NEB Mathematics: Class 9 Opt Maths: Section Formula

Section Formulae

Simply, section formulae refer to the external and internal division of a line segment by a given point.
Section formulae have two types. They are,

Section formulae for an internal division.
Section formulae for an external division.

Section Formulae for Internal Division
Let's take a line with two ends point A(x1, y2) and B(x2, y2) which are joined by the line segment AB. Consider P(x, y) be any point on AB which divides the line internally in the ratio m1:m2
i.e. AP:PB = m1:m2
The formula for the section formulae in internal division is (x, y) = ( $\frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}$ , $\frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}}$

Section Formulae for External Division
If the point P(x, y) divides AB externally in the ratio of m1:m2 then the divided segment BP is measured in opposite direction and hence m2 is taken as negative.

$∴$ The section formulae for external division is,
(x, y) = ( $\frac{m_{1} x_{2} - m_{2} x_{1}}{m_{1} - m_{2}}$ , $\frac{m_{1} y_{2} - m_{2} y_{1}}{m_{1} - m_{2}}$ )

In special case, the midpoint formulae is also used'
m1:m2 = 1:1 i.e. m1 = m2
$∴$ x = $\frac{x_{1} + x_{2}}{2}$ and y = $\frac{y_{1} + y_{2}}{2}$
Thus, co-ordinates P(x, y) are P( $\frac{x_{1} + x_{2}}{2}$ , $\frac{y_{1} + y_{2}}{2}$ ) which is called mid-point formulae.

Some problems:

1. Find the co-ordinates of the points on Y-axis which are at a distance of 10 units from the point (6, 6).
Solution: Here,
Let a point on Y –axis be A(0, y) and given point is B(6, 6) and AB = 10 units.
Then,
AB = √0-62+y-62
⇒ 10 = √-62+y2-12y+36
⇒ 10 = √y2-12y+72
Squaring on both sides,
100 = y² – 12y + 72
⇒ y² – 12y – 28 = 0
⇒ y² – 14y + 2y – 28 = 0
⇒ y(y – 14) + 2(y – 14) = 0
⇒ (y – 14) (y + 2) = 0
Either, y – 14 = 0
∴ y = 14
or, y + 2 = 0
∴ y = -2
Thus, the co-ordinates of points on Y-axis are (0, 14) or (0, -2).

2. If the distance between the points (a, 2a) and (4a, 4a) is√13 units, find the value of a.
Solution:
Given, the distance between the points A (a, 2a) and B (4a, 4a) is √13 units.
That is,
AB = √13 units
Using distance formula, we have,
√a-4a2+2a-4a2=13
⇒ 9a2+4a2=13
⇒  13a2=13
⇒  a2=1
⇒    a=±1
So, a =±1.

3. The X-coordinate of a point on X-axis is 4 and Y-ordinate of a point Q on Y-axis is -3. Find the distance of PQ.
Solution: Here,
Let, the given two points are A (4, 0) and B (0, -3)
Now, using distance formula, we get,
AB = √(4-0)2+(0+3)²
= √42+32
= √25
= 5 units.

4. The two vertices of an equilateral triangle are (2, 4) and (2, 6) respectively, find the semi perimeter of the triangle.
Solution:
Here,
The two vertices of an equilateral triangle are A(2, 4) and B(2, 6).
Using distance formula, we have,
Length of the sides of the triangle (AB) = Sqrt ((2-2)2+(4-6)2)
= √02+-22
= √22
= 2 units.
Now, semi-perimeter of the equilateral triangle = 32AB=32×2=3 units
Hence, the required semi-perimeter of the equilateral triangle is 3 units.

5. If the distance between the points P(x, 0) and Q(8, 3) is 5 units, find the value of x.
Solution: Here,
The given points are P(x, 0) and Q (8, 3).
So,
PQ = 5 units
⇒   √x- 82+0-32 =5
⇒   x- 82+0-32=52
⇒  x2-16x+64+9=25
⇒  x2-16x+48=0
⇒  x2- 12x-4x+48=0
⇒  x(x-12)-4(x-12)=0
⇒  (x
- 12) (x - 4) =0
∴x=12 or 4.

6. In the figure, ABCD is a square. If the co-ordinates of A and C are (2,3) and(5,7) respectively, find the length of diagonal BD.

Solution:
Here, in the square ABCD,
Co-ordinates of A and C are (2, 3) and (5, 7)
We have,
BD = AC [Diagonals of the same square are equal]
BD = √(2 -5)2+(3 -7)2 [Using distance of formula.
= √(-3)2+(-4)2
= √25
= 5 units.

7. Find the co-ordinates of the points on Y-axis which are at a distance of 10 units from the points (6, 6).
Solution: Let, A(0, a) be the point on the Y-axis and B(6,6) be the given point.
Then, by the question,
AB = 10 units
⇒
√(0-6)2+(a – 6)2=10
Squaring both side, we get,
⇒   36+a2-12a+36=100
⇒   a2-12a-28=0
⇒   a2-14a+2a-28=0
⇒   a(a-14)+2(a-14)=0
⇒   (a+2)(a-14)=0
∴ a = -2 or 14.
Hence, the required co-ordinates are (0, -2) or (0, 14).

8. The two continuous vertices of a square are (2, 0) and (5, 0) respectively. Find the area of the square.
Solution:
Here the two continuous vertices of a square are P(2, 0) and Q(5, 0).
Hence, the length of the square = PQ
Now, the area of the square = PQ²
= (5-2)2+02
= 3²
= 9 square units.
Hence, the required area of the square is 9 square units.

9. (2, 3) is the mid- point between (a, 4) and (2, b). Find the values of a and b.
Solution:
Given, (2, 3) is the mid- point between (a, 4) and (2, b).
Now, using mid-point formula, we have,
(2, 3)= (a+22,4+b2 )
Now, equating corresponding coordinates, we have,
i. 2=a+22
⇒  a+2=4
⇒    a=4-2=2
ii.3=4+b2
⇒  6=4+b
⇒    b=6-4=2.
Hence, the required value of a and b are 2 and 2.

10. In what ratio does X-axis divide line joining the points (2,-4) and (-3, 6)?
Solution:
We know, any point on the x-axis is (x, 0).
Let, point (x, 0) divides the line joining the points (2,-4) and (-3, 6) in a ratio m: 1.
Using section formula, we get,
(x, 0)= (2-4mm+1, -4+6mm+1 )
Equating corresponding elements, we get,
0=-4+6mm+1
⇒   0=-4+6m
⇒   4=6m
∴ m=46=23
Hence, required ratio is 2:3.

11. Find the co-ordinates of the point which divides externally the line joining the points (1, 3) and (2, 7) in the ratio 3:4.
Solution:
Here, the line joining points (x_1,y_1 )=(1,3)and (x_2,y_2 )=(2,7)is externally divided in the ratio m: n = 3:4.
Then, using section formula, the co-ordinates of the point (x, y) is given by,
(x, y)= (mx2-nx1m-n,my2-ny1m-n )
= (3×2-4×14-3,3×7-4×34-3 )
= (6-41,21-121 )
= (2, 9).
Hence, the required point is (2, 9).

12. Find the equation of the locus of a point, which is equidistant from (1, 2) and (2, -3).
Solution: Here,
Let the moving point be P (x, y).

According to the question,
AP = BP
⇒   AP² = BP²
⇒   (x - 1) ² + (y - 2) ² = (x - 2) ² + (y + 3) ²
⇒    x² - 2x + 1 + y² - 4y + 4 = x² - 4x + 4 + y² + y² + 6y + 9
⇒   2x -10y = 8
⇒   x - 5y = 4 is the required equation of locus.

13. A point P moves so that its distances from the two points (3, 4) and (5, -2) are equal to one another. Find the equation of the locus of P.
Solution: Here,
Let the moving point be P (x, y) and A be (3, 4) & B is (5, -2).

According to the question,
AP = BP
⇒   AP² = BP²
⇒   (x - 3) ² + (y - 4) ² = (x - 5) ² + (y + 2) ²
⇒   x² - 6x + 9 + y² - 8y + 16 = x² - 10x + 25 + y² + 4y + 4
⇒   4x -12y = 4
⇒   x - 3y = 1 is the required equation of locus.

14. Prove that the locus of a point P(x, y) which is equidistant from the point (a + b , b-a) and (a-b, b + a) is bx = ay.
Solution: Here,
Let the point be P (x, y) And, A is (a + b, a - b) & B is (a - b, b + a)

According to the question,
AP = BP
⇒   AP² = BP²
⇒   (a + b - x) ² + (a – b - y) ² = (a – b - x) ² + (a + b - y) ²
⇒   (a + b)² - 2 (a + b)x + x ² + (a - b) ² - 2(a - b)y + y² = (a - b) ²- 2(a - b)x + x² + (a + b) ² - 2 (a + b)y + y ²
⇒   -2(a + b)x - 2(a - b)y = -2(a - b)x - 2 (a + b)y
⇒   (a + b – a + b)x = (a + b – a + b)y
⇒   2bx = 2ay
⇒   bx = ay .
Proved.

15. What do you mean by the locus of a moving point? If the distance of a moving point from (2, -3) is twice its distance from the point (0, 2), find the equation of the locus of the moving point.
Solution:
Locus is a path traced by a moving point.

Let P(x, y) be the moving point,

Then,
2AP = BP
4AP² = BP²
4 [(x - 0) ² + (y - 2) ² = (x - 2) ² + (y + 3) ²
⇒  4x² + 4y² - 16y + 16 = x² - 4x + 4 + y² + 6y + 9
⇒ 3x² + 3y² + 4x - 22y + 3 = 0 is the required equation of locus.

16. Find the coordinates of two points which trisect the line segment joining points (5, 3) and (8, 6).
Solution: Here,
Let the given point be P(5, 3) & Q (8, 6)

For the trisection,
PB: AQ = 1:2. = m₁: m₂
∴ Co- ordinate of A is = m 1 x 2 + m 2 x 1m 1+m 2 m 1 y 2 + m 2 y 2m 1+m 2
= ( 1 × 8 + 2 × 5 ,1 + 2 1 × 6+2 × 31 + 2 )
= (6, 4)
But B is midpoint of A & Q.
So, co- ordinate of B is = (6+82, 6+42)
= (7, 5).

17. If the point (x, 14) divides the line joining the points (7, 11) and (-18, 16), find the value of x.
Solution: Here,

Let the given points P (x, 14) divides the line AB in the ratio, m₁: m₂.
∴ (x, y) = (m 1 x 2 + m 2 x1m 1+m 2, m 1 y 2 + m 2 y 2m 1+m 2 )
(x, 4) = ( m 1 × -18 + m 2 × 7m 1+ m 2 m 1× 16 + m 2 × 11m 1+ m 2 )
i.e. x = -18m + 7m 2m 1+ m 2 ……..(i)
and 14 = 16m 1+ 11m 2m 1+ m 2
Taking 14 = 16m 1+11m 2m 1+m 2
⇒  14 m ₁ + 14 m ₂ = 16 m ₁ + 11 m ₂
⇒  14 m ₂ - 11 m ₂ = 16 m ₁ - 14 m ₁
⇒  3 m ₂ = 2 m ₁
⇒   m 1m 2 = 32
Put m 1 = 3 & m 1 = 2 in (i)
x = -18 × 2 + 7 × 32 + 3
= 155
= 3
∴ x = 3

18. A variable point moves so that the ratio of its distance from the point (-5,0) and (5,0) is 4:5. Find the equation of locus of the point.
Solution: Here,
Let P (x, y) be the moving point & A is (-5, 0) & B is (5, 0).

Given, PA : PB = 4 : 5
⇒   PAPB = 45
⇒   (5PA)² = (4PB)²
⇒   25(PA) ² = 16(PB)²
⇒  25[(x + 5)² + (4 - 0) ²] = 16[(x - 5) ² + (4 - 0) ²]
⇒  25[x² + 10x + 25 + y ²] = 16[x² - 10x +25 + y ²]
⇒  25x² + 250x + 625 + 25y² = 16x² - 160x + 400 + 16y ²]
⇒  9x² + 9y² + 410x + 225 = 0 is the required equation of Locus.

19. Three continuous corners of parallelogram are (8, 5), (-7,-5) and (-5,-5 ). Find the co- ordinates of the remaining corner.
Solution: Here,
Let A (8, 5) B (-7, -5) & C (-5, -5) and D (x₄, y₄) are the vertices of a parallelogram.

Then midpoint of AC
= (8 – 52, 5 -52) = 32, 0
Midpoint of BD = (-7 + x 42, -5 + y 42)
In parallelogram,
Midpoint of AC = mid -point of BD
i,e (32, 0) = (-7 + x 42, -5 + y 42)

Equating corresponding components, we get
32 = -7 + x 42
⇒  x₄ = 10
& 0 = -5 + y 42
⇒  y ₄ = 5.

∴ 4^th vertex of parallelogram is (10, 5) .

20. If the points (-1, 3), (1,-1) and (5,1) are the vertices of a triangle then find the length of median drawn from (-1,3).
Solution:

Here,
AD is median and D is mid- point of BC,
∴ coordinate of D = (-1 + 52, 1 + 12)
= (2, 1)
∴ Length of AD = √x 2-x 12+y 2-y 12
= √(2—1) 2+(1-3) 2
= √32+-22
= √13 Units.

21. A variable point moves so that the ratio of its distance from the points (-7,0) and (7,0) is 4:5 , find the equation of the locus of the point.
Solution:
Let P (x, y) be the moving point & A (-7, 0), B (7, 0).

According to question,
PA: PB = 4:5
⇒   PAPB = 45
⇒  5PA = 4PB
⇒  25(PA)² = 16(PB) ² [∵ Squaring both sides)
⇒  25[ (x + 7)² + (y - 0)² ] = 16 [ (x -7)² + (y - 0)² ]
⇒  25[ (x² + 14x + 49 + y² ] = 16 [ x² - 14x + 49 + y² ]
⇒  25x² + 350x + 1225 + 25y² = 16x² - 224x + 736 + y²
⇒  9x² + 9y² + 574x + 341 = 0 is the required equation of locus.