4)=(7,5)\). Then \(\overrightarrow{CD}=-\overrightarrow{AB}=(-4,-2)\), so \(D=C+\overrightarrow{CD}=(7-4,5-2)=(3,3)\). Then \(\overrightarrow{DA}=-\overrightarrow{BC}=(-2,4)\), and \(A=D+\overrightarrow{DA}=(3-2,3 + 4)=(1,7)\), which is the start point.

But a simpler way on the grid:
The given side has a slope of \(\frac{9 - 7}{5 - 1}=\frac{2}{4}=\frac{1}{2}\). The other side should have a slope of \(-2\) (negative reciprocal for perpendicular).
From \((5,9)\), moving with slope \(-2\): if we move 2 units right and 4 units down (since slope \(-2=\frac{-4}{2}\)), we get to \((7,5)\). From \((7,5)\), moving with slope \(\frac{1}{2}\) (negative of the original slope's negative reciprocal? Wait, no, the opposite side should be parallel to the first side. So from \((7,5)\), moving with slope \(\frac{1}{2}\) (same as the first side) but in the opposite direction. The first side goes from \((1,7)\) to \((5,9)\) (right 4, up 2). So from \((7,5)\), we go left 4, down 2 to \((3,3)\). Then from \((3,3)\), moving with slope \(-2\) (up 4, left 2) to \((1,7)\), which is the start.

So the vertices are \((1,7)\), \((5,9)\), \((7,5)\), \((3,3)\) (but \((3,3)\) is on the grid? Wait, maybe a better way:

The given side is between \((1,7)\) and \((5,9)\). The mid - point of the diagonals of a rhombus is the same. The mid - point of the given side is \((\frac{1 + 5}{2},\frac{7+9}{2})=(3,8)\).

We can also use the property that in a rhombus, the sides are equal in length. The length of the given side is \(\sqrt{(5 - 1)^{2}+(9 - 7)^{2}}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\).

Another approach:
1. Start at \((1,7)\) (the first point of the given side).
2. Go to \((5,9)\) (the second point of the given side).
3. To find the third point, we can use the fact that the vector from \((1,7)\) to \((5,9)\) is \((4,2)\). A perpendicular vector (for the rhombus side) can be \((-2,4)\) (rotated 90 degrees counter - clockwise) or \((2,-4)\) (rotated 90 degrees clockwise). Let's take \((2,-4)\). So from \((5,9)\), add \((2,-4)\) to get \((5 + 2,9-4)=(7,5)\).
4. Then from \((7,5)\), add \((-4,-2)\) (the negative of the first vector \((4,2)\)) to get \((7-4,5-2)=(3,3)\).
5. Then from \((3,3)\), add \((-2,4)\) (the negative of the second vector \((2,-4)\)) to get \((3-2,3 + 4)=(1,7)\), which is the start point.

So the rhombus has vertices at \((1,7)\), \((5,9)\), \((7,5)\), \((3,3)\). But since this is a drawing problem, we can also use the grid to find a more intuitive rhombus.

The horizontal change from \((1,7)\) to \((5,9)\) is 4 units right and 2 units up. So to make a rhombus, from \((5,9)\), we can go 4 units right and 2 units down (slope \(-\frac{1}{2}\) but wait, no, the side should be equal in length and adjacent sides should be perpendicular. Wait, the slope of the given side is \(\frac{9 - 7}{5 - 1}=\frac{2}{4}=\frac{1}{2}\). The slope of the adjacent side should be \(-2\) (negative reciprocal) for perpendicularity.

If we consider the given side as one side, we can also use the following method:

- The first point: \(A(1,7)\)
- The second point: \(B(5,9)\)
- To find the third point \(C\): The vector \(\overrightarrow{AB}=(4,2)\). A vector perpendicular to \(\overrightarrow{AB}\) is \((-2,4)\) (because \(4\times(-2)+2\times4=-8 + 8 = 0\)). So \(C=B+\overrightarrow{AB}^{\perp}=(5-2,9 + 4)=(3,13)\) (too high). Or \(\overrightarrow{AB}^{\perp}=(2,-4)\), so \(C=(5 + 2,9-4)=(7,5)\)
- Then the fourth point \(D=A+\overrightarrow{AB}^{\perp}=(1 + 2,7-4)=(3,3)\)

So the rhombus is formed by connecting \(A(1,7)\), \(B(5,9)\), \(C(7,5)\), \(D(3,3)\) and back to \(A(1,7)\).

To draw it using the polygon tool:
1. Click on \((1,7)\) (start point).
2. Click on \((5,9)\) (second point).
3. Click on \((7,5)\) (third point).
4. Click on \((3,3)\) (fourth point).
5. Click on \((1,7)\) (to close the polygon).

(Note: There are other possible rhombuses, but this is one valid construction based on the grid and the properties of a rhombus - four equal - length sides, opposite sides parallel, adjacent sides perpendicular (in the case of a rhombus that is a square - like rhombus, but here it's a non - square rhombus).)

Answer:

The rhombus is drawn with vertices at \((1,7)\), \((5,9)\), \((7,5)\), \((3,3)\) (and back to \((1,7)\)) using the polygon tool. (Since it's a drawing problem, the key is to identify the correct vertices based on the properties of a rhombus - equal - length sides and parallel/ perpendicular adjacent sides, and then use the polygon tool to connect them in order and close the shape.)