Question

(17, -13), (17, 8)\

1. (19, 3), (20, 3)\

(3, 0), (-11, -15)\

1. (19, -2), (-11, 10)

Explanation:

It seems you want to find the distance between the given pairs of points or maybe the slope? Since the problem isn't specified, I'll assume we need to find the distance between each pair of points using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) or the slope using \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's solve for each pair:

Pair 1: \((17, - 13)\) and \((17,8)\)

Step 1: Identify \(x_1,y_1,x_2,y_2\)

\(x_1 = 17,y_1=-13,x_2 = 17,y_2 = 8\)

Step 2: Calculate distance (or slope)

• Distance:

Using \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(x_2 - x_1=17 - 17 = 0\)
\(y_2 - y_1=8-(-13)=21\)
\(d=\sqrt{0^2 + 21^2}=\sqrt{441}=21\)

• Slope:

Using \(m=\frac{y_2 - y_1}{x_2 - x_1}\), but \(x_2 - x_1 = 0\), so the slope is undefined (vertical line)

Pair 2: \((19,3)\) and \((20,3)\)

Step 1: Identify \(x_1,y_1,x_2,y_2\)

\(x_1 = 19,y_1 = 3,x_2=20,y_2 = 3\)

Step 2: Calculate distance (or slope)

• Distance:

\(x_2 - x_1=20 - 19=1\)
\(y_2 - y_1=3 - 3 = 0\)
\(d=\sqrt{1^2+0^2}=\sqrt{1}=1\)

• Slope:

\(m=\frac{3 - 3}{20 - 19}=\frac{0}{1}=0\) (horizontal line)

Pair 3: \((3,0)\) and \((-11,-15)\)

Step 1: Identify \(x_1,y_1,x_2,y_2\)

\(x_1 = 3,y_1 = 0,x_2=-11,y_2=-15\)

Step 2: Calculate distance (or slope)

• Distance:

\(x_2 - x_1=-11 - 3=-14\)
\(y_2 - y_1=-15 - 0=-15\)
\(d=\sqrt{(-14)^2+(-15)^2}=\sqrt{196 + 225}=\sqrt{421}\approx20.52\)

• Slope:

\(m=\frac{-15 - 0}{-11 - 3}=\frac{-15}{-14}=\frac{15}{14}\approx1.07\)

Pair 4: \((19,-2)\) and \((-11,10)\)

Step 1: Identify \(x_1,y_1,x_2,y_2\)

\(x_1 = 19,y_1=-2,x_2=-11,y_2 = 10\)

Step 2: Calculate distance (or slope)

• Distance:

\(x_2 - x_1=-11 - 19=-30\)
\(y_2 - y_1=10-(-2)=12\)
\(d=\sqrt{(-30)^2+12^2}=\sqrt{900 + 144}=\sqrt{1044}\approx32.31\)

• Slope:

\(m=\frac{10-(-2)}{-11 - 19}=\frac{12}{-30}=-\frac{2}{5}=-0.4\)

If you wanted a different operation (like midpoint), let me know and I can recalculate.

Answer:

It seems you want to find the distance between the given pairs of points or maybe the slope? Since the problem isn't specified, I'll assume we need to find the distance between each pair of points using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) or the slope using \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's solve for each pair:

Pair 1: \((17, - 13)\) and \((17,8)\)

Step 1: Identify \(x_1,y_1,x_2,y_2\)

\(x_1 = 17,y_1=-13,x_2 = 17,y_2 = 8\)

Step 2: Calculate distance (or slope)

• Distance:

Using \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(x_2 - x_1=17 - 17 = 0\)
\(y_2 - y_1=8-(-13)=21\)
\(d=\sqrt{0^2 + 21^2}=\sqrt{441}=21\)

• Slope:

Using \(m=\frac{y_2 - y_1}{x_2 - x_1}\), but \(x_2 - x_1 = 0\), so the slope is undefined (vertical line)

Pair 2: \((19,3)\) and \((20,3)\)

Step 1: Identify \(x_1,y_1,x_2,y_2\)

\(x_1 = 19,y_1 = 3,x_2=20,y_2 = 3\)

Step 2: Calculate distance (or slope)

• Distance:

\(x_2 - x_1=20 - 19=1\)
\(y_2 - y_1=3 - 3 = 0\)
\(d=\sqrt{1^2+0^2}=\sqrt{1}=1\)

• Slope:

\(m=\frac{3 - 3}{20 - 19}=\frac{0}{1}=0\) (horizontal line)

Pair 3: \((3,0)\) and \((-11,-15)\)

Step 1: Identify \(x_1,y_1,x_2,y_2\)

\(x_1 = 3,y_1 = 0,x_2=-11,y_2=-15\)

Step 2: Calculate distance (or slope)

• Distance:

\(x_2 - x_1=-11 - 3=-14\)
\(y_2 - y_1=-15 - 0=-15\)
\(d=\sqrt{(-14)^2+(-15)^2}=\sqrt{196 + 225}=\sqrt{421}\approx20.52\)

• Slope:

\(m=\frac{-15 - 0}{-11 - 3}=\frac{-15}{-14}=\frac{15}{14}\approx1.07\)

Pair 4: \((19,-2)\) and \((-11,10)\)

Step 1: Identify \(x_1,y_1,x_2,y_2\)

\(x_1 = 19,y_1=-2,x_2=-11,y_2 = 10\)

Step 2: Calculate distance (or slope)

• Distance:

\(x_2 - x_1=-11 - 19=-30\)
\(y_2 - y_1=10-(-2)=12\)
\(d=\sqrt{(-30)^2+12^2}=\sqrt{900 + 144}=\sqrt{1044}\approx32.31\)

• Slope:

\(m=\frac{10-(-2)}{-11 - 19}=\frac{12}{-30}=-\frac{2}{5}=-0.4\)

If you wanted a different operation (like midpoint), let me know and I can recalculate.