Question

for items 3 - 4, use the coordinates ( j(7,8) ), ( k(1,2) ) and ( l(5,2) ) for ( \triangle jkl ).
the centroid for ( \triangle jkl ) is at point ( m ). what is ( jm ) rounded to the nearest tenth?
a. 2.4
b. 4.3
c. 4.8
d. 7.2

Explanation:

Step1: Find the centroid formula

The centroid \(M(x,y)\) of a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\) is given by \(x=\frac{x_1 + x_2+x_3}{3}\), \(y=\frac{y_1 + y_2+y_3}{3}\). Here \(J(7,8)\), \(K(1,2)\), \(L(5,2)\). So \(x=\frac{7 + 1+5}{3}=\frac{13}{3}\approx4.3\), \(y=\frac{8 + 2+2}{3}=4\). Thus, \(M(\frac{13}{3},4)\).

Step2: Use the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(J(7,8)\) and \(M(\frac{13}{3},4)\), \(x_1 = 7,y_1 = 8,x_2=\frac{13}{3},y_2 = 4\). Then \(d=\sqrt{(7-\frac{13}{3})^2+(8 - 4)^2}=\sqrt{(\frac{21-13}{3})^2+16}=\sqrt{(\frac{8}{3})^2+16}=\sqrt{\frac{64}{9}+16}=\sqrt{\frac{64 + 144}{9}}=\sqrt{\frac{208}{9}}=\frac{\sqrt{208}}{3}\approx\frac{14.42}{3}\approx4.8\)

Answer:

C. 4.8