Question
In Exercises 27-32 find and then compare lengths of segments.Triangles $J A N$ and $R F K$ have vertices $J(-2,-2), A(4,-2), N(2,2),$ $R(8,1), F(8,4),$ and $K(6,3)$ . Show that $\triangle J A N$ is similar to $\triangle R F K$ .
Step 1
We can do this by using the distance formula, which is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. For $\triangle JAN$, the lengths of the segments are: $JA = \sqrt{(4-(-2))^2+((-2)-(-2))^2} = \sqrt{6^2+0^2} = 6$ $AN = \sqrt{(2-4)^2+(2-(-2))^2} = \sqrt{(-2)^2+4^2} = Show more…
Show all steps
Your feedback will help us improve your experience
Amrita Bhasin and 74 other Geometry educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
In Exercises 27-32 find and then compare lengths of segments. Show that the triangle with vertices $A(-3,4), M(3,1),$ and $Y(0,-2)$ is isosceles.
Coordinate Geometry
The Distance Formula
In Exercises 27-32 find and then compare lengths of segments. The vertices of $\triangle K A T$ and $\triangle I E S$ are $K(3,-1), A(2,6), T(5,1),$ $I(-4,1), E(-3,-6),$ and $S(-6,-1),$ What word best describes the relationship between $\triangle K A T$ and $\triangle I E S ?$
Discover and prove two things about the triangle with vertices $K(-3,4)$, $M(3,1),$ and $J(-6,-2) .$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD