Question

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

Explanation:

Step1: Calculate horizontal distance

The formula for horizontal distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(|x_2 - x_1|\).
For points \((1,1)\) and \((-3,6)\), \(| - 3-1|=| - 4| = 4\)

Step2: Calculate vertical distance

The formula for vertical distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(|y_2 - y_1|\).
For points \((1,1)\) and \((-3,6)\), \(|6 - 1|=5\)

Step3: Calculate distance using distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
Substitute \(x_1 = 1,y_1 = 1,x_2=-3,y_2 = 6\) into the formula: \(d=\sqrt{(-3 - 1)^2+(6 - 1)^2}=\sqrt{(-4)^2+5^2}=\sqrt{16 + 25}=\sqrt{41}\)
Also, \(5^2+4^2=25 + 16=41
eq49\)

Answer:

1. True (Since \(\sqrt{41}

eq7\))

1. True (\(d=\sqrt{( - 3-1)^2+(6 - 1)^2}=\sqrt{(-4)^2+5^2}=\sqrt{5^2+4^2}\))
2. True (Horizontal distance \(| - 3-1| = 4\))
3. True (Vertical distance \(|6 - 1|=5\))
4. True (\(5^2+4^2=25 + 16=41

eq49\))