Free Math Worksheets

- 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.

- Picture a right triangle with the flight path as the hypotenuse. - Pay attention to signs when subtracting coordinates. - The leg lengths are the absolute values of the horizontal and vertical coordinate differences.

1. The horizontal change is \(6-(-6)=12\), and the vertical change is \(1-(-4)=5\). 2. Use the Pythagorean theorem: \(KL=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\).

The flight path is \(13\) coordinate units long.

- Read the horizontal and vertical distances from the coordinate differences. - Use the Pythagorean theorem to find the longest side of the right triangle. - Pay attention to the order and signs when subtracting coordinates.

1. a) The horizontal length is \(AC=|13-1|=12\), and the vertical length is \(BC=|6-1|=5\). 2. b) Apply the Pythagorean theorem: \(AB^2=12^2+5^2=144+25=169\). Therefore, \(AB=\sqrt{169}=13\).

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.

- Find the horizontal distance between \(A\) and \(B\). - Use a right triangle or the distance formula for \(A\) and \(C\). - Compare the two lengths.

1. Points \(A\) and \(B\) have the same \(y\)-coordinate, so \(AB=|5-1|=4\). 2. Use the distance formula for \(AC\): \(AC=\sqrt{(3-1)^2+(4-1)^2}=\sqrt{13}\approx 3.61\). 3. Since \(4>3.61\), \(\overline{AB}\) is longer.

\(\overline{AB}\) is longer because \(AB=4\) and \(AC=\sqrt{13}\approx 3.61\).

- Find the horizontal and vertical changes for each segment. - Use the distance formula, then compare the results.

1. For \(\overline{AB}\), the horizontal and vertical changes are \(2\) and \(3\), so \(AB=\sqrt{2^2+3^2}=\sqrt{13}\). 2. For \(\overline{CD}\), the changes are \(7-4=3\) and \(3-1=2\), so \(CD=\sqrt{3^2+2^2}=\sqrt{13}\). 3. Therefore, the segments have the same length.

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

- Find the length of each side before adding. - An isosceles triangle has at least two congruent sides. - Use the distance formula for points that do not share an \(x\)- or \(y\)-coordinate. - A quick coordinate-plane sketch may help you check the side lengths.

1. Find \(AB\): because the points have the same \(y\)-coordinate, \(AB=|4-(-2)|=6\). 2. Use the distance formula for \(BC\): \(BC=\sqrt{(1-4)^2+(5-1)^2}=\sqrt{9+16}=5\). 3. Use the distance formula for \(AC\): \(AC=\sqrt{(1-(-2))^2+(5-1)^2}=\sqrt{9+16}=5\). 4. The perimeter is \(6+5+5=16\) units. 5. Because \(AC=BC=5\), the triangle is isosceles.

a) The perimeter is \(16\) units. b) Yes. The triangle is isosceles because \(AC=BC=5\) units.

- Use a horizontal and a vertical segment to form a right triangle with the straight-line distance as the hypotenuse. - Find the horizontal and vertical changes in coordinate units. - Apply the Pythagorean theorem to find the diagonal distance. - Convert coordinate units to meters before calculating the real-world area. - Use the area formula for a right triangle.

1. The horizontal distance from \(P\) to \(R\) is \(4-(-4)=8\) units, and the vertical distance from \(R\) to \(Q\) is \(4-(-2)=6\) units. 2. By the Pythagorean theorem, \(PQ=\sqrt{8^2+6^2}=\sqrt{100}=10\) coordinate units. Since each unit represents \(10\,\text{m}\), the actual distance is \(10\cdot10\,\text{m}=100\,\text{m}\). 3. The actual leg lengths are \(80\,\text{m}\) and \(60\,\text{m}\). The area is \(A=\frac{1}{2}\cdot80\cdot60=2400\,\text{m}^2\).

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

- Use the Pythagorean theorem on the horizontal and vertical changes. - Add the changes to the starting coordinates.

1. Let the vertical change be \(\Delta y\). The arrow length and its horizontal and vertical changes form a right triangle: \(5^2=3^2+(\Delta y)^2\). 2. Then \((\Delta y)^2=25-9=16\). Because the arrow points upward, \(\Delta y=4\). 3. Add the changes to the starting coordinates: \(B=(1+3, 2+4)=(4, 6)\).

The arrow rises \(4\) units, and its endpoint is \(B(4, 6)\).

- Use coordinate differences and the distance formula for each side. - Compare the squares of the side lengths. - What do two congruent sides tell you about the angles?

1. Use the distance formula: \(OA=\sqrt{3^2+1^2}=\sqrt{10}\), \(OB=\sqrt{1^2+2^2}=\sqrt{5}\), and \(AB=\sqrt{(3-1)^2+(1-2)^2}=\sqrt{5}\). 2. Since \(OB^2+AB^2=5+5=10=OA^2\), the triangle is right with the right angle at \(B\). 3. Because \(OB=AB\), it is an isosceles right triangle. Its two acute angles are each \(45^\circ\), so \(\angle AOB=45^\circ\).

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

- Substitute the known coordinates and distance into the distance formula. - An equation of the form \(x^2=a\) can have two solutions. - Think about the two points on the line \(y=5\) that could be the same distance from \(A\).

1. Use the distance formula: \(5=\sqrt{(k-1)^2+(5-2)^2}\). 2. Square both sides: \(25=(k-1)^2+9\). 3. Subtract \(9\): \((k-1)^2=16\). 4. Consider both square-root cases: \(k-1=4\) or \(k-1=-4\). 5. Therefore, \(k=5\) or \(k=-3\).

The possible values are \(k=5\) and \(k=-3\).