Chapter 9, Problem 1CT

To determine

Points $P(3,-4)$ and $Q(-5,6)$ are given. To find (a) the length $PQ$ (b) the midpoint of $\overline{PQ}$ and (c) the point X such that the P is the midpoint of $\overline{QX}$

Answer to Problem 1CT

The length $PQ$ is $2\sqrt{41}$ .

The mid point of $PQ$ is $(-1,1)$

The point X is $(11,-14)$

Explanation of Solution

Given: Points $P(3,-4)$ and $Q(-5,6)$

Formula Used:

The distance formula of the pythagorus theorem is used to calculate the length of a line segment when the two co-ordinates are given:

  $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

The midpoint of a line segment is the halfway value between the two x co-ordinates and halfway value between the two y co-ordinates.

Calculation:

Given: $P(3,-4)$ and $Q(-5,6)$

Putting the values of respective x and y co-ordinates in the distance equation,

  $PQ=\sqrt{{\left(-5-3\right)}^2+{(6-(-4))}^2}$

  $PQ=\sqrt{8^2+{10}^2}$

  $PQ=\sqrt{164}=2\sqrt{41}$

Thus, the length of PQ is $2\sqrt{41}$

The midpoint of $\overline{PQ}$ is calculated as follows:

x-coordinate: $\frac{(3+(-5)}{2}=-1$

y-coordinate: $\frac{-4+6}{2}=1$

Thus, the co ordinates of the midpoint of PQ is $(-1,1)$

Calculation of the point X such that the P is the midpoint of $\overline{QX}$ :

Let the x co-ordinate be A and the y co-ordinate be B.

Since, P is the mid point of the line QX, the co-ordinates of P will be used to calculate the values of A and B. Thus,

  $3=\frac{A-5}{2}$

  $6=A-5$

  $A=11$

Again,

  $-4=\frac{6+B}{2}$

  $-8=6+B$

  $B=-14$

Thus, the co-ordinates of X is ( $(11,-14)$

Conclusion:

Hence, the solution is

The length of PQ is $2\sqrt{41}$

The co ordinates of the midpoint of PQ is $(-1,1)$

The co-ordinates of X is ( $(11,-14)$