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Determine the missing information in the paragraph proof.

Given: Line PQ is rotated 90° counterclockwise to form line P’Q’. The lines are perpendicular. Line PQ contains the points (a, b) and (c, d). Line P’Q’ contains the points (–b, a) and (–d, c).

Prove: The slopes of perpendicular lines are negative reciprocals.



The slopes of lines PQ and P’Q’ can be determined using the formula

The product of these slopes is ________. This product shows that the slopes are negative reciprocals. It is given that the lines are perpendicular and we have shown that the slopes of the lines are negative reciprocals.

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Determine the missing information in the paragraph proof.

Given: Line PQ contains points (w, v) and (x, z) and line P'Q' contains points (w + a, v + b) and (x + a, z + b). Lines PQ and P'Q' are parallel.

Prove: Parallel lines have the same slope.



Since slope is calculated using the formula , the slope of both lines is equivalent to ________. It is given that the lines are parallel, and we calculated that the slopes are the same. Therefore, parallel lines have the same slopes.

WILL GIVE 50 POINTSPLZ ANSWER ASAP WILL GIVE BRAINLIEST IF SOMEONE TELLS MME HOWTHIS IS 2 QUESTIONS I will use to seperate the question so it is organizedThe fi class=
WILL GIVE 50 POINTSPLZ ANSWER ASAP WILL GIVE BRAINLIEST IF SOMEONE TELLS MME HOWTHIS IS 2 QUESTIONS I will use to seperate the question so it is organizedThe fi class=

Respuesta :

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

[tex] \frac{y_{2}-y_{1} }{x_{2}-x_{1}} [/tex]

1)

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

Slope of PQ

[tex] \frac{d-b}{c-a} [/tex]

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

Solpe of P'Q'

[tex] \frac{c-a}{-d-(-b)}=\frac{c-a}{-d+b}=\frac{c-a}{-(d-b)}=-\frac{c-a}{d-b}   [/tex]

Now slope of PQ × Slope of P'Q'

[tex] -\frac{c-a}{d-b} \frac{d-b}{c-a}=-1 [/tex]

As the product of slopes = -1.

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

2)

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

Slope of PQ

[tex] \frac{z-v}{x-w}  [/tex]

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

Slope of P'Q'

[tex] \frac{(z+b)-(v+b)}{(x+a)-(w+a)}=\frac{z+b-v-b}{x+a-w-a}=\frac{z-v}{x-w}  [/tex]

Hence slope of PQ = Slope of P'Q'

Hence PQ is parallel to P'Q'

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