EXAMPLE PROBLEMS USING SECTION FORMULA

Problem 1 :

Find the coordinates of the point which divides the line segment joining the points A(4,−3) and B(9, 7) in the ratio 3:2.

Solution :

=  (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)

=  (3(9) + 2(4))/(3 + 2), (3(7) + 2(-3))/(3 + 2)

=  (27 + 8)/5, (21 - 6))/5

=  35/5, 15/5

=  (7, 3)

Problem 2 :

In what ratio does the point P(2,−5) divide the line segment joining A(−3, 5) and B(4, −9).

Solution :

=  (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)

=  (m(4) + n(-3))/(m + n), (m(-9) + n(5))/(m + n)

(4m - 3n)/(m + n), (-9m + 5n)/(m + n)  =  (2, -5)

By equating x and y-coordinates, we get

(4m - 3n)/(m + n)  =  2

4m - 3n  =  2m + 2n

4m - 2m  =  2n + 3n

2m  =  5n

m/n  =  5/2

m : n  =  5 : 2

Hence the required ratio is 5 : 2 

Problem 3 :

Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that AP = (2/5) AB.

Solution :

AP = (2/5) AB.

AP/AB  =  2/5

AB  =  5

AP + PB  =  AB

2 + PB  =  5

PB  =  5 - 2  =  3

So, the point P divides the line segment in the ratio 2 : 3.

A(1, 2) and B(6, 7)

P  =  (2(6) + 3(1))/(2 + 3), (2(7) + 3(2))/(2 + 3)

P  =  (15/5, 20/5)

P  =  (3, 4)

Hence the required point is (3, 4).

Problem 4 :

If the points A(4, 3) and B(x, 5) are on circle with center O(2, 3), then find the value of x.

Solution :

The points A and B are two points lie on the circle, distance between center and any points on the circle which lies on the boundary will be the measure of radius. So, they will be equal.

= √(x₂ - x₁)² + (y₂ - y₁)²

√(2 - 4)² + (3 - 3)² = √(2 - x)² + (3 - 5)²

√4 = √(2 - x)² + (-2)²

2 = √(4 - 4x + x² + 4)

4 = 4 - 4x + x² + 4

0 = x² - 4x + 4

x² - 4x + 4 = 0

(x - 2)(x - 2) = 0

x = 2

So, the value of x is 2.

Problem 5 :

Find the ratio in which the line 2x + 3y = 10 divide the line segment joining the points (1, 2) and (2, 3).

Solution :

Equation of line segment joining the points (1, 2) and (2, 3).

Let the required ratio be m : n

Section formula =  (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)

= (2m + 1n)/(m + n), (3m + 2n)/(m + n)

The point lies on the line 2x + 3y = 10

2[(2m + n)/(m + n)] + 3[(3m + 2n)/(m + n)] = 10

4m + 2n + 9m + 6n = 10(m + n)

13m + 8n = 10m + 10n

13m - 10m = 10n - 8n

3m = 2n

m/n = 2/3

m : n = 2 : n

So, the required ratio is 2 : 3.

Problem 6 :

Determine the ration in which the point P(a, -2) divides the line joining of points A(-4, 3) and B(2, -4). Also find the value of a.

Solution :

From the information above, we know that the point P divides the line segment joining the points A and B.

(mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n) = (a, -2)

(m(2) + n(-4))/(m + n), (m(-4) + n(3))/(m + n) = (a, -2)

(2m - 4n)/(m + n), (-4m + 3n)/(m + n) = (a, -2)

Equating the x and y-values, we get

(2m - 4n)/(m + n) = a ----(1)

(-4m + 3n)/(m + n) = -2

-4m + 3m = 2(m + n)

-4m + 3n = 2m + 2n

-4m - 2m = 2n - 3n

-6m = -1n

m/n = 1/6

m : n = 1 : 6

So, the required ratio is 1 : 6.

Applying this ratio in (1), we get

[-4(1) + 3(6)]/(1 + 6) = a

a = (-4 + 18)/7

a = 14/7

a = 2

So, the value of a is 2.

Problem 7 :

If the point C(-1, 2) divides internally the line segment joining A(2, 5) and in the ratio 3 : 4. Find the coordinate of B.

Solution :

Let (a, b) be the required point B.

= (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)

(3a + 4(2))/(3 + 4), (3b + 4(5))/(3 + 4) = (-1, 2)

(3a + 4(2))/(3 + 4), (3b + 4(5))/(3 + 4) = (-1, 2)

(3a + 8)/7, (3b + 20)/7 = (-1, 2)

So, the required point is (-5, -2).

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