Solved Examples

SOLVED EXAMPLES Example 1. From figure 1-30, name

(i) All pairs of parallel lines (ii) Concurrent lines and their point of concurrence. (iii) Lines whose point of intersection is B. (iv) Lines whose point of intersection is E. (v) Four collinear points and the line containing them. Sol: (i) Lines l and m , lines m and n , lines l and n are three pairs of parallel lines. (ii) Lines n, p and q are concurrent lines and ‘C’ is their point of concurrence. (iii) Lines m and q . (iv) Lines l and p .

(v) One set of four collinear points is A, B, C and N and the line containing them is q . Other set of four collinear points is E, D, C and M and the line containing them is p . Example 2. Identify a point, a line segment, a ray, intersecting lines and parallel lines from the following: (i) A taut thread of a kite flying in air. (ii) The tip of a tooth-pick. (iii) Opposite edges of a notebook. (iv) Beam of a light from a flash gun. (v) Two adjacent edges of a book. Sol: (i) Line segment (ii) Point (iii) Parallel lines (iv) Rays (v) Intersecting lines Example 3. Use figure 1-31 to answer

(i) Concurrent lines and their point of concurrence. (ii) Lines intersecting at J. (iii) Lines intersecting at G. Sol: (i) Lines l , p and q are concurrent lines and S is their point of concurrence. Lines m, n and p are also concurrent lines and their point of concurrence is T. (ii) Lines n and q intersect at J. (iii) Lines m and q intersect at G. Example 4. If three lines intersect (i) What is the maximum number of their points of intersection? (ii) Minimum number of their points of intersection. Sol:

If the three lines intersect the way as shown in figure 1-32 (i), they will intersect at three points A, B and C. But if they intersect the way as shown in figure 1-32 (ii) they will have only one point of intersection O. Therefore, we conclude that (i) Maximum number of points of intersection is three, and (ii) Minimum number of points of intersection is one. Example 5. Refer to figure 1-33 and answer the following questions:

(i) Name a pair of parallel lines. (ii) Two pairs of perpendicular lines. (iii) Line containing points P, S, O and Q.

(iv) Point of intersection of lines l and m .

uuur (v) Initial point of ray BA Sol: (i) Lines n and m . (ii) Lines l and n and lines l and m . (iii) Line l . (iv) Point O. (v) Point B. Example 6. Look at the figure 1-34 and answer the following questions: (i) How many lines can pass through point A? (ii) How many lines can pass through points B and C?

Sol: (i) Infinite lines

(ii) One and only one line. Example 7. Consider figure 1-35 to name:

(i) Five points (ii) A line (iii) Five line segments (iv) Four rays Sol: (i) A, B, C, D and E (ii) Many answers are possible

suur suur suur suur Some are BD, DE , BE , EF etc (iii) Many answers are possible. Some are BD, BE , DE , EF , DA etc (iv) Several answers are possible

uuur uuur uuur uuur Some are DC , DD, DA, DE etc

Example 8. Classify the following curves as

(i) Open or (ii) Closed Sol: (a) Open (b) Closed (c) Open (d) Closed Example 9. Consider figure 1-37 and answer the following questions?

(i) Is it a curve? (ii) Is it closed? (iii) If answer (i) is yes, then say whether it is convex or concave and why? Sol: (i) Yes (ii) Yes (iii) It is a concave curve because if we take any two points in its interior, then the line segment joining them does not completely lie in it as shown in figure 1-38

Example 10. There are 15 points on a plane, 6 of which are collinear, how many lines can be drawn through them by joining them in pairs? Sol: We know that if there are ' m ' points on a plane, ' n ' of which are collinear, then the number lines drawn through them by joining in pairs is given by the formula:

m(m − 1) n(n − 1) − +1 2 2 Here m = 15 and n = 6 . Substituting these values in the above formula, we get the number lines =

=

15(15 − 1) 6(6 − 1) − +1 2 2

=

15 ×14 6 × 5 − +1 2 2

= 105 − 15 + 1 = 91lines

Example 11. Find the number of diagonals of a rectilinear figure formed by 12 sides. Sol:

We know that the number of diagonals of a rectilinear figure bounded by n(n − 1) ' n ' sides in given by the formula −n 2 Here n = 12 . By substituting n = 12 in the above formula, we get number of diagonals 12(12 − 1) = − 12 2 =

12 ×11 − 12 2

=

132 − 12 2

= 66 − 12 = 54diagonals

Example 12. Find the number of diagonals in a triangle Sol:

We know that a triangle has three sides. By substituting n = 13 in the diagonal n(n − 1) formula − n , we get 2 The number of diagonals =

3(3 − 1) −3 2

=

3× 2 −3 2

= 3−3 = 0 Exampel3: Consider figure 1-39 to name the following: (i) Vertices (ii) Pairs of adjacent vertices (iii) Sides (iv) Pairs of adjacent sides (v) Diagonals

Sol: (i) A, B, C, D and E

(ii) A and B; B and C; C and D; D and E; E and A (iii) AB, BC, CD, DE and EA (iv) AB and BC; BC and CD; CD and DE; DE and EA; EA and AB (v) AC and CE Example 14. There are four points on a plane; no three of them are collinear, what is the number of lines passing through them by joining them in pairs? Sol: Since no three points are collinear, we will apply the formula for number of lines m(m − 1) 2 Since the number of points = 4 Therefore, m = 4 Substituting this value of m = 4 in the above formula, we get the total number of lines passing through them =

4(4 − 1) 4 × 3 12 = = = 6 lines 2 2 2

Let us verify this by actually plotting the points and passing lines through them

suur suur suur sur suur suur The lines drawn are PQ, QR, RS , SP, PR and QR

These are six in numbers. Hence verified. LET US REVISE

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