Introduction
- Locus is defined as the path, area or volume traced out by a point, line or region as it moves according to some given laws
- In construction the opening between the pencil and the point of the compass is a fixed distance, the length of the radius of a circle.
- The point on the compass determines a fixed point.
- If the length of the radius remains the same or unchanged, all of the point in the plane that can be drawn by the compass from a circle and any points that cannot be drawn by the compass do not lie on the circle.
- Thus the circle is the set of all points at a fixed distance from a fixed point. This set is called a locus.
Common Types of Loci
Perpendicular Bisector Locus
- The locus of a point which are equidistant from two fixed points is the perpendicular bisector of the straight line joining the two fixed points. This locus is called the perpendicular bisector locus.
- So to find the point equidistant from two fixed points you simply find the perpendicular bisector of the two points as shown below.
- Q is the mid-point of M and N.
In three Dimensions
- In three dimensions, the perpendicular bisector locus is a plane at right angles to the line and bisecting the line into two equal parts. The point P can lie anywhere in the line provided its in the middle.
The Locus of Points at a Given Distance from a Given Straight Line.
In two Dimensions
- In the figure below each of the lines from the middle line is marked a centimeters on either side of the given line MN.
- The ‘a’ centimeters on either sides from the middle line implies the perpendicular distance.
- The two parallel lines describe the locus of points at a fixed distance from a given straight line.
In three Dimensions
- In three dimensions the locus of point ‘a’ centimeters from a line MN is a cylindrical shell of radius ‘a’ c, with MN as the axis of rotation.
Locus of Points at a Given Distance from a Fixed Point.
In two Dimension
- If O is a fixed point and P a variable point‘d’ cm from O,the locus of p is the circle O radius ‘d’ cm as shown below.
- All points on a circle describe a locus of a point at constant distance from a fixed point. In three dimesion the locus of a point ‘d’ centimetres from a point is a spherical shell centre O and radius d cm.
Angle Bisector Locus
- The locus of points which are equidistant from two given intersecting straight lines is the pair of perpendicular lines which bisect the angles between the given lines.
- Conversely ,a point which lies on a bisector of given angle is equidistant from the lines including that angle.
Line PB bisects angle ABC into two equal parts.
Example
Construct triangle PQR such that PQ= 7 cm, QR = 5 cm and angle PQR = 300.Construct the locus L of points equidistant from RP and RQ.
Solution
L is the bisector of Angle PRQ.
Constant Angle Loci
A line PQ is 5 cm long, Construct the locus of points at which PQ subtends an angle of 700.
Solution
- Draw PQ = 5 cm
- Construct TP at P such that angle QPT = 700
- Draw a perpendicular to TP at P( radius is perpendicular to tangent)
- Construct the perpendicular bisector of PQ to meet the perpendicular in (iii) at O
- Using O as the centre and either OP or OQ as radius, draw the locus
- Transfer the centre on the side of PQ and complete the locus.
- Transfer the centre on the opposite sides of PQ and complete the locus as shown below.
- To O1 are of the same radius,
- Angle subtended by the same chord on the circumference are equal ,
- This is called the constant angle locus.
Intersecting Loci
Example
- Construct triangle PQR such that PQ =7 cm, QR = 5 cm and angle PQR = 300
- Construct the locus L1 of points equidistant from P and Q to meet the locus L2 of points equidistant from Q and R at M .Measure PM
Solution
In the figure below
- L1 is the perpendicular bisector of PQ
- L2 is the perpendicular bisector of PQ
- By measurement, PM is equal to 3.7 cm
Loci of Inequalities
- An inequality is represented graphically by showing all the points that satisfy it.The intersection of two or more regions of inequalities gives the intersection of their loci.
- Remember we shade the unwanted region
Example
Draw the locus of point ( x, y) such that x + y < 3 , y – x ≤ 4 and y > 2.
Solution
Draw the graphs of x + y = 3 ,y –x =4 and y = 2 as shown below.
The unwanted regions are usually shaded. The unshaded region marked R is the locus of points ( x ,y ), such that x + y < 3 , y – X ≤ 4 and y > 2.
The lines of greater or equal to ad less or equal to ( ≤ , ≥ ) are always solid while the lines of greater or less (<>) are always broken.
Example
P is a point inside rectangle ABCD such that AP≤PB and Angle DAP ≥ Angle BAP. Show the region on which P lies.
Solution
Draw a perpendicular bisector of AP=PB and shade the unwanted region. Bisect ∠DAB (∠ DAP = ∠ BAP) and shade the unwanted region lies in the unshaded region.
Example
Draw the locus of a point P which moves that AP ≤ 3 cm.
Solution
- Draw a circle, centre A and radius 3 cm
- Shade the unwanted region.
Locus Involving Chords
The following properties of chords of a circle are used in construction of loci
- Perpendicular bisector of any chord passes through the centre of the circle.
- The perpendicular drawn from a centre of a circle bisects the chord.
- If chords of a circle are equal, they are equidistant from the centre of the circle and vice -versa
- In the figure below, if chord AB intersects chord CD at O, AO = x ,BO = y, CO = m and DO =n then m×n = x×y
Past KCSE Questions on the Topic.
- Using a ruler and a pair of compasses only,
- Construct a triangle ABC such that angle ABC = 135o, AB = 8.2cm and BC = 9.6cm
- Given that D is a position equidistant from both AB and BC and also from B and C
- Locate D
- Find the area of triangle DBC.
-
- Using a ruler, a pair of compasses only construct triangle XYZ such that XY = 6cm, YZ = 8cm and ∠XYZ = 75o
- Measure line XZ and ∠XZY
- Draw a circle that passes through X, Y and Z
- A point M moves such that it is always equidistant from Y and Z. construct the locus of M and define the locus
-
- Construct a triangle ABC in which AB=6cm, BC = 7cm and angle ABC = 75o
Measure:-- Length of AC
- Angle ACB
- Locus of P is such that BP = PC. Construct P
- Construct the locus of Q such that Q is on one side of BC, opposite A and angle BQC = 30o
-
- Locus of P and locus of Q meet at X. Mark x
- Construct locus R in which angle BRC 120o
- Show the locus S inside triangle ABC such that XS ≥SR
- Construct a triangle ABC in which AB=6cm, BC = 7cm and angle ABC = 75o
- Use a ruler and compasses only for all constructions in this question.
- Construct a triangle ABC in which AB=8cm, and BC=7.5cm and ∠ABC=112½°
- Measure the length of AC
- By shading the unwanted regions show the locus of P within the triangle ABC such that
- AP ≤ BP
- AP >3cm
Mark the required region as P
- Construct a normal from C to meet AB produced at D
- Locate the locus of R in the same diagram such that the area of triangle ARB is ¾ the area of the triangle ABC.
- On a line AB which is 10 cm long and on the same side of the line, use a ruler and a pair of compasses only to construct the following.
- Triangle ABC whose area is 20 cm2 and angle ACB = 90o
- The locus of a point P such that angle APB = 45o.
- Locate the position of P such that triangle APB has a maximum area and calculate this area.
- A garden in the shape of a polygon with vertices A, B, C, D and E. AB = 2.5m, AE = 10m, ED = 5.2M and DC=6.9m. The bearing of B from A is 030º and A is due to east of E while D is due north of E, angle EDC = 110º,
- Using a scale of 1 cm to represent 1 m construct an accurate plan of the garden
- A foundation is to be placed near to CD than CB and no more than 6m from A,
- Construct the locus of points equidistant from CB and CD.
- Construct the locus of points 6m from A
- shade and label R ,the region within which the foundation could be placed in the garden
- Construct the locus of points in the garden 3.4m from AE.
- Is it possible for the foundation to be 3.4m from AE and in the region?
-
- Using a ruler and compasses only construct triangle PQR in which QR= 5cm, PR = 7cm and angle PRQ = 135°
- Determine ∠PQR
- At P drop a perpendicular to meet QR produced at T.
- Measure PT
- Locate a point A on TP produced such that the area of triangle AQR is equal to one- and - a - half times the area of triangle PQR
- Complete triangle AQR and measure angle AQR
- Use ruler and a pair of compasses only in this question.
- Construct triangle ABC in which AB = 7 cm, BC = 8 cm and ∠ABC = 600.
- Measure (i) side AC (ii) ∠ ACB
- Construct a circle passing through the three points A, B and C. Measure the radius of the circle.
- Construct ∆ PBC such that P is on the same side of BC as point A and ∠ PCB = ½ ∠ ACB,∠ BPC = ∠ BAC measure ∠ PBC.
- Without using a set square or a protractor:-
- Construct triangle ABC in which BC is 6.7cm, angle ABC is 60o and ∠BAC is 90o.
- Mark point D on line BA produced such that line AD =3.5cm
- Construct:-
- A circle that touches lines AC and AD
- A tangent to this circle parallel to line AD
- Use a pair of compasses and ruler only in this question;
- Draw acute angled triangle ABC in which angle CAB = 37½o, AB = 8cm and CB = 5.4cm. Measure the length of side AC (hint 37½o = ½ x 75o)
- On the triangle ABC below:
- On the same side of AC as B, draw the locus of a point X so that angle AXC = 52½o
- Also draw the locus of another point Y, which is 6.8cm away from AC and on the same side as X
- Show by shading the region P outside the triangle such that angle APC ≥52½o and P is not less than 6.8cm away from AC
Download Loci - Mathematics Form 4 Notes.
Tap Here to Download for 50/-
Get on WhatsApp for 50/-
Why download?
- ✔ To read offline at any time.
- ✔ To Print at your convenience
- ✔ Share Easily with Friends / Students