Question

use the coordinates j(7, 8), k(1, 2) and l(5, 2) for △jkl. the orthocenter for △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: Identify coordinates of K and N (orthocenter)

First, we need to find the orthocenter \( N \) of \( \triangle JKL \). Let's analyze the triangle:

• Coordinates: \( J(7, 8) \), \( K(1, 2) \), \( L(5, 2) \)
• Notice that \( K \) and \( L \) have the same y - coordinate (\( y = 2 \)), so \( KL \) is a horizontal line. The altitude from \( J \) to \( KL \) is vertical (since \( KL \) is horizontal). The x - coordinate of \( J \) is 7, and the altitude from \( J \) to \( KL \) will have the equation \( x = 7 \) (vertical line) and intersect \( KL \) at \( (7, 2) \), but wait, let's check the other sides.

• The side \( KL \): \( K(1, 2) \), \( L(5, 2) \), length \( KL=5 - 1=4 \), horizontal line.
• The side \( JL \): \( J(7, 8) \), \( L(5, 2) \). The slope of \( JL \) is \( m_{JL}=\frac{8 - 2}{7 - 5}=\frac{6}{2}=3 \). The altitude from \( K \) to \( JL \) will be perpendicular to \( JL \), so its slope is \( m=-\frac{1}{3} \). The equation of the altitude from \( K(1, 2) \) is \( y - 2=-\frac{1}{3}(x - 1) \)

• The altitude from \( J \) to \( KL \): since \( KL \) is horizontal, the altitude is vertical, \( x = 7 \)

• Find the intersection of \( x = 7 \) and \( y - 2=-\frac{1}{3}(x - 1) \):

Substitute \( x = 7 \) into \( y - 2=-\frac{1}{3}(7 - 1) \)

\( y - 2=-\frac{1}{3}(6)=- 2 \)

\( y=0 \)

Wait, that can't be right. Wait, maybe I made a mistake. Wait, \( K(1,2) \), \( L(5,2) \), so \( KL \) is horizontal. The side \( JK \): \( J(7,8) \), \( K(1,2) \), slope \( m_{JK}=\frac{8 - 2}{7 - 1}=\frac{6}{6}=1 \). The altitude from \( L \) to \( JK \) is perpendicular to \( JK \), so slope \( m=-1 \). Equation: \( y - 2=-1(x - 5) \), \( y=-x + 7 \)

The altitude from \( J \) to \( KL \): \( x = 7 \) (since \( KL \) is horizontal, altitude is vertical). Substitute \( x = 7 \) into \( y=-x + 7 \), we get \( y = 0 \). So orthocenter \( N \) is at \( (7,0) \)

Wait, now we need to find the distance between \( K(1,2) \) and \( N(7,0) \)

Step2: Apply 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 \( K(1,2) \) and \( N(7,0) \):

\( x_1 = 1,y_1 = 2,x_2 = 7,y_2 = 0 \)

\( d=\sqrt{(7 - 1)^2+(0 - 2)^2}=\sqrt{6^2+(- 2)^2}=\sqrt{36 + 4}=\sqrt{40}\approx6.3 \)

Wait, but maybe the orthocenter is different. Wait, maybe I misread the problem. Wait, the problem says "the orthocenter for \( \triangle JKL \) is at point \( N \)". Wait, maybe the orthocenter is \( (5,0) \)? No, let's re - examine.

Wait, \( K(1,2) \), \( L(5,2) \), \( J(7,8) \)

The side \( KL \) is horizontal. The altitude from \( J \) to \( KL \) is vertical (since \( KL \) is horizontal), so it's the line \( x = 7 \) (because \( J \) has x - coordinate 7). The side \( JK \): slope \( \frac{8 - 2}{7 - 1}=1 \), so the altitude from \( L \) to \( JK \) has slope - 1, equation \( y-2=-1(x - 5)\Rightarrow y=-x + 7 \). Intersection of \( x = 7 \) and \( y=-x + 7 \) is \( y=-7 + 7 = 0 \), so orthocenter \( N(7,0) \)

Now, distance between \( K(1,2) \) and \( N(7,0) \):

Using distance formula \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}=\sqrt{(7 - 1)^2+(0 - 2)^2}=\sqrt{36 + 4}=\sqrt{40}\approx6.3 \)

Step3: Calculate the distance

\( d=\sqrt{(7 - 1)^2+(0 - 2)^2}=\sqrt{36 + 4}=\sqrt{40}\approx6.3 \)

Answer:

\( \boxed{6.3} \) (corresponding to option D)