Question

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:  The required values are  x = 2  and  y = 2.

Step-by-step explanation:  Given that the points A, B and C are collinear on AC and AB : BC = 3 : 4.

Also, A is located at (x, y), B is located at (4, 1) and C is located at (12,5).

We are to find the values of x and y.

We know that

the co-ordinates of a point that divides a line segment with endpoints (a, b) and (c, d) in the ratio m : n are given by

$\left(\dfrac{mc+na}{m+n},\dfrac{md+nb}{m+n}\right).$

According to the given information, we get

$4=\dfrac{3\times 12+4\times x}{3+4}\\\\\\\Rightarrow 4=\dfrac{36+4x}{7}\\\\\Rightarrow 36+4x=28\\\\\Rightarrow 4x=8\\\\\Rightarrow x=\dfrac{8}{4}\\\\\Rightarrow x=2$

and

$1=\dfrac{3\times5+4\times y}{3+4}\\\\\\\Rightarrow 1=\dfrac{15+4y}{7}\\\\\Rightarrow 15+4y=7\\\\\Rightarrow 4y=8\\\\\Rightarrow y=\dfrac{8}{4}\\\\\Rightarrow y=2.$

Thus, the required values are  x = 2  and  y = 2.

Answer 2

We have two unknowns: x and y. Now, we have to formulate 2 equations. The first would come from the use of the given ratio:

We use the distance formula to find the distance between coordinates:

3/4 = √[(x-4)²+(y-1)²] / √[(4-12)²+(1-5)²]
√[(x-4)²+(y-1)²]  = 3√5
(x-4)²+(y-1)² = 45
x² - 8x + 16 + y² - 2y + 1 = 45
x² - 8x + y² - 2y = 28  --> eqn 1

The second equation must come from the equation of a line:
y = mx +b
m = (5-1)/(12-4) = 1/2

Substitute y=5 and x=12 for point (12,5)
5 = (1/2)(12) + b
b = -1

So, the second equation is

y = 1/2x -1 or x = 2 + 2y --> eqn 2

Solving the equations simultaneously:
(2 + 2y)² - 8(2 + 2y) + y² - 2y = 28 
Solving for y, 
y = -2
x = 2+2(-2) = -2

Therefore, the coordinates of point A is (-2,-2).