4 4. The locus of a point is the path traced out by the point moving under given geometrical condition (or conditions). lternatively, the locus is the set of all those points which satisfy the given geometrical condition (or conditions). For example:. Let a point move in a plane such that its distance from a fixed point (in the plane), say, is always equal to r. The point will trace out a circle with centre (the fixed point) and radius r. Thus, the locus of a point (in a plane) equidistant from a fixed point (in the plane) is a circle with the fixed point as centre. 2. Let a point move such that its distance from a fixed line (on one side of the line) is always equal to d. The point will trace out a straight line parallel to the fixed line. Thus, the locus of a point is the straight line (shown in the adjoining diagram). Remarks every point which satisfies the given geometrical condition (or conditions) lies on the locus. point which does not satisfy the given geometrical condition (or conditions) cannot lie on the locus. every point which lies on the locus satisfies the given geometrical condition (or conditions). point which does not lie on the locus cannot satisfy the given geometrical condition (or conditions). The locus of a point moving in a plane under a given geometrical condition (or conditions) is always a straight or curved line (or lines). To find the locus of a moving point, plot some points satisfying the given geometrical condition (or conditions), and then join these points. In a theorem on locus, we have to prove the theorem and its converse. The plural of locus is loci and is read as losai. r Fixed point d Fixed line
4.2 Theorems on Theorem 4.. The locus of a point, which is equidistant from two fixed points, is the perpendicular bisector of the line segment joining the two fixed points. Given. Two fixed points and, is a moving point such that =. To prove. lies on the perpendicular bisector of the line segment. onstruction. Join, let be mid-point of and join. roof. Statements In s and. =. Given. Reasons 2. = 2. is mid-point of (construction). 3. = 3. ommon. 4. 4. S.S.S. axiom of congruency. 5. Α = 5. c.p.c.t.. 6. Α + = 80 6. is a straight line. 7. Α = 90 7. From 5 and 6. is the perpendicular bisector of. Hence, lies on the perpendicular bisector of. onversely, any point on the perpendicular bisector of a line segment joining two fixed points is equidistant from the fixed points. Given. Two fixed points and, Q is perpendicular bisector of and is any point on Q. Q To prove. =. onstruction. Join and. roof. Statements Reasons In s and. =. is mid-point of (given). 2. = 2. (given) = 90 =. 3. = 3. ommon. 4. 4. S..S. axiom of congruency. 5. = 5. c.p.c.t.. 332 Understanding ISE mathematics x
onclusion : From the above theorem and its converse, it follows that the locus of a point, which is equidistant from two fixed points, is the perpendicular bisector of the line segment joining the two fixed points. Theorem 4.2. The locus of a point, which is equidistant from two intersecting straight lines, consists of a pair of N N straight lines which bisect the angles between the two given lines. O Given. Two straight lines and intersecting at O. is a moving point such that O, N O and = N. To prove. lies on the bisector of O. onstruction. Join O. roof. Statements Reasons In s O and ON. = N. Given. 2. O = ON 2. O, N O (given) O = 90, ON = 90. 3 O = O 3. ommon. 4. O ON 4. R.H.S. axiom of congruency. 5. O = O 5. c.p.c.t.. lies on the bisector of O. Similarly, if is a moving point such that O, N O and = N, then lies on the bisector of O. onversely, any point on the bisector of an angle is equidistant from the arms of the angle. Given. Two straight lines and intersecting at O. is a point on the bisector of O, O and N O. N O N To prove. = N. roof. Statements Reasons In s O and ON. O = NO. lies on bisector of O (given). 2. O = ON 2. O, N O (given) O = 90, ON = 90. 3. O = O 3. ommon. 4. O ON 4...S. axiom of congruency. 5. = N 5. c.p.c.t.. Similarly, if is a point on the bisector of O and O, N O, then = N. 333
onclusion : From the above theorem and its converse, it follows that the locus of a point, which is equidistant from two intersecting straight lines, consists of a pair of straight lines which bisect the angles between the two given lines. 4.3 in some standard cases. The locus of a point, which is equidistant from two fixed points, is the perpendicular bisector of the line segment joining the two fixed points. 2. The locus of a point, which is equidistant from two intersecting straight lines, consists of a pair of straight lines which bisect the angles between the two given lines. 3. The locus of a point, which is equidistant from two parallel straight lines, is a straight line parallel to the given lines and midway between them. 4. The locus of a point, which is at a given distance from a given straight line, consists of a pair of straight lines parallel to the given line and at a given distance from it. 5. The locus of the centre of a wheel, which moves on a straight horizontal road, is a straight line parallel to the road and at a distance equal to the radius of the wheel. Road 6. The locus of a point, which is inside a circle and is equidistant from two points on the circle, is the diameter of the circle which is perpendicular to the chord of the circle joining the given points. 7. The locus of the mid-points of all parallel chords of a circle is the diameter of the circle which is perpendicular to the given parallel chords. 334 Understanding ISE mathematics x
8. The locus of a point (in a plane), which is at a given distance r from a fixed point (in the plane), is a circle with the fixed point as its centre and radius r. r Fixed point 9. The locus of a point which is equidistant from two given concentric circles of radii r and r 2 is the circle of radius r + r 2 concentric with the given circles. It 2 lies midway between them. 0. The locus of a point which is equidistant from a given circle consists of a pair of circles concentric with the given circle.. If, are fixed points, then the locus of a point such that = 90 is the circle with as diameter. 90 2. The locus of the mid-points of all equal chords of a circle is the circle concentric with the given circle and of radius equal to the distance of equal chords from the centre of the given circle. 3. The locus of centres of circles touching a given line Q at a given point T on it is the straight line perpendicular to Q at T. T Q Remark The problems on locus, concerning circle, should be attempted after learning chapter 5 on circles. 335
Illustrative Examples Example. Find a point in a given line which is equidistant from two fixed points and. Solution. Let the given line and the fixed points, be as shown in the figure alongside. Join and construct perpendicular bisector of. s is equidistant from the points and, it lies on the perpendicular bisector of. lso the point lies on, therefore, the required point is the point of intersection of the perpendicular bisector of and the line. Example 2. Find a point which is equidistant from three given non-collinear points. Solution. Let, and be three given noncollinear points as shown in the figure alongside. s the point is equidistant from the points and, it lies on the perpendicular bisector of. lso as the point is equidistant from the points and, it lies on the perpendicular bisector of. onstruct perpendicular bisectors of and. Then the required point is the point of intersection of the perpendicular bisectors of and. Example 3. onstruct a triangle in which = 5 cm, = 4 6 cm and = 3 8 cm. Find by construction a point which is equidistant from and, and also equidistant from and. Solution. onstruct with the given data. s the point is equidistant from the intersecting lines and, it lies on the bisector of. lso as the point is equidistant from the point and, it lies on the perpendicular bisector of the line segment. onstruct bisector of and the perpendicular bisector of. Then the required point is the point of intersection of the bisector of and the perpendicular bisector of. Example 4. onstruct O = 60. ark a point equidistant from O and O such that its distance from another given line is 2 5 cm. 336 Understanding ISE mathematics x
Solution. onstruct O = 60 as shown in the figure. s the point is equidistant from the intersecting lines O and O, it lies on the bisector of O. onstruct OE, the bisector of O. Let be the other given line. Take any point N on and draw a perpendicular N to. ut off NF = 2 5 cm, and through F draw a straight line GH parallel to. Let GH meet OE at point. Then is a required point which is equidistant from the lines O and O, and is also at a distance 2 5 cm from another given line. Remark If we draw GH on the other side of, then GH will intersect OE at some other point, say Q. Thus, we get one more point satisfying the given conditions. Example 5. onstruct triangle, where = 5 cm, = 4 cm, = 45. omplete the rectangle such that (i) is equidistant from and ; and (ii) is equidistant from and. easure and record the length of. (2007) Solution. Steps of construction.. onstruct with the given data. Q R 2. Since is equidistant from and, lies on the bisector of. ut = 45, therefore, lies on the perpendicular to at. onstruct Q. 3. onstruct R. 4. Since is equidistant from and, =. With as centre, radius equal to, draw an arc to meet R at. 5. With as centre, radius equal to, draw an arc to meet Q at. Join. Length = 5 7 cm approximately. Example 6. onstruct a triangle such that = 5 cm, = 3 cm and = 30. omplete rhombus such that is equidistant from and. Locate the point Q on the line such that Q is equidistant from and. Solution. Steps of construction. Q. onstruct with the given data. 2. Since is equidistant from and, lies on the bisector of. ut = 30, therefore, construct R = 60. ut off = 5 cm from R. 3. omplete rhombus. 4. Since Q is equidistant from and, draw perpendicular bisector of. The point of intersection of the right bisector of and the line is the required point Q. Example 7. Use graph paper for this question. Take cm = unit on both axes. (i) lot the points (, ), (5, 3) and (2, 7). (ii) onstruct the locus of points equidistant from and. O G N R F E H 337
(iii) onstruct the locus of points equidistant from and. (iv) Locate the point such that = and is equidistant from and. (v) easure and record the length in cm. Solution. Take cm = unit on both axes. (i) lot the given points (, ), (5, 3) and (2, 7). (ii) s the locus of points equidistant from and is the right bisector of, so construct the right bisector of segment. (iii) s the locus of points equidistant from and is the bisector of, so construct the bisector of. (iv) is the point of intersection of the right bisector of the segment and the bisector of. (v) Length = 2 5 cm approx. Example 8. Using ruler and compasses only : (i) onstruct a triangle with = 6 cm, = 20 and = 3 5 cm. (ii) In the above figure, draw a circle with as diameter. Find a point on the circumference of the circle which is equidistant from and. (iii) easure. (203, 05) Solution. Steps of construction.. onstruct with the given data. 2. raw right bisector of to meet it at. With as centre and radius, draw the circle. 3. s is equidistant from and. raw bisector Q of. 4. Let Q meet the circle at, then is the required point. On measuring, we find that = 30. Y 7 6 5 4 3 2 (, ) (2, 7) (5, 3) O 2 3 4 5 6 Q X Example 9. Use ruler and compasses only for the following question. ll construction lines and arcs must be clearly shown. (i) onstruct a Δ in which = 6 5 cm, = 60, = 5 cm. (ii) onstruct the locus of points at a distance of 3 5 cm from. (iii) onstruct the locus of points equidistant from and. (iv) ark two points X and Y which are at distance of 3 5 cm from and also equidistant from and. easure XY. (206) 338 Understanding ISE mathematics x
Solution. Steps of construction. (i) onstruct Δ with the given data. (ii) raw a circle with centre and of radius 3 5 cm, which is the locus of points at a distance of 3 5 cm from. (iii) raw the bisector of, which is the locus of points equidistant from and. X Y (iv) Let the bisector of meet the circle with centre and of radius 3 5 cm at points X and Y (as shown in fig.), then X and Y are the required points which are at a distance of 3 5 cm from and also equidistant from and. Length of XY = 4 9 (approx.) Example 0. is a fixed point on the circumference of a circle of radius 5 cm with centre O. is mid-point of a variable chord. State the locus of and justify your answer. Solution. raw a circle of radius 5 cm with centre O. is fixed point and is a chord. is mid-point of. of is a circle with O as diameter. Justification Join O, O and O. In s O and O, = O = O and O is common O O O = O but O + O = 80 ( is mid-point of ) (radii of circle) O = 90 lies on a circle with O as diameter. ( is a straight line) Example. If the diagonals of a quadrilateral bisect each other at right angles, prove that the quadrilateral is a rhombus. O 339
Solution. Let be a quadrilateral in which the diagonals and bisect each other at right angles. Since lies on the perpendicular bisector of, = (i) Similarly, = (ii) lso lies on the perpendicular bisector of, = (iii) From (i), (ii) and (iii), we get = = = is a rhombus. Example 2. If the bisectors of and of a quadrilateral intersect each other at the point, prove that is equidistant from and. Solution. From, draw, N and L. Since lies on the bisector of, = L lso lies on the bisector of, = N From (i) and (ii), we get L = N is equidistant from and. (i) (ii) L N Exercise 4. point moves such that its distance from a fixed line is always the same. What is the relation between and the path travelled by? 2. point moves so that its perpendicular distances from two given lines and are equal. State the locus of the point. 3. is a fixed point and a point Q moves such that the distance Q is constant. What is the locus of the path traced out by the point Q? 4. (i) is a fixed line. State the locus of the point so that = 90. (ii), are fixed points. State the locus of so that = 60. 5. raw and describe the locus in each of the following cases : (i) The locus of points at a distance 2 5 cm from a fixed line. (ii) The locus of vertices of all isosceles triangles having a common base. (iii) The locus of points inside a circle and equidistant from two fixed points on the circle. (iv) The locus of centres of all circles passing through two fixed points. (v) The locus of a point in rhombus which is equidistant from and. (vi) The locus of a point in the rhombus which is equidistant from points and. 6. escribe completely the locus of points in each of the following cases : (i) mid-point of radii of a circle. (ii) centre of a ball, rolling along a straight line on a level floor. (iii) point in a plane equidistant from a given line. (iv) point in a plane, at a constant distance of 5 cm from a fixed point (in the plane). (v) centre of a circle of varying radius and touching two arms of. 340 Understanding ISE mathematics x