A, b, and c are collinear, and B is between a and c. The ratio of AB to AC is 1:2. If A is at (7,-1) and B is at (2,1) what are the coordinates of point C

Respuesta :

Answer:

C(-3,3)

Step-by-step explanation:

Given

A = (7,-1)

B = (2,1)

AB:AC = 1:2

Required

Determine the coordinates of C

Since, B is between A and C; we need to determine ratio BC as follows;

[tex]AB:AC = 1:2[/tex]

Convert to division

[tex]\frac{AB}{AC} = \frac{1}{2}[/tex]

AC = AB + BC;

[tex]\frac{AB}{AB + BC} = \frac{1}{2}[/tex]

Cross Multiply

[tex]2 * AB = 1 * (AB + BC)[/tex]

[tex]2 AB = AB + BC[/tex]

[tex]2AB - AB = BC[/tex]

[tex]AB = BC[/tex]

Divide both sides by BC

[tex]\frac{AB}{BC} = 1[/tex]

Rewrite as

[tex]\frac{AB}{BC} = \frac{1}{1}[/tex]

Write as ratio

[tex]AB:BC = 1:1[/tex]

Next is to determine the coordinates of C as follows;

Because B is between both points. we have:

[tex]B(x,y) = (\frac{mx_2 + nx_1}{m+n},\frac{my_2 + ny_1}{m+n})[/tex]

Where

[tex]m:n = AB:BC = 1:1[/tex]

[tex]B(x,y) = B(2,1)[/tex]

[tex]A(x_1,y_1) = A(7,-1)[/tex]

So; we're solving for x2 and y2

[tex]B(2,1) = (\frac{mx_2 + nx_1}{m+n},\frac{my_2 + ny_1}{m+n})[/tex]

Where

Solving for x2;

[tex]x = \frac{mx_2 + nx_1}{m+n}[/tex]

[tex]2 = \frac{1 * x_2 + 1 * 7}{1+1}[/tex]

[tex]2 = \frac{x_2 + 7}{2}[/tex]

Cross Multiply

[tex]2 * 2 = x_2 + 7[/tex]

[tex]4 = x_2 + 7[/tex]

[tex]x_2 = 4 - 7[/tex]

[tex]x_2 = -3[/tex]

Solving for y2;

[tex]y = \frac{my_2 + ny_1}{m+n}[/tex]

[tex]1 = \frac{1 * y_2 + 1 * -1}{1+1}[/tex]

[tex]1 = \frac{y_2- 1}{2}[/tex]

Cross Multiply

[tex]2 * 1 = y_2 - 1[/tex]

[tex]2 = y_2 - 1[/tex]

[tex]y_2 = 2 + 1[/tex]

[tex]y_2 = 3[/tex]

Hence, the coordinates of C are: C(-3,3)

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