**GSEB Solutions for Class 9 Mathematics – Some Primary Concepts in Geometry : 1 (English Medium)**

**Exercise 7:**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5(1):**

**Solution 5(2):**

**Solution 5(3):**

**Solution 5(4):**

**Solution 5(5):**

**Solution 6:**

**Solution 7:**

**Solution 8(1):**

**Solution 8(2):**

**Solution 8(3):**

**Solution 8(4):**

**Solution 8(5):**

**Solution 8(6):**

**Solution 8(7):**

**Solution 8(8):**

**Solution 8(9):**

b. 2

Every line has at least 2 distinct points.

**Exercise 7.1:**

**Solution 1:**

**Solution 2:**

**Exercise 7.2:**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Exercise 7.3:**

**Solution 1:**

- Distance postulate: With each pair of points, there corresponds one and only one non-negative real number called the distance between these points.
- X-Y-Z symbolises that the distinct points X, Y and Z are collinear and point Y lies between the points X and Z.

Also if x, y and z are the real numbers corresponding to X, Y and Z respectively, then x > y > z or x

< y < z.

Also, X-Y-Z gives XY + YZ = XZ. - Ruler Postulate:
- Corresponding to each point on a line, there is one and only one real number.
- Corresponding to each real number, there is one and only one point on the line.
- There is a one-to-one correspondence between points on a line and real numbers such that for every pair of distinct points on the line, the positive difference of corresponding real numbers is the distance between them.

- Condition of ‘betweeness‘: If p, q, and r are real numbers corresponding to the points P, Q and R respectively, and p > q > r or p < q < r then Q is between P and R. So, we can write it as P-Q-R or R-Q-P.

**Solution 2:**

Let X ↔ x, Y ↔ y and Z ↔ z.

Then, x = 6, y = -3 and z = -1.

XY = |x – y| = |6 – (-3)| = |9| = 9

YZ = |y – z| = |(-3) – (-1) | = |-2| = 2

ZX = |z – x| = |(-1) – 6 | = |-7| = 7

**Solution 3:**

Let P ↔ p, Q ↔ q and R ↔ r.

Then, p = 7, q = -3 and r = 3.

Now, we know, -3 < 3 < 7.

∴ q < r < p

Hence, point R lies between the points P and Q,

i.e. Q-R-P or P-R-Q.

OR

Le P ↔ p, Q ↔ q and R ↔ r.

Then, p = 7, q = -3 and r = 3.

PQ = |p – q| = |7 – (-3)| = |10| = 10

QR = |q – r| = |(-3) – 3| = |-6| = 6

RP = |r – p| = |3 – 7 | = |-4| = 4

Thus, we have PQ = QR + RP

Hence, point R lies between the points P and Q,

i.e. Q-R-P or P-R-Q.

**Solution 4:**

Let A ↔ a and B ↔ b.

Then, a = -3 and AB = 5

AB = |a – b|

∴ 5 = |-3 – b|

∴ -3 – b = 5 or -3 – b = -5

∴ -b = 5 + 3 or -b = -5 + 3

∴ -b = 8 or -b = -2

∴ b = -8 or b = 2

Thus, the number corresponding to B is -8 or 2.

**Solution 5:**

Given: A-B-C, BC = 3 and AC = 9

∴ AB + BC = AC

∴ AB + 3 = 9

∴ AB = 9 – 3

∴ AB = 6

**Solution 6:**

**Solution 7:**

**Exercise 7.4:**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**