Loci - Mathematics Form 4 Notes

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

1. Draw PQ = 5 cm
2. Construct TP at P such that angle QPT = 700
3. Draw a perpendicular to TP at P( radius is perpendicular to tangent)
4. Construct the perpendicular bisector of PQ to meet the perpendicular in (iii) at O
5. Using O as the centre and either OP or OQ as radius, draw the locus
6. Transfer the centre on the side of PQ and complete the locus.
7. 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

1. Construct triangle PQR such that PQ =7 cm, QR = 5 cm and angle PQR = 300
2. 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

1. L1 is the perpendicular bisector of PQ
2. L2 is the perpendicular bisector of PQ
3. 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 APPB 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

1. Draw a circle, centre A and radius 3 cm

Locus Involving Chords

The following properties of chords of a circle are used in construction of loci

1. Perpendicular bisector of any chord passes through the centre of the circle.
2. The perpendicular drawn from a centre of a circle bisects the chord.
3. If chords of a circle are equal, they are equidistant from the centre of the circle and vice -versa
4. 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.

1. Using a ruler and a pair of compasses only,
1. Construct a triangle ABC such that angle ABC = 135o, AB = 8.2cm and BC = 9.6cm
2. Given that D is a position equidistant from both AB and BC and also from B and C
1. Locate D
2. Find the area of triangle DBC.
2.
1. Using a ruler, a pair of compasses only construct triangle XYZ such that XY = 6cm, YZ = 8cm and ∠XYZ = 75o
2. Measure line XZ and ∠XZY
3. Draw a circle that passes through X, Y and Z
4. A point M moves such that it is always equidistant from Y and Z. construct the locus of M and define the locus
3.
1. Construct a triangle ABC in which AB=6cm, BC = 7cm and angle ABC = 75o
Measure:-
1. Length of AC
2. Angle ACB
2. Locus of P is such that BP = PC. Construct P
3. Construct the locus of Q such that Q is on one side of BC, opposite A and angle BQC = 30o
4.
1. Locus of P and locus of Q meet at X. Mark x
2. Construct locus R in which angle BRC 120o
3. Show the locus S inside triangle ABC such that XS ≥SR
4. Use a ruler and compasses only for all constructions in this question.
1. Construct a triangle ABC in which AB=8cm, and BC=7.5cm and ∠ABC=112½°
2. Measure the length of AC
1. By shading the unwanted regions show the locus of P within the triangle ABC such that
1. AP ≤ BP
2. AP >3cm
Mark the required region as P
2. Construct a normal from C to meet AB produced at D
3. Locate the locus of R in the same diagram such that the area of triangle ARB is ¾ the area of the triangle ABC.
5. 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.
1. Triangle ABC whose area is 20 cm2 and angle ACB = 90o
1. The locus of a point P such that angle APB = 45o.
2. Locate the position of P such that triangle APB has a maximum area and calculate this area.
6. 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,
1. Using a scale of 1 cm to represent 1 m construct an accurate plan of the garden
2. A foundation is to be placed near to CD than CB and no more than 6m from A,
1. Construct the locus of points equidistant from CB and CD.
2. Construct the locus of points 6m from A
1. shade and label R ,the region within which the foundation could be placed in the garden
2. Construct the locus of points in the garden 3.4m from AE.
3. Is it possible for the foundation to be 3.4m from AE and in the region?
7.
1. Using a ruler and compasses only construct triangle PQR in which QR= 5cm, PR = 7cm and angle PRQ = 135°
2. Determine ∠PQR
3. At P drop a perpendicular to meet QR produced at T.
4. Measure PT
5. 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
6. Complete triangle AQR and measure angle AQR
8. Use ruler and a pair of compasses only in this question.
1. Construct triangle ABC in which AB = 7 cm, BC = 8 cm and ABC = 600.
2. Measure (i) side AC (ii) ACB
3. Construct a circle passing through the three points A, B and C. Measure the radius of the circle.
4. Construct ∆ PBC such that P is on the same side of BC as point A and PCB = ½ ACB,BPC = BAC measure PBC.
9. Without using a set square or a protractor:-
1. Construct triangle ABC in which BC is 6.7cm, angle ABC is 60o and ∠BAC is 90o.
2. Mark point D on line BA produced such that line AD =3.5cm
3. Construct:-
1. A circle that touches lines AC and AD
2. A tangent to this circle parallel to line AD
10. Use a pair of compasses and ruler only in this question;
1. 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)
2. On the triangle ABC below:
1. On the same side of AC as B, draw the locus of a point X so that angle AXC = 52½o
2. Also draw the locus of another point Y, which is 6.8cm away from AC and on the same side as X
3. 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

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