Question

for items 3 - 4, use the coordinates ( j(7,8) ), ( k(1,2) ) and ( l(5,2) ) for ( \triangle jkl ).
the orthocenter for ( \triangle jkl ) is at point ( n ). what is ( kn ) rounded to the nearest tenth?
a 2.7
b 4.3
c 6.0
d 6.3

Explanation:

Step1: Find the slope of \(KL\)

The formula for slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For \(K(1,2)\) and \(L(5,2)\), \(m_{KL}=\frac{2 - 2}{5 - 1}=0\). The altitude from \(J\) to \(KL\) is a vertical line. Since \(J(7,8)\), the equation of the altitude from \(J\) is \(x = 7\).

Step2: Find the slope of \(JK\)

For \(J(7,8)\) and \(K(1,2)\), \(m_{JK}=\frac{8 - 2}{7 - 1}=\frac{6}{6}=1\). The slope of the altitude from \(L\) to \(JK\) is \(m=- 1\) (negative reciprocal). Using the point - slope form \(y - y_1=m(x - x_1)\) with \(L(5,2)\), \(y - 2=-1(x - 5)\), which simplifies to \(y=-x + 7\).

Step3: Find the orthocenter \(N\)

Substitute \(x = 7\) into \(y=-x + 7\). Then \(y=-7 + 7=0\). So \(N(7,0)\).

Step4: Calculate the distance \(KN\)

The distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(K(1,2)\) and \(N(7,0)\), \(d=\sqrt{(7 - 1)^2+(0 - 2)^2}=\sqrt{36 + 4}=\sqrt{40}\approx6.3\).

Answer:

D. \(6.3\)