Endpoints of segment MN have coordinates (0, 0), (5, 1). The endpoints of segment AB have coordinates (1 1/22 , 2 1/4 ) and (−2 1/4 , k). What value of k makes these segments perpendicular?

Respuesta :

Answer: [tex]k=18\dfrac{8}{11}[/tex].

Step-by-step explanation:

If a line passing through two points, then

[tex]Slope=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

Endpoints of segment MN have coordinates (0, 0) and (5, 1).

Slope of MN [tex]=\dfrac{1-0}{5-0}=\dfrac{1}{5}[/tex]

The endpoints of segment AB have coordinates [tex]\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)[/tex] and  [tex]\left(-2\dfrac{1}{4} , k\right)[/tex].

[tex]A=\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)=\left(\dfrac{23}{22} ,\dfrac{9}{4}\right)[/tex]

[tex]B=\left(-2\dfrac{1}{4} , k\right)=\left(-\dfrac{9}{4} , k\right)[/tex].

Slope of AB [tex]=\dfrac{k-\frac{9}{4}}{-\frac{9}{4}-\dfrac{23}{22}}[/tex]

[tex]=\dfrac{\frac{4k-9}{4}}{\frac{-99-46}{44}}[/tex]

[tex]=\dfrac{4k-9}{4}\times \dfrac{44}{-145}[/tex]

[tex]=4k-9\times \dfrac{11}{-145}[/tex]

[tex]=\dfrac{44k-99}{-145}[/tex]

Product of slopes of two perpendicular segments is -1.

Slope of MN × Slope of AB = -1

[tex]\dfrac{1}{5}\times \dfrac{44k-99}{-145}=-1[/tex]

[tex]\dfrac{44k-99}{-725}=-1[/tex]

[tex]44k-99=725[/tex]

[tex]44k=725+99[/tex]

[tex]k=\dfrac{824}{44}[/tex]

[tex]k=\dfrac{206}{11}[/tex]

[tex]k=18\dfrac{8}{11}[/tex]

Therefore, the value of k is [tex]k=18\dfrac{8}{11}[/tex].

Answer:

k=21

Step-by-step explanation:

Find the slope of MN which is 1/5, (use the slope formula.)

Use the slope formula for AB using k as y_2

Substitute and you get k=21

21-2.25/-2.25-1.5=-5

The lines are perpendicular because the slopes are the negative reciprocal of one another.