Point Sections Or Ratio Formulae: 41 Critical Solutions


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 (x1,y1) and (x2,y2) respectively. Then Coordinates of P are 

[latex]\textbf{}\textbf{x}\textbf{=}\frac{\textbf{m}\textbf{x}_{2}\textbf{+}\textbf{n}\textbf{x}_{1}}{\textbf{m}\textbf{+}\textbf{n}}[/latex]

and

[latex]\textbf{}\textbf{y}\textbf{=}\frac{\textbf{m}\textbf{y}_{2}\textbf{+}\textbf{n}\textbf{y}_{1}}{\textbf{m}\textbf{+}\textbf{n}}[/latex]

(See formulae chart)

Using this formula we can say , (x1,y1) ≌(1,1) i.e.   x1=1, y1=1   ;

(x2,y2)≌(4,1)   i.e.   x2=4, y2=1   

and

m:n  ≌ 1:2     i.e   m=1,n=2

Graphical Representation

Therefore,       

x  =[latex]\mathbf{\frac{\left ( 1\times x2\right )+\left ( 2\times x1 \right )}{1+2}}[/latex] ( putting values of m & n in     [latex]\textbf{}\textbf{x}\textbf{=}\frac{\textbf{m}\textbf{x}_{2}\textbf{+}\textbf{n}\textbf{x}_{1}}{\textbf{m}\textbf{+}\textbf{n}}[/latex] )

Or, x  =[latex]\mathbf{\textbf{}\tfrac{1×4+2×1}{3}}[/latex] ( putting values of x1 &  x2 too )

Or, x  = [latex]\mathbf{\tfrac{4+2}{3}}[/latex]

Or, x  = [latex]\mathbf{\textbf{}\tfrac{6}{3}}[/latex]

 Or, x = 2

Similarly we get,  

y =[latex]\mathbf{\frac{\left ( 1\times y2 \right )+\left ( 2\times y1 \right )}{1+2}}[/latex] ( putting values of m & n in     y  =[latex]\mathbf{\frac{my2+ny1}{m+n}}[/latex])

Or, y =[latex]\mathbf{\frac{\left ( 1\times 1 \right )+\left ( 2\times 1 \right )}{3}}[/latex] ( putting values of y1 &  y2 too )

Or, y = [latex] \mathbf{\frac{\left ( 1\times 1+2 \right )}{3}}[/latex]

Or, y  =  [latex]\mathbf{\frac{3}{3}}[/latex]

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 (x1,y1) and (x2,y2) respectively,then the coordinates of the point P are 

[latex]\textbf{}\textbf{x}\textbf{=}\frac{\textbf{m}\textbf{x}_{2}\textbf{-}\textbf{n}\textbf{x}_{1}}{\textbf{m}\textbf{-}\textbf{n}}[/latex]

and

[latex]\textbf{}\textbf{y}\textbf{=}\frac{\textbf{m}\textbf{y}_{2}\textbf{-}\textbf{n}\textbf{y}_{1}}{\textbf{m}\textbf{-}\textbf{n}}[/latex]

(See formulae chart)

Using this formula we can say ,  (x1,y1) ≌(2,1)  i.e.  x1=2, y1=1   ;

                                                    (x2,y2)≌(6,1)  i.e.   x2=6, y2=1    and   

                                                    m:n  ≌ 3:1    i.e.    m=3,n=1   

Point sections
Graphical Representation

Therefore, 

x =[latex]\mathbf{\frac{\left ( 3\times x2 \right )-\left ( 1\times x1 \right )}{3-1}}[/latex] ( putting values of m & n in     x  =[latex]\mathbf{\frac{mx2-nx1}{m-n}}[/latex])

Or, x =[latex]\mathbf{\frac{\left ( 3\times 6 \right )-\left ( 1\times 2 \right )}{2}}[/latex] ( putting values of x1 &  x2 too )

Or, x[latex] \mathbf{\frac{18-2}{2}}[/latex]

Or, x  =  [latex]\mathbf{\frac{16}{2}}[/latex]

Or, x = 8

Similarly we get,  

y =[latex]\mathbf{\frac{\left ( 3\times y2 \right )-\left ( 1\times y1 \right )}{3-1}}[/latex] ( putting values of m & n in     y  =[latex]\mathbf{\frac{my2-ny1}{m-n}}[/latex])

Or, y  =[latex]\mathbf{\frac{\left ( 3\times 1 \right )-\left ( 1\times 1 \right )}{2}}[/latex] ( putting values of y1 &  y2 too )

Or, y = [latex] \mathbf{\frac{3-1}{2}}[/latex]

Or, y  =  [latex]\mathbf{\frac{2}{2}}[/latex]

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(x1,y1) and B(x2,y2) .Then coordinates of R are

[latex]\textbf{}\textbf{x}\textbf{=}\frac{\textbf{x}_{1}\textbf{+}\textbf{x}_{2}}{\textbf{2}}[/latex]

and

[latex]\textbf{}\textbf{y}\textbf{=}\frac{\textbf{y}_{1}\textbf{+}\textbf{y}_{2}}{\textbf{2}}[/latex]

(See formulae chart)

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

Using this formula we can say ,  (x1,y1) ≌(-1,2)  i.e.  x1=-1, y1=2   and

                                                    (x2,y2)≌(1,2)  i.e.   x2=1, y2=2

Graphical Representation

Therefore,

x =[latex]\mathbf{\frac{\left ( -1 \right )+1}{2}}[/latex] ( putting values of x1 &  x2  in x =[latex]\mathbf{\frac{x1+x2}{2}}[/latex])

Or, x  =  [latex]\mathbf{\frac{0}{2}}[/latex]

Or, x = 0

Similarly we get, 

y =[latex]\mathbf{\frac{2+2}{2}}[/latex] ( putting values of y1 &  y2  in y =[latex]\mathbf{\frac{x1+x2}{2}}[/latex])

Or, y [latex]\mathbf{\frac{4}{2}}[/latex]

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 (1,32).

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)

Graphical Representation

Using this formula we can say , 

 h = 3 units and b = √ [(x2-x1)2+(y2-y1)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(x1,y1) , B(x2,y2) and C(x3,y3) be the vertices of a Triangle,

Area of the Triangle is  =|½[x1 (y2  y3) + x2 (y3  y2) + x3 (y2– y1)]|

(See formulae chart)

Using this formula we have , 

                                              (x1,y1) ≌(1,2) i.e.   x1=1, y1=2   ;

                                              (x2,y2) ≌(5,3)  i.e.   x2=5, y2=3 and

                                              (x3,y3) ≌(3,5)  i.e.    x3=3, y3=5

Graphical Representation

Therefore, the area of the triangle is = |½[x1 (y2  y3) + x2 (y3  y1) + x3 (y1-y2)]| 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

= 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

 _______________________________________________________________________________________

For more post on Mathematics, please follow our Mathematics page.

NASRINA PARVIN

I am Nasrina Parvin, Having 10 years of experience working in Ministry of communication and information technology of India . I have done Graduation in Mathematics. In my free time I love to teach, solve math problems. From my childhood Math is the only subject which fascinated me the most.

Recent Posts