# What are the length, slope and the midpoint of PQ for P(7,11) and Q(-2,4)?

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### 3 Answers

We are given two points P(7,11), Q(-2,4).

Now for the distance between P and Q we use sqrt[(-2-7)^2+(4-11)^2]

= sqrt [ 81 + 49]

= sqrt 130

= 11.40 approximately.

The slope of the line PQ is ( 4 - 11)/ (-2 -7) = -7/-9 = 7/9

The mid point of PQ is [(-2 + 7)/2 , (4 +11)/2] = (5/2 , 15/2)

**Therefore the length of PQ is sqrt 130, the slope of PQ is 7/9 and the mid point of PQ is (5/2 , 15/2)**

P(7,11), Q(-2,4)

(1) To find the length :

The distance betwen two ponits (x1,y1) and (x2,y2) is given by:

d = sqrt{(x2-x1)^2 +(y2-y1)^2} .

So the distance PQ = sqrt{(-2-7)^2+(4-11)^2} = sqrt{(-9)^2 +(-7)^2} = sqrt{81+49} = sqrt130.

2) The slope m of the line joining the two points (x1,y1)and(x2,y2) is given by:

m = (y2-y1)/(x2-x1) .

Therefore slope of the line joining the line (7,11) and (-2,4) is given by: m = (4-11)/(-2-7) = -7/-9 = 7/9. So the slope is m = 7/9.

3)

The mid point of the two points (x1,y1) and (x2, y2) is ((x1+x2)/2 , (y1+y2)/2).

Therefore the mid point of (7,11) and (-2,4) is ((7-2)/2 , (11+4)/2) = ( 5/2 , 15/2).

The length of PQ is:

[PQ] = sqrt[(xQ-xP)^2 + (yQ-yP)^2]

We'll substitute the coordinates of P and Q into the formula:

[PQ] = sqrt[(-2-7)^2 + (4-11)^2]

[PQ] = sqrt (81 + 49)

**[PQ] = sqrt 130**

The slope of the line PQ is:

mPQ = (yQ-yP)/(xQ-xP)

mPQ = (4-11)/(-2-7)

mPQ = -7/-9

**mPQ = 7/9**

The midpoint of the line PQ is :

xM = (xP + xQ)/2

xM = (7-2)/2

**xM = 5/2**

yM = (yP + yQ)/2

yM = (11+4)/2

**yM = 15/2**

**The coordinates of the midpoint are: M(5/2 ; 15/2).**