- Read both ordered pairs from the graph. - Compare the x-coordinates for the horizontal distance and the y-coordinates for the vertical distance.

1. From the graph, \(A(-2,1)\) and \(B(1,5)\). 2. The horizontal distance is \(|1-(-2)|=3\), and the vertical distance is \(|5-1|=4\).

Horizontal distance: \(3\) units; vertical distance: \(4\) units

- Treat the dashed horizontal and vertical segments as perpendicular legs. - Use the coordinate grid to determine the two leg lengths. - The straight segment \(PQ\) is the hypotenuse of that right triangle.

The dashed horizontal and vertical segments have lengths \(8\) and \(6\), so \(PQ=\sqrt{8^2+6^2}\).

\(PQ=\sqrt{8^2+6^2}\)

- Interpret the range as a maximum allowable distance. - Substitute both ordered pairs into the distance formula. - Be careful when subtracting a negative coordinate.

1. Use the distance formula: \(d=\sqrt{(-2-2)^2+(6-3)^2}\). 2. Simplify: \(d=\sqrt{(-4)^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\,\text{m}\). 3. The distance is exactly the router's maximum range, so the laptop is within range.

Yes. The laptop is exactly \(5\,\text{m}\) from the router, so it is within range.

- Find the horizontal and vertical changes from the ordered pairs. - Those two changes form perpendicular legs of a right triangle whose hypotenuse is the direct path. - Keep the square root unrounded until the requested precision.

1. The horizontal change is \(5-(-6)=11\), and the vertical change is \(2-(-4)=6\). 2. The flight path is the hypotenuse of the right triangle formed by those changes, so \(KL=\sqrt{11^2+6^2}=\sqrt{157}\approx12.530\). 3. Rounded to the nearest tenth, the flight path is \(12.5\) coordinate units.

The flight path is approximately \(12.5\) coordinate units long.

- Read each labeled point's coordinates from the graph. - Compare the x-coordinates and y-coordinates to find the two perpendicular leg lengths. - What relationship connects the two perpendicular legs to the diagonal segment?

1. From the graph, \(A(1, 1)\), \(B(13, 6)\), and \(C(13, 1)\). 2. The horizontal leg has length \(AC=|13-1|=12\), and the vertical leg has length \(BC=|6-1|=5\). 3. Then \(AB=\sqrt{12^2+5^2}=\sqrt{169}=13\).

a) \(AC=12\) and \(BC=5\) b) \(AB=13\)

- Picture the direct segment as the hypotenuse of a right triangle. - Find the horizontal and vertical changes from the coordinates. - Take the square root and round only at the end.

1. The horizontal change is \(3-(-2)=5\), and the vertical change is \(5-1=4\). 2. The distance is \(d=\sqrt{5^2+4^2}=\sqrt{41}\). 3. Since \(\sqrt{41}\approx6.4031\), the distance rounds to \(6.40\) units.

The markers are approximately \(6.40\) units apart.

- Read the coordinates of the labeled points from the graph. - A horizontal segment can be measured from its x-coordinates. - Use a right triangle or the distance formula for the oblique segment.

1. From the graph, \(A(1,1)\), \(B(5,1)\), and \(C(3,4)\). 2. Since \(A\) and \(B\) have the same y-coordinate, \(AB=|5-1|=4\). 3. Also, \(AC=\sqrt{(3-1)^2+(4-1)^2}=\sqrt{13}\approx3.61\). 4. Since \(4>3.61\), \(\overline{AB}\) is longer.

\(\overline{AB}\) is longer because \(AB=4\) and \(AC=\sqrt{13}\approx3.61\).

- Read all four endpoints from the graph. - Find the horizontal and vertical changes for each segment. - Compare the two exact distances before using decimals.

1. From the graph, \(A(0,0)\), \(B(2,3)\), \(C(4,1)\), and \(D(7,3)\). 2. For \(\overline{AB}\), the changes are \(2\) and \(3\), so \(AB=\sqrt{2^2+3^2}=\sqrt{13}\). 3. For \(\overline{CD}\), the changes are \(3\) and \(2\), so \(CD=\sqrt{3^2+2^2}=\sqrt{13}\). 4. Therefore, the segments have the same length.

Yes. Both segments have length \(\sqrt{13}\).

- If you choose \(R\) in the wrong row, could \(PR\) still be horizontal? - Can you check your corner by tracing a vertical path from \(R\) toward \(Q\)? - How can you read a segment's length from the grid without measuring the picture?

1. The grid shows \(P(-3,2)\) and \(Q(2,-2)\). 2. Since \(R\) lies on the horizontal line through \(P\), it has \(y=2\). Since it lies on the vertical line through \(Q\), it has \(x=2\). Thus \(R(2,2)\). 3. \(PR=5\) units horizontally and \(RQ=4\) units vertically.

\(R(2,2)\), \(PR=5\) units, \(RQ=4\) units.

- Where could you put a third corner so that two grid-aligned segments meet at a right angle? - Could you place the right-angle corner in a different spot? Would that change the lengths of the two legs?

1. The coordinates are \(A(-1,1)\) and \(B(3,4)\). 2. The horizontal leg is \(3-(-1)=4\) units and the vertical leg is \(4-1=3\) units. 3. Therefore, \(AB^2=4^2+3^2\).

The legs are \(4\) and \(3\) units; \(AB^2=4^2+3^2\).

- One side is horizontal, so its length can be read from the x-coordinates. - For each slanted side, compare both the horizontal and vertical coordinate changes. - An isosceles conclusion needs evidence that two side lengths are equal.

1. Since \(A\) and \(B\) have the same y-coordinate, \(AB=|4-(-2)|=6\). 2. The other two side lengths are \(AC=\sqrt{(1-(-2))^2+(6-1)^2}=\sqrt{34}\) and \(BC=\sqrt{(1-4)^2+(6-1)^2}=\sqrt{34}\). 3. Therefore, the perimeter is \(6+2\sqrt{34}\) units. 4. Since \(AC=BC=\sqrt{34}\), triangle \(ABC\) is isosceles.

a) \(6+2\sqrt{34}\) units b) Yes. The triangle is isosceles because \(AC=BC=\sqrt{34}\).

- A horizontal segment's length is the difference of its x-coordinates. - Use the horizontal and vertical changes as the legs of a right triangle. - A triangle with two equal side lengths has a special classification.

1. Segment \(\overline{AB}\) is horizontal, so \(AB=8-2=6\) units. 2. From \(A\) to \(C\), the horizontal change is \(3\) and the vertical change is \(4\), so \(AC=\sqrt{3^2+4^2}=5\) units. From \(B\) to \(C\), the horizontal change has magnitude \(3\) and the vertical change is \(4\), so \(BC=5\) units. 3. Since \(AC=BC\), the triangle is isosceles.

a) \(AB=6\) units b) \(AC=5\) units and \(BC=5\) units c) The triangle is isosceles.

- Read the coordinates of the labeled points from the graph first. - Use the horizontal and vertical changes as the legs of a right triangle. - Convert coordinate units to meters before calculating the real-world area. - Use the area formula for a right triangle.

1. From the graph, \(P(-4, -2)\), \(Q(4, 4)\), and \(R(4, -2)\). The horizontal and vertical leg lengths are \(8\) and \(6\) coordinate units, so \(PQ=\sqrt{8^2+6^2}=10\) coordinate units. Since each unit represents \(10\,\text{m}\), the actual distance is \(100\,\text{m}\). 2. The actual leg lengths are \(80\,\text{m}\) and \(60\,\text{m}\). Thus, \(A=\frac{1}{2}\cdot80\cdot60=2400\,\text{m}^2\).

a) \(100\,\text{m}\) b) \(2400\,\text{m}^2\)

- Read the marked starting coordinates. - The horizontal change, vertical change, and arrow length form a right triangle. - Choose a positive vertical change because the arrow points upward.

1. The grid places the start at \(A(1,2)\). 2. For upward change \(\Delta y\), \((\sqrt{34})^2=3^2+(\Delta y)^2\), so \((\Delta y)^2=25\) and \(\Delta y=5\). 3. The endpoint is \(B(1+3,2+5)=(4,7)\).

The arrow rises \(5\) units; \(B(4,7)\).

- Express the horizontal and vertical changes symbolically before introducing the distance. - Think of those two changes as perpendicular legs of a right triangle. - After writing the squared relationship, use the fact that a distance cannot be negative.

1. The horizontal change from \(P\) to \(Q\) is \(x_2-x_1\), and the vertical change is \(y_2-y_1\). 2. These changes can be used as the legs of a right triangle whose hypotenuse has length \(d\). 3. By the Pythagorean theorem, \(d^2=(x_2-x_1)^2+(y_2-y_1)^2\). 4. Distance is nonnegative, so \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). 5. If a coordinate difference is negative, squaring it gives the same value as squaring its positive magnitude, so the formula is unchanged.

\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). Negative coordinate differences do not change the distance because they are squared.

- Find both the horizontal and vertical changes between the points. - What type of triangle can you use to find a slanted segment? - What changes if you subtract the coordinates in the opposite order?

1. From \(P\) to \(Q\), the horizontal change is \(3\) and the vertical change is \(4\). 2. By the Pythagorean theorem, \(PQ=\sqrt{3^2+4^2}=5\) units. 3. Reversing direction changes both signs but not the squared changes, so \(QP=5\) units too.

Horizontal change: \(1-(-2)=3\); vertical change: \(5-1=4\). Thus \(PQ=\sqrt{3^2+4^2}=5\) units. Reversing direction changes the signs of the differences but not their squares, so \(QP=5\) units too.

- Look at the sign of each \(y\)-coordinate. - Subtract the starting coordinate from the ending coordinate with parentheses. - Recompute the diagonal using the corrected legs.

1. The horizontal change is \(1-(-2)=3\). 2. The vertical change is \(3-(-1)=4\), not \(2\); subtracting a negative number adds its opposite. 3. The distance is \(\sqrt{3^2+4^2}=5\) units.

The vertical change is \(4\), not \(2\), because \(3-(-1)=4\). The distance is \(5\) units.

- The unknown x-coordinate controls only the horizontal change between the points. - Use the stated distance together with the known vertical change. - Why should there be one possible point to the left and another to the right?

1. The horizontal change is \(k-1\), and the vertical change is \(6-2=4\). 2. The distance condition gives \(7^2=(k-1)^2+4^2\). 3. Thus \(49=(k-1)^2+16\), so \((k-1)^2=33\). 4. Therefore, \(k-1=\sqrt{33}\) or \(k-1=-\sqrt{33}\), giving \(k=1+\sqrt{33}\) or \(k=1-\sqrt{33}\).

\(k=1+\sqrt{33}\) or \(k=1-\sqrt{33}\).

- Read the three coordinates from the graph before calculating. - Compare the squares of the side lengths to test for a right angle. - Equal side lengths imply equal opposite angles. - Use the angle sum of a triangle after identifying the right angle.

1. From the graph, \(O(0, 0)\), \(A(3, 1)\), and \(B(1, 2)\). 2. The side lengths are \(OA=\sqrt{3^2+1^2}=\sqrt{10}\), \(OB=\sqrt{1^2+2^2}=\sqrt{5}\), and \(AB=\sqrt{(3-1)^2+(1-2)^2}=\sqrt{5}\). 3. Since \(OB^2+AB^2=5+5=10=OA^2\), the triangle is right with the right angle at \(B\). 4. Also, \(OB=AB\), so the two acute angles are equal. They sum to \(90^\circ\), so \(\angle AOB=45^\circ\).

a) \(OA=\sqrt{10}\), \(OB=\sqrt{5}\), and \(AB=\sqrt{5}\) b) \(\angle AOB=45^\circ\)