HOW TO FIND THE RATIO IN WHICH A POINT DIVIDES A LINE

Example 1 :

In what ratio does the point P(-2, 3) divide the line segment joining the points A(-3, 5) and B (4, -9) internally?

Solution :

Given points are (-3 , 5) and B (4 ,- 9).

Let P (-2, 3) divide AB internally in the ratio l : m.

By the section formula,

[(lx₂ + mx₁)/(l + m),  (ly₂ + my₁)/(l + m)] = (-2, 3)

Substitute (x₁, y₁) = (-3, 5) and (x₂, y₂) = (4, -9).

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

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

Equating the coefficients of x, we get

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

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

4l - 3m = -2l - 2m

Add 2l and 3m on both sides

6l = m

l/m = 1/6

l : m = 1 : 6

Hence the point P divides the line segment joining the points in the ratio 1 : 6.

Example 2 :

Let A (-6, -5) and B (-6, 4) be two points such that a point P on the line AB satisfies AP = (2/9)AB. Find the point P.

Solution :

AP = (2/9)AB

9AP = 2AB

9AP = 2(AP + PB)

9AP = 2AP + 2PB

9AP – 2AP = 2PB

7AP = 2PB

AP/AB = 2/7

AP: PB = 2 : 7

So P divides the line segment in the ratio 2 : 7.

Section formula internally

= [(lx₂ + mx₁)/(l + m), (ly₂ + my₁)/(l + m)]

Substitute l = 2 and m = 7.    

= [(2(-6)) + (7(-6)]/(2+7), [(2(4)) + (7(-5)]/(2+7)

= [(-12-42)/9, (8 - 35)/9]

= (-54/9, -21/7)

= (-6, -3)

Example 3 :

Find the ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, -7).

Solution :

Let l : m be the ratio of the line segment joining the points (6, 4) and (1, -7) and let p(x, 0) be the point on the x axis.

= (lx₂ + mx₁)/(l + m), (ly₂ + my₁)/(l + m)

(x, 0) = [l(1) + m(6)]/(l + m), [l(-7) + m(4)]/(l + m)

(x, 0) = [l + 6 m]/(l + m), [-7l + 4m]/(l + m)

Equating y-coordinates, we get

[-7l + 4m]/(l + m) = 0

-7l + 4m = 0

-7l = -4m

l/m = 4/7

l : m = 4 : 7

Hence x-axis divides the line segment in the ratio 4 : 7. 

Example 4 :

In what ratio is the line joining the points (-5, 1) and (2, 3) divided by y-axis? Also, find the point of intersection.

Solution :

Let L : m be the ratio of the line segment joining the points (-5 , 1) and (2 ,3) and let p(0,y) be the point on the y axis

Section formula internally

= (lx₂ + mx₁)/(l + m) , (ly₂ + my₁)/(l + m)

(0 , y)  =  [L(2) + m(-5)]/(L + m) , [L(3) + m(1)]/(L + m)

(0 , y)  =  [2L - 5 m]/(L + m) , [3L + m]/(L + m)

[2L - 5 m]/(L + m) = 0

2 l - 5 m = 0

2 l = 5 m

l/m = 5/2

l : m = 5 : 2 

To find the required point we have to apply this ratio in the formula

(0, y) = [2(5) – 5(2)]/(5 + 2), [3(5) + 2]/(5 + 2)

(0 , y) = [10 – 10]/7, [15 + 2]/7

(0 , y) = (0 , 17/7)

Hence the required point is (0, 17/7).

Example 5 :

If (a/3, 4) is the midpoint of the line segment joining the points Q(-6, 5) and R(-2, 3), then find the value of a.

Solution :

It can be done using section formula or midpoint formula. Since the point (a/3, 4) is the midpoint,  the line segment divided by this point will be in the ratio of 1 : 1 .

Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

(-6 + (-2))/2, (5 + 3)/2 = (a/3, 4)

(-8)/2, 8/2 = (a/3, 4)

(-4, 4) = (a/3, 4)

Equating the x and y-coordinates, we get

a/3 = -4

a = -12

So, the value of a is -12.

Example 6 :

A line intersects y-axis and x-axis at the points P and Q respectively. If (2, -5) is the midpoint of PQ, then find the coordinates of P and Q respectively.

Solution :

Let (x, 0) be the point of x-intercept

Let (0, y) be the point of y-intercept.

(x + 0)/2, (0 + y)/2 = (2, -5)

(x/2, y/2) = (2, -5)

x/2 = 2 and y/2 = -5

x = 4 and y = -10

So, the required points are Q(4, 0) and P(0, -10).

Example 7 :

Find the rational number which y-axis divides the line segment joining (-3, 6) and (12, -3).

Solution :

Let (0, y) be the point on y-axis and l : m is the required ratio

(lx₂ + mx₁)/(l + m), (ly₂ + my₁)/(l + m) = (0, y)

(12l - 3m)/(l + m), (-3l + 6m)/(l + m) = (0, y)

12l - 3m = 0

12l - 3m = 0

3m = 12l

3/12 = l/m

1/4 = l/m

l : m is 1 : 4.

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