distance formula

[lor15]

Heres the question...

Determine whether triangle DEF is congruent to triangle PQR given the corrdinates of the vertices. Explain.

Heres the coordinates...

D(-6,1), E(4,-1), F(-2,-5), P(2,-2), Q(5,-4), R(0,-5)

Now how do i plug these in th distance formula?¿?

ok never mind...my friend taught me how to do it

 

[Guest]

so is this a closed question is do you still need help?

 

[pka]

\(\displaystyle d[(a,b),(p,q)] = \sqrt {\left( {a - p} \right)^2 + \left( {b - q} \right)^2 }
\\)

 

[soroban]

Hello, lor15!

Have you never ever used the Distance Formula before?

Given two points: $\displaystyle P(x_1,y_1)$ and $\displaystyle Q(x_2,y_2)$, the distance PQ is given by:

. . . . . . . . . . \(\displaystyle PQ\;=\;\sqrt{(x_2\,-\,x_1)^2\:+\y_2\,-\,y_1)^2}\)

Determine whether triangle DEF is congruent to triangle PQR, given the coordinates of the vertices. Explain.

D(-6,1), E(4,-1), F(-2,-5), P(2,-2), Q(5,-4), R(0,-5)

Now how do i plug these in the distance formula?

Given: $\displaystyle D(-6,1),\;E(4,-1)$

. . $\displaystyle DE\;=\;\sqrt{[4\,-\,(-6)]^2\,+\,[-1\,-\,1]^2}\:=\:\sqrt{10^2\,+\,2^2} \:=\:\sqrt{100\,+\,4}\:=\;\sqrt{104}$


Given: $\displaystyle E(4,-1),\;F(-2,-4)$

. . $\displaystyle EF\;=\;\sqrt{[-2\,-\,4]^2\,+\,[-5\,-\,(-1)]^2} \:=\:\sqrt{(-6)^2\,+\,(-4)^2}\:=\:\sqrt{36\,+\,16}\:=\:\sqrt{52}$


Given: $\displaystyle D(-6,1),\;F(-2,-5)$

. . $\displaystyle DF\;=\;\sqrt{[-2\,-\,(-6)^2\,+\,[-5\,-\,1]^2}\:=\:\sqrt{4^2\,+\,(-6)^2}\:=\:\sqrt{16\,+\,36} \:=\:\sqrt{52}$


Now do the same for triangle PQR.

[You might note that triangle DEF is isoceles.
If you look further, you'll see that DEF is a right triangle.]