8.2.1 Coordinates, PT3 Focus Practice

8.2.1 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}

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}

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}

Question 4:

Diagram shows a straight line PQ on a Cartesian plane.

Calculate the length, in unit, of PQ.

Solution:

PS = 15 – 3 = 12 units

SQ = 8 – 3 = 5 units

By Pythagoras’ theorem,

PQ²= PS² + SQ²

= 12²+ 5²

PQ = √169

= 13 units

Question 5:

The diagram shows an isosceles triangle STU.

Given that ST = 5 units, the coordinates of point S are

Solution:

For an isosceles triangle STU, M is the midpoint of straight line TU.

\begin{array}{l} x - coordinate of M \\ = \frac{- 2 + 4}{2} = 1 \end{array}

Point M = (1, 0)

MT = 4 – 1 = 3 units

By Pythagoras’ theorem,

SM²= ST² – MT²

= 5² – 3²

= 25 – 9

= 16

SM = √16

= 4

Therefore, point S = (1, 4).