## myCBSEguide App

CBSE, NCERT, JEE Main, NEET-UG, NDA, Exam Papers, Question Bank, NCERT Solutions, Exemplars, Revision Notes, Free Videos, MCQ Tests & more.

Install Now

**Exercise 12.2**

**1. Find the distance between the following pairs of points:**

**(i) (2, 3, 5) and (4, 3, 1) **

**(ii) and **

**(iii) and **

**(iv) and **

**Ans. (i)** Let A (2, 3, 5) and B (4, 3, 1) be two points, then

AB =

= = units

**(ii)** Let A (–3, 7, 2) and B (2, 4, –1) be two points, then

AB =

= = units

**(iii)** Let A (–1, 3, –4) and B (1, –3, 4) be two points, then

AB =

= = units

**(iv)** Let A (2, –1, 3) and B (–2, 1, 3) be two points, then

AB =

= = units

**2. Show that the points **** and **** are collinear.**

**Ans. **Let A (–2, 3, 5), B (1, 2, 3) and C (7, 0, –1) be three points, then

AB =

= = units

BC =

= = units

AC = =

= = units

Here, AC = AB + BC

Therefore A, B and C are collinear.

**3. Verify the following:**

**(i) **** and **** are the vertices of an isosceles triangle.**

**(ii) **** and **** are the vertices of right angled triangle.**

**(iii) **** and **** are the vertices of a parallelogram.**

**Ans. (i)** Let A (0, 7, –10) B (1, 6, –6) and C (4, 9, –6) be three vertices of , then

AB =

= = units

BC =

= = units

AC = =

= = units

Here, AB = BC

Therefore is an isosceles triangle.

**(ii)** Let A (–1, 2, 1), B (–1, 6, 6) and C (–4, 9, 6) be three vertices of , then

AB =

= = units

BC =

= = units

AC = =

= = units

Here, AC^{2} = AB^{2} + BC^{2}

Therefore is a right angled triangle.

**(iii)** Let A (0, 7, 10), B (1, –2, 5), C (4, –7, 8) and D (2, –3, 4) be four vertices of a quadrilateral ABCD, then

AB =

= = units

BC =

= = units

CD =

= = units

AD =

= = units

AC =

= = units

BD =

= = units

Here, AB = CD, BC = AD and AC BD

Therefore A, B, C and are the vertices of a parallelogram ABCD.

**4. Find the equation of the set of points which are equidistant from the point (1, 2, 3) and **

**Ans. **Let A be any point which is equidistant from points B (1, 2, 3) and C Then

According to question, AB = AC

Squaring both sides, we get

**5. Find the equation of the set of points P, the sum of whose distance from A (4, 0, 0) and B**** is equal to 10.**

**Ans. **Let P be any point, then

According to question, PA + PB = 10

Squaring both sides, we get

Again squaring both sides, we get

This is the required equation.

Thank you

But show us some other process of 5 ex12.2

This is a fentastic website.

I am very thankful to u

Thanks once again

???????

Thanks uu so much

Thanks

Thanx sir

very helpful for me

because i have problem in 5 question

Easily understoodable…