FIND THE RATIO THAT X AXIS DIVIDES THE LINE SEGMENT JOINING THE POINTS

The coordinates of the point P(x, y) which divides the line segment joining the points A (x1, y1) and B (x2, y2) internally in the ratio l : m

lx2 + mx1l + m, ly2 + my1l + m
line-segment-divides-x-axis

Problem 1 :

Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution:

Given, A(1, -5) and B(-4, 5)

Let the x axis divide the line segment at point (x, 0) in the ratio k : 1.

Using section formula,

(x,0)=k(-4)+1k+1,k(5)+(-5)k+1On comparing ordinates, we get0=k(5)-5k+15k-5=05k=5k=1

So, the required ratio is 1 : 1.

Here, the point of division is P(x, 0).

On comparing x coordinate,

x=k(-4)+1k+1=1(-4)+11+1=-32P(x,0)=-32,0

Hence, the required ratio is 1 : 1 and the coordinates of the point of division is (-3/2, 0).

Problem 2 :

In what ratio is the line segment joining A(6, 3) and B(-2, -5) is divided by the x-axis. Also find the coordinates of the point of intersection of AB and the x-axis.

Solution:

Using the section formula, if a point (x, y) divides the line joining the points (x1, y1) and (x2, y2) in the ratio of l : m, then

(x,y)=lx2+mx1l+m,ly2+my1l+m

Let point P on the x axis divides the line segment joining the points A and B the ratio l : m

Consider P lies on x axis having coordinates (x, 0).

(x,0)=l(-2)+m(6)l+m,l(-5)+m(3)l+mx=-2l+6ml+mx(l+m)=-2l+6mlx+mx=-2l+6mlx+2l=6m-mxl(x+2)=m(6-x)Equating the y coefficient0=-5l+3ml+m-5l+3m=0-5l=-3mlm=35l:m=3:5

Then,

3(x + 2)=5(6 - x)

3x + 6 = 30 - 5x

8x = 24

x = 3

Hence, the coordinates are (3, 0).

Problem 3 :

In what ratio is the line segment joining A(2, -3) and B(5, 6) is divided by the x-axis. Also find the coordinates of the point of intersection of AB and the x-axis.

Solution:

Using the section formula, if a point (x, y) divides the line joining the points (x1, y1) and (x2, y2) in the ratio of l : m, then

(x,y)=lx2+mx1l+m,ly2+my1l+m

Let point P on the x axis divides the line segment joining the points A and B the ratio l : m

Consider P lies on x axis having coordinates (x, 0).

(x,0)=l(5)+m(2)l+m,l(6)+m(-3)l+mx=5l+2ml+mxl+xm=5l+2m(x-5)l=(2-x)mEquating the coefficient of y, we get0=6l-3ml+m6l-3m=06l=3mlm=36lm=12

l : m = 1 : 2

Then,

(x - 5)(1) = (2 - x)(2)

x - 5 = 4 - 2x

3x = 9

x = 3

Hence, the coordinates are (3, 0).

Problem 4 :

Find the point P on the x-axis which is equidistant from the points A(5, 4) and B(-2, 3). Also find the area of ΔPAB.

Solution:

Let the point of x axis be P(x, 0) = (x1, y1)

Given, A(5, 4) and B(-2, 3) are equidistant from P.

PA = PB

PA2 = PB2 --> (1)

Distance between two points is

(x2-x1)2+(y2-y1)2PA = (5-x)2+(4-0)2=x2+25-10x+16=x2-10x+41PA2=x2-10x+41Here P(x,0)=(x1,y1),B(-2,3)=(x2,y2)Similarly PB=(-2-x)2+(3-0)2=x2+4+4x+9=x2+4x+13PB2=x2+4x+13

From equation (1)

x2 - 10x + 41 = x2 + 4x + 13

-10x - 4x = 13 - 41

-14x = -28

x = 2

Therefore, coordinates of P is (2, 0).

Now, Area of  ΔPAB

=12|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|=12|2(4-3)+5(3-0)+(-2)(0-4)|=12|2+15+8|=252=12.5

Hence, point P is (2,0) and area of ΔPAB is 12.5 square unit. 

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