Question

worksheet level 3:
goals:
use the pythagorean theorem to find distances on a coordinate grid
concept #
practice #1

1. greenhouse and stadium
2. police station and art museum
3. greenhouse and hospital
4. city hall and gas station

practice #2
a. find the coordinates of j and k. (1,1) (4,7)
b. use the coordinates to find the distance from j to k. explain your method.
c. use your method from part (b) to find the distance from l to m.
concept 15 pythagorean theorem

Explanation:

Step1: Find coordinates of \(J\) and \(K\)

From the graph, \(J=(1,1)\) and \(K=(4,7)\)

Step2: Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) (derived from Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a=\vert x_2 - x_1\vert\), \(b=\vert y_2 - y_1\vert\) and \(c = d\))

For points \(J(1,1)\) and \(K(4,7)\):
\(x_1 = 1,y_1=1,x_2 = 4,y_2 = 7\)
\(d_{JK}=\sqrt{(4 - 1)^2+(7 - 1)^2}=\sqrt{3^{2}+6^{2}}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5}\approx6.71\)

Step3: Assume coordinates of \(L\) and \(M\) (from the graph, assume \(L=(6,8)\) and \(M=(9,10)\))

Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
\(x_1 = 6,y_1 = 8,x_2=9,y_2 = 10\)
\(d_{LM}=\sqrt{(9 - 6)^2+(10 - 8)^2}=\sqrt{3^{2}+2^{2}}=\sqrt{9+4}=\sqrt{13}\approx3.61\)

Answer:

a. \(J=(1,1)\) and \(K=(4,7)\)
b. Distance from \(J\) to \(K\): \(d_{JK}=\sqrt{(4 - 1)^2+(7 - 1)^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5}\approx6.71\) (using the distance formula based on Pythagorean theorem)
c. Assuming \(L=(6,8)\) and \(M=(9,10)\), \(d_{LM}=\sqrt{(9 - 6)^2+(10 - 8)^2}=\sqrt{9 + 4}=\sqrt{13}\approx3.61\)