If square pqrs is rotated 180° about the origin to create square p'q'r's', which set of sides would be parallel in the resulting image?

i. p'q' and q'r'
ii. q'r' and p's'
iii. p'q' and s'p'
iv. q'r' and r's'

a.
ii only
b.
iii only
c.
i and iv only
d.
i and ii only

A. II only is correct.

Step-by-step explanation:

We are given the square PQRS with co-ordinates P(2,-5), Q(4,-3), R(6,-5) and S(4,-7).

As, we know that,

Reflection about 180° changes the co-ordinates (x,y) to (-x,-y).

So, the co-ordinates of P'Q'R'S' are given by,

P'(-2,-5), Q'(-4,-3), R'(-6,-5) and S'(-4,-7).

So, from the figure below, we see that,

P'Q' is parallel to S'R'

Q'R' is parallel to P'S'.

So, according to the options, we have,

Option II i.e. Q'R' is parallel to P'S' is correct.

Thus, option A is correct.

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form

explanation:

step-by-step explanation: