Question

each empty box to prove that $(-3,6)$ is not on circle $t$.
statements
$(1,1)$ is not 7 units away from $(-3,6)$.
the distance between $(1,1)$ and $(-3,6)$ is equal to $sqrt{5^{2}+4^{2}}$.
the horizontal distance between $(1,1)$ and $(-3,6)$ is 4 units.
the vertical distance between $(1,1)$ and $(-3,6)$ is 5 units.
$5^{2}+4^{2}$ is not equal to 49.

Explanation:

Step1: Determine the order of statements

The correct logical order for proving that a point is not on a circle (by showing the distance from the center to the point is not equal to the radius) is:

1. Find the vertical distance.
2. Find the horizontal distance.
3. Use the distance formula (which involves the square of vertical and horizontal distances).
4. Show that \(5^{2}+4^{2}

eq49\).

1. Conclude that the distance is not \(7\) (since if \(d = \sqrt{x^{2}+y^{2}}\), and \(x = 4,y = 5\), \(d=\sqrt{4^{2}+5^{2}}=\sqrt{16 + 25}=\sqrt{41}

eq7\) as \(41
eq49\)).

Step2: Match the statements to the order

• The vertical distance between \((1,1)\) and \((-3,6)\) is \(5\) units (calculated as \(|6 - 1|=5\)).
• The horizontal distance between \((1,1)\) and \((-3,6)\) is \(4\) units (calculated as \(|1-(-3)| = 4\)).
• The distance between \((1,1)\) and \((-3,6)\) is equal to \(\sqrt{5^{2}+4^{2}}\) (by the distance formula \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\), here \(x_1 = 1,y_1 = 1,x_2=-3,y_2 = 6\), so \(d=\sqrt{(-3 - 1)^{2}+(6 - 1)^{2}}=\sqrt{(-4)^{2}+5^{2}}=\sqrt{4^{2}+5^{2}}\)).
• \(5^{2}+4^{2}=25 + 16=41

eq49\).

• \((1,1)\) is not \(7\) units away from \((-3,6)\) (since if \(d=\sqrt{x^{2}+y^{2}}\), and \(x^{2}+y^{2}

eq49\), then \(d
eq7\)).

Answer:

1. The vertical distance between \((1,1)\) and \((-3,6)\) is \(5\) units.
2. The horizontal distance between \((1,1)\) and \((-3,6)\) is \(4\) units.
3. The distance between \((1,1)\) and \((-3,6)\) is equal to \(\sqrt{5^{2}+4^{2}}\).
4. \(5^{2}+4^{2}\) is not equal to \(49\).
5. \((1,1)\) is not \(7\) units away from \((-3,6)\).