Answer Details

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Construction of Triangle PQR and the Required Loci

(a)(i) Construct triangle PQR

1. Using a ruler, draw a line segment PQ = 10 cm.
2. At point Q, construct a perpendicular to PQ using a ruler and compasses. This gives a line making an angle of 90° with PQ.
3. Using the compasses set to 7 cm, measure from Q along the perpendicular line and mark the point R.
4. Join P to R. The required triangle PQR is formed.

Therefore, PQ = 10 cm, QR = 7 cm, and ∠PQR = 90°.

(a)(ii) Construct locus l<sub>1</sub>

The locus of points equidistant from Q and R is the perpendicular bisector of QR.

1. With centres Q and R, draw arcs of equal radius greater than half of QR.
2. Let the arcs intersect at two points.
3. Join the two intersection points with a straight line.

This line is labelled l<sub>1</sub>.

(a)(iii) Construct locus l<sub>2</sub>

The locus of points equidistant from P and Q is the perpendicular bisector of PQ.

1. With centres P and Q, draw arcs of equal radius greater than half of PQ.
2. Join the two points where the arcs intersect.

This line is labelled l<sub>2</sub>.

(b) Locate point O

Point O, which is equidistant from P, Q, and R, is the point of intersection of l<sub>1</sub> and l<sub>2</sub>.

Since triangle PQR is right-angled at Q, the point O is also the midpoint of the hypotenuse PR. It is the centre of the circle passing through P, Q, and R.