suppose R(3,-5)is the midpoint of the line seg PQ and the coordinates of P are(7,-2).Find the coordinates of Q

Question

a. (2,1.5)
b. (5,-3.5)
c. (6,-10)
d. (-1,-8)

Answer 1

this is simple addition:
3-7 =-4
-5 --2 = -3

this shows the midpoint is 4 units left and 3 units below point P

Q is therefore at -1, -8

Answer 2

If ‘R( 3,-5 )’ is the midpoint of ‘P( 7,-2 ) ’ and ‘Q( x ,y ). Then we have to find x and y.
The procedure is;
3 = (7+x)/2
=> 6 = 7 + x
=> x = -1
And
-5 = (-2+y)/2
=> -10 = -2 + y
=> y = -8
So the coordinates of 'Q' are (-1,-8).

Answer 3

use the distance formula between points R and P

square root ((x2 - x1)^2 + (y2 - y1)^2)

sqr (( 7 - 3)^2 + (-2 + 5)^2)
sqr (4^2 + 3^2)
sqr (16 + 9)
sqr (25)
distance = 5
now double that distance to find the length of the whole segment (10)

now see which answer has a distance of 10 from point P

d.

sqr ((-1 - 7)^2 + (-8 + 2)^2)
sqr ((-8)^2 + (-6)^2)
sqr (64 + 36)
sqr (100)
distance = 10

so choice d is your answer

Answer 4

Whats the square root of the main variable. You have to equate that before you go any further. The coordinates are parallel and the dirigible is in the barn.

Answer 5

Answer: d. (-1,-8)

Let Q(x,y)

3=(x+7)/2
x=-1


-5=(y-2)/2
y=-8

Q(-1,-8)