By the distance formula 

\(\displaystyle \sqrt{(x_2-x_1)^2+(y_2-y_1)^2 }\)

where \(\displaystyle (x_1,y_1)=(1,2)\) and \(\displaystyle (x_2,y_2)=(4,6)\)

we have

\(\displaystyle \sqrt{(4-1)^2+(6-2)^2 }=\sqrt{(3)^2+(4)^2}= \sqrt{9+16} = \sqrt{25} = 5\)

To find the distance, first we have to find the specific coordinate pairs that we're finding the distance between. We know the x-values, so to find the y-values we can plug these endpoint x-values into the line:

\(\displaystyle \small y = \frac{1}{5}(-2)-2\) first multiply

\(\displaystyle \small y = -0.4 - 2\) then subtract

\(\displaystyle \small y = -2.4\)

\(\displaystyle \small y = \frac{1}{5}(5)-2\) first multiply 

\(\displaystyle \small y = 1 - 2\) then subtract

\(\displaystyle \small y = -1\)

Now we know that we're finding the distance between the points \(\displaystyle \small \small (5, -1)\) and \(\displaystyle \small (-2, -2.4)\). We can just plug these values into the distance formula, using the first pair as \(\displaystyle \small (x_{1}, y_{1})\) and the second pair as \(\displaystyle \small (x_{2}, y_{2})\). It would work either way since we are squaring these values, this just makes it easier.

\(\displaystyle \small \sqrt{(x_{1}-x_{2})^2 + (y_{1}- y_{2})^2}\)

\(\displaystyle \small \sqrt{(5--2)^2 + (-1 - -2.4)^2 }\)

\(\displaystyle \small \sqrt{(5+2)^2 + (-1 + 2.4)^2}\)

\(\displaystyle \small \sqrt{7^2 + 1.4^2}\)

\(\displaystyle \small \sqrt{49+1.96} = \sqrt{50.96}\)

To find the distance, first we have to find the specific coordinate pairs that we're finding the distance between. We know the x-values, so to find the y-values we can plug these endpoint x-values into the line:

\(\displaystyle \small \small y = -3(0)+4\) first multiply

\(\displaystyle \small \small y = 0 +4\) then add

\(\displaystyle \small \small y = 4\)

\(\displaystyle \small \small y =-3(5)+4\) first multiply 

\(\displaystyle \small \small y = -15 +4\) then add

\(\displaystyle \small \small y = -11\)

Now we know that we're finding the distance between the points \(\displaystyle \small (0,4)\) and \(\displaystyle \small (5, -11)\). We can just plug these values into the distance formula, using the first pair as \(\displaystyle \small (x_{1}, y_{1})\) and the second pair as \(\displaystyle \small (x_{2}, y_{2})\). Note that it would work either way since we are squaring these values anyway.

\(\displaystyle \small \sqrt{(x_{1}-x_{2})^2 + (y_{1}- y_{2})^2}\)

\(\displaystyle \small \small \sqrt{(0-5)^2 + (4 - -11)^2 }\)

\(\displaystyle \small \small \sqrt{(-5)^2 + (15)^2}\)

\(\displaystyle \small \small \sqrt{25 + 225} = \sqrt{250}\)

To find the distance, first we have to find the specific coordinate pairs that we're finding the distance between. We know the y-values, so to find the x-values we can plug these endpoint y-values into the line:

\(\displaystyle \small -1 = \frac{1}{2}x - 6\) add 6 to both sides

\(\displaystyle \small 5 = \frac{1}{2} x\) multiply by 2

\(\displaystyle 10 = x\) this endpoint is (10, -1)

\(\displaystyle \small 8 = \frac{1}{2} x -6\) add 6 to both sides

\(\displaystyle \small 14 = \frac{1}{2}x\) multiply by 2

\(\displaystyle \small 28 = x\) this endpoint is (28, 8)

Now we can plug these two endpoints into the distance formula:

\(\displaystyle \small \sqrt{(x_{1}- x_{2})^2 + (y_{1} - y_{2})^2 }\) note that it really does not matter which pair we use as \(\displaystyle \small (x_{1}, y_{1})\) and which as \(\displaystyle \small (x_{2}, y_{2})\) since we'll be squaring these differences anyway, just as long as we are consistent.

\(\displaystyle \small \sqrt{(28 - 10)^2 + (8 - - 1 )^2 }\)

\(\displaystyle \small \sqrt{18^2 + 9^2 }\)

\(\displaystyle \small \sqrt{324 + 81 } = \sqrt{405}=9\sqrt{5}\)

To find this length, we need to know the y-coordinates for the endpoints.

First, plug in -5 for x:

\(\displaystyle y = \frac{1}{5}(-5) + 7 = -1 + 7 = 6\)

Next, plug in 10 for x:

\(\displaystyle y = \frac{1}{5} (10 ) + 7 = 2 + 7 = 9\)

So we are finding the distance between the points \(\displaystyle (-5, 6)\) and \(\displaystyle (10, 9 )\)

We will use the distance formula, \(\displaystyle \sqrt{(x_1 - x_2)^2 + (y_1 - y_2 ) ^2 }\). We could assign either point as \(\displaystyle (x_1, y_1)\) and it would still work, but let's choose \(\displaystyle (10, 9 )\):

\(\displaystyle \sqrt{(10 - - 5 )^2 + (9 - 6 )^ 2 } = \sqrt{15^2 + 3^2 } = \sqrt{225 + 9 } = \sqrt{234} \approx 15.297\)

First, we need to figure out the x-coordinates of the endpoints so that we can use the distance formula, \(\displaystyle \sqrt{(x_1 - x_2 )^2 + (y_1 - y_2 ) ^ 2 }\)

Plug in -10 for y and solve for x:

\(\displaystyle -10 = -2x + 3\) subtract 3 from both sides

\(\displaystyle -13 = -2x\) divide both sides by -2

\(\displaystyle 6.5 = x\)

Plug in 15 for y and solve for x:

\(\displaystyle 15 = -2x + 3\) subtract 3 from both sides

\(\displaystyle 12 = -2x\) divide both sides by -2 

\(\displaystyle -6 = x\)

The endpoints are \(\displaystyle (6.5, -10)\) and \(\displaystyle (-6, 15 )\). We could choose either point to be \(\displaystyle (x_1, y_1)\). Let's choose \(\displaystyle (6.5, -10 )\).

\(\displaystyle \sqrt{(6.5 -- 6 )^2 + (-10 - 15 )^2 }= \sqrt { 12.5^2 + (-25)^2 } = \sqrt{156.25 + 625}= \sqrt{781.25} \approx 27.951\)

To calculate the distance, first find the y-coordinates of the endpoints by plugging the x-coordinates into the equation.

First plug in -5

\(\displaystyle 2(-5) - 4y + 10 = 0\)

\(\displaystyle -10 -4y + 10 = 0\)  combining like terms, we get -10 + 10 is 0 

\(\displaystyle -4 y = 0\) divide by -4

\(\displaystyle y = 0\)

Now plug in 0

\(\displaystyle 2(0 ) - 4 y + 10 = 0\)

\(\displaystyle -4y + 10 = 0\) subtract 10 from both sides

\(\displaystyle -4y = -10\) divide by -4

\(\displaystyle y = 2.5\)

The endpoints are \(\displaystyle (-5, 0 )\) and \(\displaystyle (0, 2.5)\), and now we can plug these points into the distance formula:

\(\displaystyle \sqrt{(-5 - 0 )^2 + (0 - 2.5)^2 } = \sqrt{(-5)^2 + (-2.5)^2 } = \sqrt{25 + 6.25} \approx 5.590\)

To find the length, we need to first find the y-coordinates of the endpoints.

First, plug in -8 for x:

\(\displaystyle y = \frac{1}{4} (-8) + 3 = -2 + 3 = 1\)

Now plug in 12 for x:

\(\displaystyle y = \frac{1}{4} (12 ) + 3 = 3 + 3 = 6\)

Our endpoints are \(\displaystyle (-8, 1)\) and \(\displaystyle (12, 6 )\).

To find the length, plug these points into the distance formula:

\(\displaystyle \sqrt{(6-1)^2 + (12--8)^2 } = \sqrt{5^2 + 20^2 } = \sqrt{25 + 400 } \approx 20.616\)

When walking north and then taking a left west, a 90 degree angle is formed. When Jose returns home going in a straight line, this will now form the hypotenuse of a right triangle. The legs of the triangle are 120 ft and 50 ft respectively. 

To solve, use the pythagorean formula. 

\(\displaystyle 50^{2}+120^{2}=c^{2}\)

\(\displaystyle 2500+14400=c^{2}\)

\(\displaystyle 16900=c^{2}\)

\(\displaystyle \sqrt{16900}=\sqrt{c^{2}}\)

\(\displaystyle 130=c\)

130 ft is the straight line distance home. 

The distance formula could also be used to solve this problem. 

We will assume that home is at the point (0,0)

\(\displaystyle Distance=\sqrt{(50-0)^{2}+(120-0)^{2}}\)

Distance = 130 ft. 

We find the exact length of lines using their endpoints and the distance formula.

\(\displaystyle d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\)

Given the endpoints,

\(\displaystyle (x_1,y_1)=(8,4)\)

\(\displaystyle (x_2,y_2)=(5, 10)\)

the distance formula becomes,

\(\displaystyle \sqrt{(5-8)^2+(10-4)^2}=\sqrt{9+36}=\sqrt{45}=\sqrt{9\cdot 5}=\mathbf{3\sqrt{5}}\).