Basic Concepts In Geometry

To find the distance between two points, consider their co-ordinates and subtract smaller co-ordinate from the larger.

(i) Now, the co-ordinates of B and E are 2 and 5 respectively. We know that 5 > 2.

So, d (B, E) = 5 – 2 = 3

⇒ d (B, E) = 3

(ii) The co-ordinates of J and A are -2 and 1 respectively. We know that 1 > -2.

So, d (J, A) = 1 – (-2) = 1 + 2 = 3

⇒ d (J, A) = 3

(iii) The co-ordinates of P and C are -4 and 3 respectively. We know that 3 > -4.

So, d (P, C) = 3 – (-4) = 3 + 4 = 7

⇒ d (P, C) = 7

(iv) The co-ordinates of J and H are -2 and -1 respectively. We know that -1 > -2.

So, d (J, H) = (-1) – (-2) = -1 + 2 = 1

⇒ d (J, H) = 1

(v) The co-ordinates of K and O are -3 and 0 respectively. We know that 0 > -3.

So, d (K, O) = 0 – (-3) = 0 + 3 = 3

⇒ d (K, O) = 3

(vi) The co-ordinates of O and E are 0 and 5 respectively. We know that 5 > 0.

So, d (O, E) = 5 – 0 = 5

⇒ d (O, E) = 5

(vii) The co-ordinates of P and J are -4 and -2 respectively. We know that -2 > -4.

So, d (P, J) = (-2) – (-4) = -2 + 4 = 2

⇒ d (P, J) = 2

(viii) The co-ordinates of Q and B are -5 and 2 respectively. We know that 2 > -5.

So, d (Q, B) = 2 – (-5) = 2 + 5 = 7

⇒ d (Q, B) = 7

To find the distance between two points, consider their co-ordinates and subtract smaller co-ordinate from the larger.

(i) The co-ordinates of A and B are 1 and 7 respectively. We know that 7 > 1.

So, d (A, B) = 7 – 1 = 6

⇒ d (A, B) = 6

(ii) The co-ordinates of A and B are 6 and -2 respectively. We know that 6 > -2.

So, d (A, B) = 6 – (-2) = 6 + 2 = 8

⇒ d (A, B) = 8

(iii) The co-ordinates of A and B are -3 and 7 respectively. We know that 7 > -3.

So, d (A, B) = 7 – (-3) = 7 + 3 = 10

⇒ d (A, B) = 10

(iv) The co-ordinates of A and B are -4 and -5 respectively. We know that -4 > -5.

So, d (A, B) = (-4) – (-5) = -4 + 5 = 1

⇒ d (A, B) = 1

(v) The co-ordinates of A and B are -3 and -6 respectively. We know that -3 > -6.

So, d (A, B) = (-3) – (-6) = -3 + 6 = 3

⇒ d (A, B) = 3

(vi) The co-ordinates of A and B are 4 and -8 respectively. We know that 4 > -8.

So, d (A, B) = 4 – (-8) = 4 + 8 = 12

⇒ d (A, B) = 12

For any three-distinct collinear points P, Q and R, the point Q is said to be in between P and R if

d (P, Q) + d (Q, R) = d (P, R).

(i) Given: d (P, R) = 7, d (P, Q) = 10, d (Q, R) = 3

Now, consider d (P, Q) = 10 ……….(I)

And d (P, R) + d (Q, R) = 7 + 3 = 10 ……….(II)

∴ Using (I) and (II), we have

d (P, R) + d (Q, R) = d (P, Q)

⇒ R lies between P and Q.

(ii) Given: d (R, S) = 8, d (S, T) = 6, d (R, T) = 4

Here, d (R, S) + d (S, T) = 8 + 6 = 14 ≠ 4 = d (R, T)

Also, d (R, S) + d (R, T) = 8 + 4 = 12 ≠ 6 = d (S, T)

And d (S, T) + d (R, T) = 6 + 4 = 10 ≠ 8 = d (R, S)

Hence, the points R, S and T are non-collinear.

(iii) Given: d (A, B) = 16, d (C, A) = 9, d (B, C) = 7

Now, consider d (A, B) = 16 ……….(I)

And d (C, A) + d (B, C) = 9 + 7 = 16 ……….(II)

∴ Using (I) and (II), we have

d (C, A) + d (B, C) = d (A, B)

⇒ C lies between A and B.

(iv) Given: d (L, M) = 11, d (M, N) = 12, d (N, L) = 8

Here, d (L, M) + d (M, N) = 11 + 12 = 23 ≠ 8 = d (N, L)

Also, d (L, M) + d (N, L) = 11 + 8 = 19 ≠ 12 = d (M, N)

And d (M, N) + d (N, L) = 12 + 8 = 20 ≠ 11= d (L, M)

Hence, the points L, M and N are non-collinear.

(v) Given: d (X, Y) = 15, d (Y, Z) = 7, d (X, Z) = 8

Now, consider d (X, Y) = 15 ……….(I)

And d (Y, Z) + d (X, Z) = 7 + 8 = 15 ……….(II)

∴ Using (I) and (II), we have

d (Y, Z) + d (X, Z) = d (X, Y)

⇒ Z lies between X and Y.

(vi) Given: d (D, E) = 5, d (E, F) = 8, d (D, F) = 6

Here, d (D, E) + d (E, F) = 5 + 8 = 13 ≠ 6 = d (D, F)

Also, d (D, E) + d (D, F) = 5 + 6 = 11 ≠ 8 = d (E, F)

And d (E, F) + d (D, F) = 8 + 6 = 14 ≠ 5 = d (D, E)

Hence, the points D, E and F are non-collinear.

We know that the distance between two points is obtained by subtracting the smaller co-ordinate from the larger co-ordinate.

Given: d (A, C) = 10 and d (C, B) = 8

Now, we know that

If A, B and C are three distinct collinear points, then there are three possibilities:

(i) Point B is between A and C

(ii) Point C is between A and B

(iii) Point A is between B and C

Now, let (i) holds true, i.e. Point B is between A and C, then

d (A, B) + d (B, C) = d (A, C)

⇒ d (A, B) = d (A, C) – d (B, C) = 10 – 8 = 2

⇒ d (A, B) = 2

Next, let (ii) holds true, i.e. Point C is between A and B, then

d (A, C) + d (B, C) = d (A, B)

⇒ d (A, B) = 10 + 8 = 18

⇒ d (A, B) = 18

Lastly, let (iii) holds true, i.e. Point A is between B and C, then

D (A, B) + d (A, C) = d (B, C)

⇒ d (A, B) = d (B, C) – d (A, C) = 8 – 10 = -2

⇒ d (A, B) = -2, which is not possible as distance between any two points is a non-negative real number.

Hence, the value of d (A, B) is either 2 or 18.

Given: X, Y and Z are collinear and

d (X, Y) = 17, d (Y, Z) = 8

We know that

If X, Y and Z are three distinct collinear points, then there are three possibilities:

(i) Point Y is between X and Z

(ii) Point Z is between X and Y

(iii) Point X is between Y and Z

Now, let (i) holds true, i.e. Point Y is between X and Z, then

d (X, Y) + d (Y, Z) = d (X, Z)

⇒ d (X, Z) = 17 + 8 = 25

⇒ d (X, Z) = 25

Next, let (ii) holds true, i.e. Point Z is between X and Y, then

d (X, Z) + d (Y, Z) = d (X, Y)

⇒ d (X, Z) = d (X, Y) - d (Y, Z) = 17 – 8 = 9

⇒ d (X, Z) = 9

Lastly, let (iii) holds true, i.e. Point X is between Y and Z, then

d (X, Y) + d (X, Z) = d (Y, Z)

⇒ d (X, Z) = d (Y, Z) – d (X, Y) = 8 – 17 = -9

⇒ d (X, Z) = -9, which is not possible as distance between any two points is a non-negative real number.

Hence, the value of d (X, Z) is either 25 or 9.

i. l (AC) = 11, l (BC) = 6.5We know that

For any three-distinct collinear points P, Q and R, the point Q is said to be in between P and R if

d (P, Q) + d (Q, R) = d (P, R), or

l (PQ) + l (QR) = l (PR)

So, we have l (AB) = l (AC) – l (BC) = 11 – 6.5 = 4.5

l (AB) = 4.5

ii. l (ST) = 3.7, l (RS) = 2.5

We know that

For any three-distinct collinear points P, Q and R, the point Q is said to be in between P and R if

d (P, Q) + d (Q, R) = d (P, R), or

l (PQ) + l (QR) = l (PR)

So, l (RT) = l (ST) + l (RS) = 3.7 + 2.5 = 6.2

l (RT) = 6.2

iii.

We know that

For any three-distinct collinear points P, Q and R, the point Q is said to be in between P and R if

d (P, Q) + d (Q, R) = d (P, R), or

l (PQ) + l (QR) = l (PR)

So,

Three non- collinear points always form a Triangle.

Now, we know that

The distance between the end points of a segment is called the length of the segment and is denoted by l (AB) = seg AB = AB.

i. seg DE = l (DE) = 9 – (-7) = 9 + 7 = 16

and seg AB = l (AB) = 5 – (-3) = 5 + 3 = 8

Now, we know that

Two segments are congruent if the length of two segments is equal.

But, here, we have seg DE = 16 ≠ 8 = seg AB

So, seg DE and seg AB are not congruent.

ii. seg BC = l (BC) = 5 – 2 = 3

and seg AD = l (AD) = -3 – (-7) = -3 + 7 = 4

We know that

Two segments are congruent if the length of two segments is equal.

But, here, we have seg BC = 3 ≠ 4 = seg AD

So, seg BC and seg AD are not congruent.

iii. seg BE = l (BE) = 9 – 5 = 4

and seg AD = l (AD) = -3 – (-7) = -3 + 7 = 4

We know that

Two segments are congruent if the length of two segments is equal.

And, here, we have seg BE = 4 = seg AD

So, seg BE and seg AD are congruent.

Since M is the midpoint of seg AB.

Therefore seg AM ≅ seg MB, i.e. seg AM = seg MB ……….(i)

Also, we know that

For any three distinct collinear points P, Q and R, the point Q is said to be in between P and R if

d (P, Q) + d (Q, R) = d (P, R), or

l (PQ) + l (QR) = l (PR)

So, seg AB = seg AM + seg MB

⇒ seg AB = 2 seg AM [Using (i)]

So, seg AM = 4

Since P is the midpoint of seg CD.

Therefore seg CP ≅ seg PD, i.e. seg CP = seg PD ……….(i)

Also, we know that

For any three-distinct collinear points P, Q and R, the point Q is said to be in between P and R if

d (P, Q) + d (Q, R) = d (P, R), or

l (PQ) + l (QR) = l (PR)

So, seg CD = seg CP + seg PD

⇒ seg CD = 2 seg CP [Using (i)]

⇒ seg CD = 2 × 2.5 = 5

So, seg CD = 5

AB = 5 cm, BP = 2 cm and AP = 3.4 cm

Clearly, 2 < 3.4 < 5

So, BP < AP < AB.

i. Ray RS or Ray RT

ii. Ray PQ

iii. Ray QR

iv. Ray RQ or Ray SQ or Ray TQ

v. Ray RT and Ray RQ

vi. Ray SQ, Ray SR or Ray SP

Vii. Point S

i. Points A and C or Points P and D

ii. Points U and L or Points R and P

iii. d (U, V) = 5 – (-5) = 5 + 5 = 10

d (U, V) = 10

d (P, C) = 4 – (-2) = 4 + 2 = 6

d (P, C) = 6

d (V, B) = 5 – 2 = 3

d (V, B) = 3

d (U, L) = -3 – (-5) = -3 + 5 = 2

d (U, L) = 2

The if-then forms for the above statements are as follows:

i. If a quadrilateral is a parallelogram then its opposite angles are congruent.

ii. If a quadrilateral is a rectangle then its diagonals are congruent.

iii. If a triangle is isosceles then the segment joining the vertex of the triangle and the midpoint of the base is perpendicular to the base.

We know that if the antecedent (if part) and consequent (then part) in a given conditional statement are interchanged, then the resulting statement is called the Converse.

So, the converses for the above statements are as follows:

i. If the alternate angles made by two lines and its transversal are congruent then the lines are parallel.

ii. If two parallel lines are intersected by a transversal then the interior angles so formed are supplementary.

iii. If the diagonals of a quadrilateral are congruent then the quadrilateral is a rectangle.

Since, we know that

Every segment has one and only one midpoint.

So, the correct option is A.

Since we know that intersection of two lines means the common part in both the lines.

Therefore, for two distinct lines, there can be at the most one point in common for the two.

So, the correct option is C.

Since we know that three distinct points can either be collinear or non-collinear. Further, 3 collinear points can form one line and 3 non-collinear points can form a triangle (quadrilateral with three sides) and hence three lines.

Therefore, one or three lines can be determined by three distinct points.

So, the correct option is C.

The coordinates of A and B are -2 and 5 respectively. We know that 5 > -2.

Thus, d (A, B) = 5 – (-2) = 5 + 2 = 7

So, the correct option is C.

d (P, Q) = 2 and d (P, R) = 10

Now, we know that for any three-distinct collinear points P, Q and R, the point Q is said to be in between P and R if

d (P, Q) + d (Q, R) = d (P, R).

So, d (P, Q) + d (Q, R) = d (P, R)

⇒ d (Q, R) = d (P, R) - d (P, Q) = 10 – 2 = 8

So, the correct option is B.

Since, the coordinates of P, Q and R are 3, -5 and 6 respectively. We know that 6 > 3 > -5.

Therefore, d (P, Q) = d (Q, P) = 3 – (-5) = 3 + 5 = 8

d (Q, R) = d (R, Q) = 6 – (-5) = 6 + 5 = 11

and d (R, P) = d (P, R) = 6 – 3 = 3

i. Here, consider

d (P, Q) + d (Q, R) = 8 + 11 = 19 ≠ 3 = d (P, R)

So, this statement is false.

ii. Here, consider

d (P, R) + d (R, Q) = 3 + 11 = 14 ≠ 8 = d (P, Q)

So, this statement is false.

iii. Here, consider

d (R, P) + d (P, Q) = 3 + 8 = 11 = d (R, Q)

So, this statement is true.

iv. Here, consider

d (P, Q) – d (P, R) = 8 – 3 = 5 ≠ 11 = d (Q, R)

So, this statement is false.

The distance between the given pairs is as follows:

i. 3, 6

We know that 6 > 3, so

The required distance is 6 – 3 = 3.

ii. -9, -1

We know that -1 > -9, so

The required distance is -1 – (-9) = -1 + 9 = 8.

iii. -4, 5

We know that 5 > -4, so

The required distance is 5 – (-4) = 5 + 4 = 9.

iv. x, -2

Here, we consider two cases:

Case I When x > -2, then

The required distance is x – (-2) = x + 2.

Case II When -2 > x, then

The required distance is -2 – x.

v. x + 3, x – 3

As x+3 > x - 3

Distance = (x + 3) - (x - 3)

= x + 3 - x + 3

= 6 units

Let the point 8 units away from P be Q, then d (P, Q) = 8 where the coordinate of P is -7.

So, we have either of the following two cases:

Case I- If P > Q, then

d (P, Q) = P – Q

⇒ 8 = - 7 – Q ⇒ Q = - 7 – 8 = - 15

Case II- If P < Q, then

d (P, Q) = Q – P

⇒ 8 = Q – (-7) ⇒ Q = 8 + (-7) = 8 – 7 = 1

Hence, the coordinates of the required points are -15 and 1.

i. Given: d(A,C) = 17, d(B,C) = 6.5

We know that

For any three-distinct collinear points P, Q, and R, the point Q is said to be in between P and R if

d (P, Q) + d (Q, R) = d (P, R)

So, here, d(A, C) = d(A, B) + d(B,C)

⇒ d(A, B) = d(A, C) – d(B, C) = 17 – 6.5 = 10.5

So, d(A, B) = 10.5.

ii. Given: d(P,Q) = 3.4, d(Q,R)= 5.7

We know that

For any three-distinct collinear points P, Q, and R, the point Q is said to be in between P and R if

d (P, Q) + d (Q, R) = d (P, R)

So, here, d(P, R) = d(P, Q) + d(Q, R) = 3.4 + 5.7 = 9.1

So, d(P, R) = 9.1.

Let the point 7 units away from A be B, then d (A, B) = 7 where the coordinate of A is 1.

So, we have either of the following two cases:

Case I- If A > B, then

d(A, B) = A – B

⇒ 7 = 1 – B ⇒ B = 1 – 7 = -6

Case II- If A < B, then

d(A, B) = B – A

⇒ 7 = B – 1 ⇒ B = 7 + 1 = 8

Hence, the coordinates of the required points are -6 and 8.

The conditional forms of the above statements are as follows:

i. If the given quadrilateral is a rhombus then it is a square.

ii. If a pair of angles is linear then the angles are supplementary.

iii. If a given figure is a triangle then it is formed by three segments.

iv. If a number has only two divisors then it is called a prime number.

The converses of the above statements are as follows:

i. If a figure is a triangle then the sum of measures of its angles is 180°.

ii. If two angles are complement of each other than the sum of measures of the two angles is 90°.

iii. If two lines are parallel then the corresponding angles formed by the transversal of the two lines are congruent.

iv. If a number is divisible by 3 than the sum of the digits of that number is divisible by 3.

i. If all sides of a triangle are congruent then it’s all angles are congruent.

Antecedent: All sides of a triangle are congruent.

Consequent: All angles of the triangle are congruent.

ii. The diagonals of a parallelogram bisect each other.

The conditional form of the above statement is:

If the given quadrilateral is a parallelogram then its diagonals bisect each other.

Antecedent: The given quadrilateral is a parallelogram.

Consequent: The diagonals of the quadrilateral bisect each other.

i. Two equilateral triangles are similar.

The conditional form of the above statement is:

If two triangles are equilateral then they are similar.

Antecedent: Two triangles are equilateral.

Consequence: The triangles are similar.

ii. If angles in a linear pair are congruent then each of them is a right angle.

Antecedent: Angles in a linear pair are congruent.

Consequence: Each of the angles is a right angle.

iii. If the altitudes drawn on two sides of a triangle are congruent then those two sides are congruent.

Antecedent: The altitudes drawn on two sides of a triangle are congruent.

Consequence: Those two sides are congruent.