Question

WILL GIVE 50 POINTS!!!!!PLZ ANSWER ASAP!!!!!!!!!!! WILL GIVE BRAINLIEST IF SOMEONE TELLS MME HOW!!!!!!!!!!!!!

Answer 1

If lines are parallel, slopes are equal

If lines are perpendicular, the product of the slopes is -1.

The formula to find the slope is

$\frac{y_{2}-y_{1} }{x_{2}-x_{1}}$

1)

Here P (a, b) & Q(c,d)

Slope of PQ

$\frac{d-b}{c-a}$

Also P'(-b, a) & Q'(-d,c)

Solpe of P'Q'

$\frac{c-a}{-d-(-b)}=\frac{c-a}{-d+b}=\frac{c-a}{-(d-b)}=-\frac{c-a}{d-b}$

Now slope of PQ × Slope of P'Q'

$-\frac{c-a}{d-b} \frac{d-b}{c-a}=-1$

As the product of slopes = -1.

PQ & P'Q' are perpendicular to each other.

2)

P(w,v), Q(x,z)

Slope of PQ

$\frac{z-v}{x-w}$

P'(w+a, v+b), Q' (x+a, z+b)

Slope of P'Q'

$\frac{(z+b)-(v+b)}{(x+a)-(w+a)}=\frac{z+b-v-b}{x+a-w-a}=\frac{z-v}{x-w}$

Hence slope of PQ = Slope of P'Q'

Hence PQ is parallel to P'Q'