5.2.1 Coordinates, PT3 Focus Practice - Mathematics

5.2.1 Coordinates, PT3 Focus Practice

Question 1:

In diagram below, Q is the midpoint of the straight line PR.

The value of m is

Solution:

\begin{array}{l} \frac{2 + m}{2} = 5 \\ 2 + m = 10 \\ m = 8 \end{array}

$\begin{array}{l} \frac{2 + m}{2} = 5 \\ 2 + m = 10 \\ m = 8 \end{array}$

Question 2:

In diagram below, P and Q are points on a Cartesian plane.

If M is the midpoint of PQ, then the coordinates of M are

Solution:

\begin{array}{l} P (- 4, 8), Q (6, - 2) \\ Coordinates of M \\ = (\frac{- 4 + 6}{2}, \frac{8 + (- 2)}{2}) \\ = (1, 3) \end{array}

$\begin{array}{l} P (- 4, 8), Q (6, - 2) \\ Coordinates of M \\ = (\frac{- 4 + 6}{2}, \frac{8 + (- 2)}{2}) \\ = (1, 3) \end{array}$

Question 3:

Find the distance between P (–4, 6) and Q (20, –1).

Solution:

\begin{array}{l} Distance of P Q \\ = \sqrt{{(- 4 - 20)}^{2} + {[6 - (- 1)]}^{2}} \\ = \sqrt{{(- 24)}^{2} + 7^{2}} \\ = \sqrt{576 + 49} \\ = \sqrt{625} \\ = 25 units \end{array}

$\begin{array}{l} Distance of P Q \\ = \sqrt{{(- 4 - 20)}^{2} + {[6 - (- 1)]}^{2}} \\ = \sqrt{{(- 24)}^{2} + 7^{2}} \\ = \sqrt{576 + 49} \\ = \sqrt{625} \\ = 25 units \end{array}$

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