Respuesta :
Given:
Endpoints of a line segment AB are A(2,3) and B(8,11).
To find:
(1) Slope of AB.
(2) Length of AB.
(3) Midpoint of AB.
(4) Slope of a line perpendicular to AB.
Solution:
We have, endpoints of line segment AB, A(2,3) and B(8,11).
(1)
Slope of AB is
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
[tex]m=\dfrac{11-3}{8-2}[/tex]
[tex]m=\dfrac{8}{6}[/tex]
[tex]m=\dfrac{4}{3}[/tex]
Therefore, the slope of AB is [tex]\dfrac{4}{3}[/tex].
(2)
Length of AB is
[tex]AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]AB=\sqrt{(8-2)^2+(11-3)^2}[/tex]
[tex]AB=\sqrt{(6)^2+(8)^2}[/tex]
[tex]AB=\sqrt{36+64}[/tex]
[tex]AB=\sqrt{100}[/tex]
[tex]AB=10[/tex]
Therefore, the length of AB is 10 units.
(3) Midpoint of AB is
[tex]Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)[/tex]
[tex]Midpoint=\left(\dfrac{2+8}{2},\dfrac{3+11}{2}\right)[/tex]
[tex]Midpoint=\left(\dfrac{10}{2},\dfrac{14}{2}\right)[/tex]
[tex]Midpoint=\left(5,7\right)[/tex]
Therefore, the midpoint of AB is (5,7).
(4)
Product of slopes of two perpendicular lines is -1.
Let the slope of line perpendicular to AB be m₁.
[tex]m_1\times \dfrac{4}{3}=-1[/tex]
[tex]m_1=-\dfrac{3}{4}[/tex]
So, slope of line perpendicular to AB is [tex]-\dfrac{3}{4}[/tex].