JKL has vertices at J(2, 4), K(2, -3), and L(-6, -3). What is the approximate length of line segment JL?

2 Answers
Jun 19, 2017

Answer:

#sqrt(113) " units "~~10.63 " units"#

Explanation:

To find the length of a line segment from two points, we can form a vector and find the length of the vector.

The vector from two points #A (x_1,y_1)# and #B(x_2,y_2)#, is
#vec(AB)=B-A#

#=>vec(AB)=((x_2-x_1),(y_2-y_1))#

So to find #vec(JL)# from points #J(2,4)# and #L(-6,-3)# we would do the following steps:

#vec(JL)=((-6-2),(-3-4))#

#=>vec(JL)=((-8),(-7))#

We have found the vector #vec(JL)#. Now we need to find the length of the vector. To do this, use the following:

If #vec(AB)=((x),(y))#

Then length of #vec(AB)=|vec(AB)|=sqrt(x^2+y^2)#

Hence for JL:
#|vec(JL)|=sqrt((-8)^2+(-7)^2)#

#|vec(JL)|=sqrt(64+49)#

#|vec(JL)|=sqrt(113) " units "~~10.63 " units"#

Jun 19, 2017

Answer:

#JL~~10.63" to 2 decimal places"#

Explanation:

#"to calculate the length use the "color(blue)"distance formula"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(2/2)|)))#
where # (x_1,y_1),(x_2,y_2)" are 2 points"#

#"the 2 points are " J(2,4),L(-6,-3)#

#"let " (x_1,y_1)=(2,4),(x_2,y_2)=(-6,-3)#

#d=sqrt((-6-2)^2+(-3-4)^2)#

#color(white)(d)=sqrt(64+49)#

#color(white)(d)=sqrt113larrcolor(red)" exact value"#

#color(white)(d)~~10.63" to 2 decimal places"#