We are given the points that form the hypothenuse side as:
(4,-2) and (-1, -6)
To find the distance (d) between the two points, we use the formula below:
![d\text{ = }\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://img.qammunity.org/2023/formulas/mathematics/college/pg7b3eqvs8y3dq61owmp0ak7y8f3rag9sq.png)
Where (x1,y1) and (x2, y2) are the points
By substituting the given points (4, -2) and (-1, -6), we have:
![\begin{gathered} d\text{ = }\sqrt[]{(-1-4)^2+(-6-(-2))^2} \\ d\text{ = }\sqrt[]{25\text{ +16}} \\ d\text{ = }\sqrt[]{41} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vqvd2bp35z9lwbgdpxcnujh67icsih4whe.png)
To find the length of leg 1, we should identify the points that form leg 1, which are:
(-1, -2) and (-1, -6)
Using the formula above, the length (l) of leg 1 is:
![\begin{gathered} l\text{ = }\sqrt[]{(-1-(-1))^2+(-6-(-2))^2} \\ =\text{ }\sqrt[]{0+16} \\ =\text{ 4} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p83pv5tpkhv8q7p940j5378cns24css0os.png)
To find the length of leg 2, we can apply the Pythagoras theorem since the triangle is right-angled

In this case, the length of leg 2 is adjacent. let us label it as x.
![\begin{gathered} (\sqrt[]{41)}^2=4^2+x^2 \\ 41=16+x^2 \\ x\text{ = }\sqrt[]{25} \\ x\text{ = 5} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qhi7pp8mxarqty544so2wti7x256rf43z4.png)
Answers: leg 1 = 4 units, leg 2 = 5 units, hypothenuse side = sqrt(41)