Question

Find the ratio in which P (4, m) divides the line
segment joining the points A (2, 3) and B (6,-3).
Hence find m.
[2]
AD in be​

Answer 1

Step-by-step explanation:

Given:-

P (4, m) divides the linesegment joining the points A (2, 3) and B (6,-3).

To find:-

Find the ratio in which P (4, m) divides the line

segment joining the points A (2, 3) and B (6,-3).

Hence find m.

Answer:-

Given point P(4,m)

Given points A(2,3) and B(6,-3)

Let (x1, y1)=(2,3)=>x1=2 and y1=3

(x2, y2)=(6,-3)=>x2=6 and y2=-3

we know that the point P(x,y) divides the the linesegment joining the points A(x1, y1) and B(x2, y2) in the ratio m1:m2 is

=[(m1x2+m2x1)/(m1+m2) ,(m1y2+m2y1)/(m1+m2)]

=>[(6m1+2m2)/(m1+m2) , (-3m1+3m2)/(m1+m2)]

=(4,m)

On comparing both sides then

=>(6m1+2m2)/(m1+m2)=4

=>6m1+2m2=4(m1+m2)

=>6m1+2m2=4m1+4m2

=>6m1-4m1=4m2-2m2

=>2m1=2m2

=>m1=m2

=>m1/m2=1/1

=>m1:m2=1:1

The ratio =1:1

now (-3m1+3m2)/(m1+m2)=m

Substituting the values of m1 and m2 then

=>[-3(1)+3(1)]/[1+1]=m

=>(-3+3)/2=m

=>0/2=m

=>0=m

=>m=0

Answer:-

The required ratio =1:1

The value of m=0

Used formula:-

the point P(x,y) divides the the linesegment joining the points A(x1, y1) and B(x2, y2) in the ratio m1:m2 is

[(m1x2+m2x1)/(m1+m2) ,

(m1y2+m2y1)/(m1+m2)]

Answer 2

Section formula if a point P(x,y) divides the line segment joining A( x1 ,y1 )B(x2 ,y2) in ratio of m:n

Then,

$\sf \: (x,y ) = ( \frac{mx1 + nx2}{m + n} , \frac{my1 + ny2}{m + n} )$

Let P divides AB in ratio of k:1

$\sf \: 4 = \frac{6k + 2}{k + 1}$

$\sf4k + 4 = 6k + 2$

$\sf \: k = \color{blue}1$

Again using section formula;

$\sf \: m = \frac{ - 3.1 + 3.1}{1 + 1}$

$\sf \: m = \color{purple}0$