Unit 11: Practical Geometry - Locus

Exercise 11.2 Solutions

Complete solved exercises demonstrating standard 2D locus constructions (distance from points, lines, segments) and complex real-world modeling applications.

Questions 1 - 4 Basic Locus Concepts
Q.1 Two points A and B are 8.2cm apart. Construct the locus of points 5 cm from point A.
A B 8.2cm r = 5cm
Steps of Construction
  • Using a ruler, draw a line segment AB of length 8.2cm.
  • With centre at point A and radius 5cm, draw a complete circle.
  • The boundary of this circle represents the locus of all points exactly 5cm away from point A.
Q.2 Construct a locus of point 2.2cm from line segment CD of measure 5.7cm.
C D 5.7cm 2.2cm
Steps of Construction
  • Using a ruler, draw a line segment CD of length 5.7cm.
  • Draw perpendicular dotted lines at points C and D.
  • With centre C, draw a semicircle on the left of radius 2.2cm.
  • With centre D, draw another semicircle on the right of radius 2.2cm.
  • Draw parallel lines above and below segment CD at a distance of 2.2cm, joining the endpoints of the semicircles.
  • The resulting stadium-shaped boundary is the required locus of points 2.2cm from segment CD.
Q.3 Construct an angle ABC = 105°. Construct a locus of a point P which moves such that it is equidistant from →BA and →BC.
B C A P 105°
Steps of Construction
  • Draw ray →BC. At vertex B, construct an angle of 105° using compass/ruler.
  • Draw the angle bisector ray →BP of ∠ABC.
  • The ray →BP is the required locus of points equidistant from the arms →BA and →BC.
Q.4 Two points E and F are 5.4 cm apart. Construct a locus of a point P which moves such that it is equidistant from E and F.
E F l 5.4cm
Steps of Construction
  • Draw line segment EF of length 5.4cm.
  • With centers at E and F and radius more than half of EF, draw arcs intersecting above and below EF.
  • Draw a line l passing through these intersection points.
  • Line l is the right bisector of segment EF, representing the locus of points equidistant from points E and F.
Questions 5 - 8 Triangles & Intersections
Q.5 An island has two main cities A and B 8 km apart. Kashif lives on the island exactly 6.8km from city A and exactly 7.3 km from city B. Mark with a cross the points on the island where Kashif could live.
A B P P' 8 km
Steps of Construction
  • Scale selection: Let 1km = 1cm. So, distance AB = 8cm, AP = 6.8cm, BP = 7.3cm.
  • Draw segment AB of length 8cm.
  • With centre A, draw an arc of radius 6.8cm above and below AB.
  • With centre B, draw another arc of radius 7.3cm above and below AB.
  • The intersection points of these arcs, marked as P and P', are the two possible locations where Kashif could live.
Q.6 Construct a triangle CDE with mCD = 7.6cm, m∠D = 45° and mDE = 5.9cm. Draw the locus of points equidistant from C and D, and equidistant from CD and CE. Mark the point X where they intersect.
C D E X 7.6cm 5.9cm 45°
Steps of Construction
  • Construct ΔCDE with the given measurements: CD = 7.6cm, ∠D = 45°, and DE = 5.9cm.
  • Draw the right bisector of CD, which is the locus of points equidistant from C and D.
  • Draw the angle bisector of ∠C, which is the locus of points equidistant from CD and CE.
  • Mark the intersection of these two bisectors as point X.
Q.7 Construct a triangle LMN with mLM = 7cm, m∠L = 70° and m∠M = 45°. Find a point within the triangle LMN which is equidistant from L and M and 3cm from L.
L M N 7cm r = 3cm 70° 45° No Intersection!
Analysis of Results:
The locus of points equidistant from L and M is the right bisector of LM, which passes through the midpoint (3.5cm from L). The locus of points 3cm from L is a circle of radius 3cm. Since 3cm < 3.5cm, the arc of radius 3cm from L can never reach the perpendicular bisector. Therefore, no such point exists.
Q.8 Construct a right angled triangle RST with mRS = 6.8 cm, m∠S = 90° and mST = 7.5cm. Find a point within the triangle RST which is equidistant from RS and RT and 4.5cm from R.
S R T P 6.8cm 7.5cm
Steps of Construction
  • Construct the right-angled triangle ΔRST with base SR = 6.8cm, ∠S = 90°, and height ST = 7.5cm.
  • Draw the angle bisector ray of ∠R, which represents the locus of points equidistant from the sides RS and RT.
  • With centre at R and radius 4.5cm, draw an arc intersecting the angle bisector at point P.
  • Point P is the required point.
Questions 9 - 13 Practical Applications
Q.9 Construct a rectangle UVW with measure mUV = 7.2cm and mVW = 5.6 cm. Draw the locus of points at a distance of 2 cm from UV and 3.5cm from W.
U V W X A 7.2cm 5.6cm
Steps of Construction
  • Construct rectangle UVWX of dimensions 7.2cm × 5.6cm.
  • Draw a line segment CD parallel to UV at a distance of 2cm inside the rectangle. This represents the first locus.
  • With centre W and radius 3.5cm, draw an arc inside the rectangle. This represents the second locus.
  • The intersection point of the line CD and the arc is the required point A.
Q.10 Imagine two cell towers located at points A and B on a coordinate plane. The GPS-enabled device, positioned somewhere on the plane, receives signals from both towers. To ensure accurate navigation, the device is placed equidistant from both towers to estimate its position. Draw this locus of navigation.
Theoretical Justification

For any two fixed coordinates representing positions on a plane, the locus of points equidistant from them is the perpendicular bisector of the line segment joining the two coordinates. This bisector represents all potential coordinates where a device receives equal strength signals (under line-of-sight propagation assumptions).

Steps of Construction
  • Mark points A and B representing the tower locations.
  • Draw segment AB.
  • Draw the perpendicular bisector line l of segment AB.
  • Line l is the locus of navigation.
Q.11 Quarantine zone within 10km of the infection source S. Draw locus defining the quarantine area.
Steps of Construction
  • Mark point S to represent the outbreak source.
  • Choose a scale (e.g. 2km = 1cm, meaning a 10km radius is drawn as 5cm).
  • With centre S, draw a circle of radius 5cm.
  • The shaded interior of this circle represents the 10km quarantine containment zone.
Q.12 Treasure buried 24 km from A and equidistant from B and C. Scale: 1cm = 10km.
Steps of Construction
  • Scale conversion: 24km = 2.4cm.
  • Draw segment BC and construct its perpendicular bisector l (equidistant locus).
  • With centre A, draw a circle of radius 2.4cm (representing 24km distance).
  • The points of intersection T and T' where the circle cuts the line l represent the two possible locations of the buried treasure.
Q.13 Mango tree M is 64 metres from apple tree A and between 54 and 82 metres from banana tree B, where AB = 90m. Scale: 1cm = 10m.
A B M (Region) 90m 64m 54m 82m
Steps of Construction
  • Scale conversion: AB = 9cm, AM = 6.4cm, boundaries from B are 5.4cm and 8.2cm.
  • Draw segment AB = 9cm.
  • With centre A, draw a semicircle of radius 6.4cm.
  • With centre B, draw concentric arcs of radii 5.4cm and 8.2cm.
  • The arc segment of radius 6.4cm centered at A that lies inside the region between the two concentric arcs of radii 5.4cm and 8.2cm is the suitable place for planting the mango tree M.
Interactive Locus Sandbox Visual Locus Explorer
Select any question from the dropdown list below. Adjust the parameters using the sliders and observe how the geometric shapes, circles, parallel lines, and intersection points change dynamically on the canvas!