Question

4. Given the points M(-3, -4) and T(5,0), find the coordinates of the point Q on directed line segment MT that partitions MT in the ratio 2:3.
5. Points A,B and C are collinear on AC, and AB:BC = 3/4. A is located at (x,y), B is located at (4, 1) and C is located at (12, 5). What are the values of x and y?

Answer 1

Answer:

4 : Coordinate of Q = $(\frac{1}{5} ,\frac{-12}{5} )$

5 :  Coordinate of point A = (-2,-2)

Explanation :

4)  Given the points M(-3, -4) and T(5,0)

x₁= -3, y₁ = -4 and

x₂ = 5, y₂ = 0

m = 2, n = 1

Now apply the section formula,

{(mx₂+nx₁)/(m+n) , (my₂+ny₁)/(m+n)}

Coordinate of point Q = $(\frac{1}{5} ,\frac{-12}{5} )$

5)   AB:BC = 3:4

A = (x,y),   end point

B =  (4, 1)  

C = (12, 5)      end point

So, m = 3   &   n  = 4

x₁ = x ,  y₁ = y   , x₂=12  & y₂ = 5

Apply section formula we get

4 = (36+4x)/7                              &       1 = (15 + 4y)/7

28= 36 + 4x                                        7 = 4y + 15

4x =-8                                                         4y =-8

x =  -2                                                           y = -2