By the distance formula, the sum of the squares of each point must add up to 10 squared. The only point that doesn't fufill this requirement is (5,9)

Using the distance formula, \(\displaystyle \sqrt{8^{2}+15^{2}} = 17\)

In general, the distance formula is given by:  \(\displaystyle d=\sqrt{(x_{2}-x_{1})^{2}+{(y_{2}-y_{1})^{2}}}\) and is based on the Pythagorean Theorem.

Let \(\displaystyle P_{1}=(-2,-3)\) and \(\displaystyle P_{2}= (6,12)\)

So the equation to soolve becomes \(\displaystyle d=\sqrt{(6-(-2))^{2}+{(12-(3))^{2}}}\) or \(\displaystyle d=17\)

First, you must figure out where the x and y intercepts lie. To do this we begin by plugging in \(\displaystyle y=0\) to our equation, giving us \(\displaystyle -2x+4=0\). Thus \(\displaystyle x=2\). So our x-intercept is the point \(\displaystyle (2,0)\). We then plug in \(\displaystyle x=0\), giving us \(\displaystyle y=4\), so we know our y-intercept is the point \(\displaystyle (0,4)\). We then use the distance formula \(\displaystyle d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}\) and plug in our points, giving us \(\displaystyle d=\sqrt{(2-0)^2+(0-4)^2}=\sqrt{20}=2\sqrt{5}\)

By the distance formula 

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

where \(\displaystyle (x_1,y_1)=(0,0)\) and \(\displaystyle (x_2,y_2)=(5,12)\)

we have

\(\displaystyle \sqrt{(5-0)^2+(12-0)^2 }=\sqrt{(5)^2+(12)^2}= \sqrt{25+144} = \sqrt{169} = 13\)

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 \(\displaystyle (10, 9 )\):

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