Points a, b,c are colinear on line segment 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?

Given: Points A,B,C are co-linear on line segment AC , and

AB : BC = m : n= 3 : 4.

Coordinates of A = (x,y) ,

Coordinates of B = (x₁, y₁) = (4,1) , and

Coordinates of C = (x₂, y₂ ) = (12,5).

To find: the values of x and y

Now, we shall apply intersection (internally) formula for point B,

x₁ = (mx₂ + nx ) ÷ (m + n)

4 = (3×12 + 4× x ) ÷ (3 +4)  

or,       x = - 2  and,

y₁ = (my₂ + ny ) ÷ (m + n)

1 = (3×5 + 4× y ) ÷ (3 +4)  

or,       y = - 2

Hence, the required point A (x, y) will be (-2, -2)

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )