**Basic Examples on the Formulae “Point sections or Ratio”**

**Case-I**

**Problems 21: Find the coordinates of the point P(x, y) which internally divides the line segment joining the two points (1,1) and (4,1) in the ratio 1:2.**

**Solution: **We already know,

If a point **P(x, y)** divides the line segment AB** internally** in the ratio **m:n,**where coordinates of **A** and **B** are **(x _{1},y_{1})** and

**(x**respectively. Then Coordinates of P are

_{2},y_{2})and

**(See formulae chart)**

Using this formula we can say , **(x _{1},y_{1})** ≌(1,1) i.e.

**x**=1,

_{1}**y**1 ;

_{1}=**(x _{2},y_{2})**≌(4,1) i.e.

**x**=4,

_{2}**y**=1

_{2}and

**m:n** ≌ 1:2 i.e **m=1,n=2**

Therefore,

**x =**

**(** putting values of m & n in

Or, **x =**1*4+2*1/3 **( **putting values of **x _{1}** &

**x**too

_{2}**)**

Or, **x =** 4+2/3

Or, **x =** 6*3

** **Or**,** **x = 2**

**Similarly we get,**

**y** **=**

**(** putting values of m & n in ** y =**

Or, **y =**(1*1+2*1)/3 **(** putting values of **y _{1}** &

**y**too

_{2}**)**

Or, **y = **1*1+2/3

Or, **y = **3/3

Or, **y = 1**

**Therefore, **x=2 and y=1 are the coordinates of the point P i.e. (2,1). **(Ans)**

**More answered problems are given below for further practice using the procedure described in above problem 21:-**

**Problem 22: **Find the coordinates of the point which internally divides the line segment joining the two points (0,5) and (0,0) in the ratio 2:3.

**Ans. (0,2)**

**Problem 23: **Find the point which internally divides the line segment joining the points (1,1) and (4,1) in the ratio 2:1.

**Ans. (3,1)**

**Problem 24: **Find the point which lies on the line segment joining the two points (3,5,) and (3,-5,) dividing it in the ratio 1:1

**Ans. (3,0)**

**Problem 25: **Find the coordinates of the point which internally divides the line segment joining the two points (-4,1) and (4,1) in the ratio 3:5

**Ans. **(-1,1)

**Problem 26: **Find the point which internally divides the line segment joining the two points (-10,2) and (10,2) in the ratio **1.5** **: 2.5**.

**_____________________________**

**Case-II**

**Problems 27:**** Find the coordinates of the point Q(x,y) which externally divides the line segment joining the two points (2,1) and (6,1) in the ratio 3:1.**

**Solution: **** **We already know,

If a point **Q(x,y)** divides the line segment AB** externally** in the ratio **m:n,**where coordinates of **A** and **B** are **(x _{1},y_{1})** and

**(x**respectively,then the coordinates of the point P are

_{2},y_{2})and

**(See formulae chart)**

Using this formula we can say , ** (x**_{1}**,y**_{1}**)** ≌(2,1) i.e. ** x**** _{1}**=2,

**y**

_{1}**=**1 ;

**(x**_{2}**,y**_{2}**)**≌(6,1) i.e. ** x**** _{2}**=6,

**y**

**=1 and**

_{2}** m:n** ≌ 3:1 i.e. **m=**3**,n=**1

Therefore,

**x** **=**

**(** putting values of m & n in ** x =**

Or, ** x =**(3*6)-(1*2)/2

**(**putting values of

**&**

**x**_{1}**too**

**x**_{2}**)**

Or, ** x = **18-2/2

Or, **x** =16/2

Or, **x** = 8

**Similarly we get,**

**y** **=**

**(** putting values of m & n in ** y =**

Or, **y =**

**(** putting values of **y _{1}** &

**y**too

_{2}**)**

Or, **y = **3-1/2

Or, **y = **2/2

Or, **y = 1**

**Therefore,** x=8 and y=1 are the coordinates of the point Q i.e. **(8,1)**. **(Ans)**

**More answered problems are given below for further practice using the procedure described in above problem 27:-**

**Problem 28: **Find the point which externally divides the line segment joining the two points (2,2) and (4,2) in the ratio **3** **: 1**.

**Ans. (5,2)**

**Problem 29: **Find the point which externally divides the line segment joining the two points (0,2) and (0,5) in the ratio **5:2**.

**Ans. (0,7)**

**Problem 30: **Find the point which lies on the extended part of the line segment joining the two points (-3,-2) and (3,-2) in the ratio **2** **: 1**.

**Ans. (9,-2)**

**________________________________**

**Case-III**

**Problems 31: **** Find the coordinates of the midpoint of the line segment joining the two points (-1,2) and (1,2).**

**Solution: **We already know,

If a point **R(x,y) **be the midpoint of the line segment joining **A(x _{1},y_{1})** and

**B(x**Then coordinates of

_{2},y_{2}) .**R**are

and

**(See formulae chart)**

**Case-III is the form of case-I while m=1 and n=1**

Using this formula we can say , ** (x _{1}**

**,y**

_{1}**)**≌(-1,2) i.e.

**x**=-1,

_{1}**y**

_{1}**=**2 and

**(x _{2},y_{2})**≌(1,2) i.e.

**x**=1,

_{2}**y**=2

_{2}Therefore,

**x** **=**

**(** putting values of **x _{1}** &

**x**in

_{2 }**x**

**=**

Or, ** x = **0/2

Or, ** x =** 0

**Similarly we get,**

**y** **=**2+2/2 **(** putting values of ** y_{1}** &

**in**

**y**_{2 }**y**

**=**

Or, **y****= **4/2

Or, **y****= 2**

**Therefore,** x=0 and y=2 are the coordinates of the midpoint R i.e. (0,2). **(Ans)**

**More answered problems are given below for further practice using the procedure described in above problem 31:-**

**Problem 32: **Find the coordinates of the midpoint of the line joining the two points (-1,-3) and (1,-4).

**Ans. (0,3.5)**

**Problem 33:** Find the coordinates of the midpoint which divides the line segment joining the two points (-5,-7) and (5,7).

**Ans. (0,0)**

**Problem 34: **Find the coordinates of the midpoint which divides the line segment joining the two points (10,-5) and (-7,2).

**Ans. (1.5, -1.5)**

**Problem 35: **Find the coordinates of the midpoint which divides the line segment joining the two points (3,√2) and (1,3**√**2).

**Ans. (2,2√2)**

**Problem 36: **Find the coordinates of the midpoint which divides the line segment joining the two points (2+3i,5) and (2-3i,-5).

**Ans. (2,0)**

**Note: How to check if a point divides a line (length=d units) internally or externally by the ratio m:n **

**If ( m×d)/(m+n) + ( n×d)/(m+n) = d , then internally dividing** **and**

**If ( m×d)/(m+n) – ( n×d)/(m+n) = d , then externally dividing**

**____________________________________________________________________________**

**Basic Examples on the Formulae “Area of a Triangle”**

**Case-I**

**Problems 37: **What is the area of the triangle with two vertices **A(1,2) **and** B(5,3) **and** **height with respect to **AB** be **3** units** **in the coordinate plane ?

Solution: ** **We already know,

If** “h”** be the height and **“b”** be the base of Triangle, then **Area of the Triangle is = ½ × b × h**

**(See formulae chart)**

Using this formula we can say ,

**h **= 3 units and **b** = √ [(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2 }] i.e ** **√ [(5-1)^{2}+(3-2)^{2 }]

Or, **b** = √ [(4)^{2}+(1)^{2 }]

Or, **b** = √ [(16+1^{ }]

Or, **b** = √ 17 units

Therefore, the required area of the triangle is ** = ½ × b × h i.e**

**= ½ × (√ 17 ) × 3** **units**

**= 3⁄2 × (√ 17 ) units (Ans.)**

**______________________________________________________________________________________**

**Case-II**

**Problems 38:**What is the area of the triangle with vertices **A(1,2), B(5,3) and C(3,5) **in the coordinate plane ?

Solution: ** **We already know,

If ** A(x _{1},y_{1}) ,**

**B(x**and

_{2},y_{2})**C(x**be the vertices of a Triangle,

_{3},y_{3})**Area of the Triangle is** **=****|****½****[****x**_{1}** (y**_{2}**–**_{ }** y**_{3}**) + x**_{2}** (y**_{3}**–**_{ }** y**_{2}**) + x**_{3}** (y**_{2}**– y**_{1}**)****]****|**

**(See formulae chart)**

Using this formula we have ,

**(x**_{1}**,y**_{1}**)** ≌(1,2) i.e. ** x**** _{1}**=1,

**y**

_{1}**=**2 ;

**(x _{2}**

**,y**

_{2}**)**≌(5,3) i.e.

**x**=5,

_{2}**y**=3 and

_{2}** (x**_{3}**,y**_{3}**) **≌(3,5) i.e. **x**** _{3}**=3,

**y**

**=5**

_{3}Therefore, the area of the triangle is = **|½[x _{1} (y_{2}–_{ } y_{3}) + x_{2} (y_{3}–_{ } y_{1}) + x_{3} (y_{1}-y_{2})]| **i.e

= **|½[1 (3-5) + 5 (5-3) + 3 (3-2)]| **sq.units** **

= **|½[ 1x (-2) + (5×2) + (3×1)]| **sq.units

= **|½[-2 + 10 + 3]| **sq.units

= **|½ **x 11**| **sq.units

= **11⁄2 ** sq.units

=** 5.5** sq.units **(Ans.)**

**More answered problems are given below for further practice using the procedure described in above problems :-**

**Problem 39: **Find the area of the triangle whose vertices are (1,1), (-1,2) and (3,2).

**Ans. 2 **sq.units

**Problem 40: **Find the area of the triangle whose vertices are (3,0), (0,6) and (6,9).

**Ans. 22.5 **sq.units

**Problem 41: **Find the area of the triangle whose vertices are (-1,-2), (0,4) and (1,-3).

**Ans. 6.5 **sq.units

**Problem 42: **Find the area of the triangle whose vertices are (-5,0,), (0,5) and (0,-5). **Ans. 25 **sq.units

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