Respuesta :
The correct answer is: Option (A) [tex] \frac{v-z}{w-x} [/tex]
Explanation:
If both lines are parallel, then their slopes must be equal.
For the line PQ, the slope is:
Slope of PQ = [tex] \frac{v-z}{w-x} [/tex] --- (1)
For the line P'Q', the slope is:
Slope of P'Q' = [tex] \frac{v+b-z-b}{w+a-x-a} [/tex]
Slope of P'Q' = [tex] \frac{v-z}{w-x} [/tex] --- (2)
As (1) = (2), hence the lines are parallel and their slope is [tex] \frac{v-z}{w-x} [/tex] (Option A).
Explanation:
If both lines are parallel, then their slopes must be equal.
For the line PQ, the slope is:
Slope of PQ = [tex] \frac{v-z}{w-x} [/tex] --- (1)
For the line P'Q', the slope is:
Slope of P'Q' = [tex] \frac{v+b-z-b}{w+a-x-a} [/tex]
Slope of P'Q' = [tex] \frac{v-z}{w-x} [/tex] --- (2)
As (1) = (2), hence the lines are parallel and their slope is [tex] \frac{v-z}{w-x} [/tex] (Option A).
Answer:
Slope of both the lines are same.
Step-by-step explanation:
Line PQ contains points P(x,z) and Q(w,v)
similarly line P'Q' contains points P'(x+a, z+b) and Q'(w+a, v+b)
As we know slope of parallel lines are same, so to prove PQ and P'Q' we will show that these lines have same slope.
Slope of PQ = [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{v-z}{w-x}[/tex]
Slope of P'Q' = [tex]\frac{y_{2-y_{1} } }{x_{2}-x_{1}}[/tex]
= [tex]\frac{(v+b)-(z+b)}{(w+a)-(x+a)}[/tex]
= [tex]\frac{v+b-z-b}{w+a-x-a}=\frac{v-z}{w-x}[/tex]
So slope of both the lines are same, and both the lines have slope equivalent to [tex]\frac{(v-z)}{(w-x)}[/tex]