Question

Find the ratio in which point (a,7) divides AB.Given A(-5,6) and B(4,10).Also fined value of a.

Answer 1

the formula for internal division is x= mx(2)+nx(1)/m+n , y= my(2)+ny(1)/m+n
here, m and n are unknown 
x=a, y=7
x(1)=-5 y(1)=6
x(2)=4 y(2)=10
by substituting in the y =my(2)+ny(1)/m+n formula ( as we dont know a )
we get, 
7=10m+6n/m+n
7m+7n=10m+6n
3m=n
m/n= 1/3= 1:3
now let m=1 and n=3, 
substitute in terms of x formula
a=1*4+3*-5/1+3
a=4+(-15)/4
a=-11/4
hope this helped

Answer 2

if A(x1,y1) = (-5,6) ,
B(x2,y2) = (4,10)
(a,7) dividing joining of the AB in the ratio of k:1
by section formula

i) a= (k*4 +1*(-5) /(k+1)      since a = (kx2+1*x1)/(k+1)
a= (4k -5)/(k+1)----(1)

ii) 7 = (k*10 +1*6) /(k+1)

7(k+1) = (10k+6)
⇒7k+7 = 10k+6
⇒7k-10k =6-7
∴-3k =-1
k= -1/3
ratio = k:1 = -1:3
ii) put k= 1/3 in (1)
a= (4*1/3-5)/(1/3+1)
a = (4-15) / (1+3)
a= -11/4