Respuesta :

Answer:

(a) A translation is a rigid transformation. How does this statement support line s being parallel to line r?

Because ALL of the points of line r move the same distance in the same direction, then the inclination of the line s doesn't change with respect line r.

(b) Write and expression for the slope of line r:

(k-n)/(j-m)

(c) Write and expression for the slope of line s:

(k-n)/(j-m)

(d) Line q is a vertical is a vertical translation of line s 3 units down. P'' is the image of P'. What are the coordinates of P''?

The coordinates of P'' are:

P''=(m,n+ϴ-3)

Step-by-step explanation:

(a) A translation is a rigid transformation. How does this statement support line s being parallel to line r?

Because ALL of the points of line r move the same distance in the same direction, then the inclination of the line s doesn't change with respect line r.


(b) Write and expression for the slope of line r.

Line r goes through the points:

P=(m,n)=(xp,yp)→xp=m, yp=n

Q=(j,k)=(xq,yq)→xq=j, yq=k

Slope of line r is:

mr=(yq-yp)/(xq-xp)

Replacing the coordinates, the expression for the slope of line r is:

mr=(k-n)/(j-m)


(c) Write and expression for the slope of line s.

Line s goes through the points:

P'=(m,n+ϴ)=(xp',yp')→xp'=m, yp'=n+ϴ

Q'=(j,k+ϴ)=(xq',yq')→xq'=j, yq'=k+ϴ

Slope of line s is:

ms=(yq'-yp')/(xq'-xp')

Replacing the coordinates, the expression for the slope of line s is:

ms=[(k+ϴ)-(n+ϴ)]/(j-m)

Eliminating the parentheses in the numerator:

ms=[k+ϴ-n-ϴ]/(j-m)

Simplifyng:

ms=(k-n)/(j-m)


(d) Line q is a vertical is a vertical translation of line s 3 units down. P'' is the image of P'. What are the coordinates of P''?

3 units down, then change only the ordinate y:

P''=(xp',yp'-3)

Replacing xp' and yp':

P''=(m,n+ϴ-3)

Answer:

1) In the following given figure we have that line segment P'Q' is parallel to line segment PQ.

so, slope of P'Q' where coordinates of P' is (m,n+[tex]\theta[/tex]) and Q' is (j,k+[tex]\theta[/tex])

So, slope of P'Q' [tex]=\dfrac{(k+\theta)-(n+\theta)}{j-m}=\dfrac{k-n}{j-m}[/tex]

The coodinates of P are (m,n) and that of Q are (j,k)

slope of PQ [tex]=\dfrac{k-n}{j-m}[/tex]

Hence, slope of both the lines are equal.

(a) Since, by translation we mean there is a fixed number 's' such that every point on one line is translated by 's' units on the second line.

i.e. each corresponding points between the two lines are translated by a fixed unit.

(b) the slope of line 'r' is same as slope of the line segment PQ

slope of line r [tex]=\dfrac{k-n}{j-m}[/tex]

(c)The slope of line 's' is same as slope of line segment P'Q'

Hence, slope of line s [tex]=\dfrac{k-n}{j-m}[/tex]

(d) q is a vertical translation of line 's' 3 units down

P'' is a image of P' .

so the coordinates of P'' is given by: (m,n+[tex]\theta[/tex]-3)




 

Otras preguntas