Question

Suppose A, R, and B are collinear on AB, and AR:AB= 1/5. What are the coordinates of R?

Answer 1

Answer:

R(1.6, 0)

Step-by-step explanation:

Use the formula, $(x, y) = (x_1 + k(x_2 - x_1), y_1 + k(y_2 - y_1))$ to find the coordinates of point R, that partition the lie segment AB into the ratio 1/5.

Let,

$A(1, -1) = (x_1, y_1)$

$B(4, 4) = (x_2, y_2)$

Thus, plug in the values as follows:

$R(x, y) = (1 + \frac{1}{5}(4 - 1), -1 + \frac{1}{5}(4 -(-1))$

$R(x, y) = (1 + \frac{1}{5}(3), -1 + \frac{1}{5}(4 + 1)$

$R(x, y) = (1 + \frac{1}{5}(3), -1 + \frac{1}{5}(5)$

$R(x, y) = (1 + \frac{3}{5}, -1 + \frac{5}{5})$

$R(x, y) = (\frac{5 + 3}{5}, -1 + 1)$

$R(x, y) = (\frac{8}{5}, 0)$

$R(x, y) = (1.6, 0)$

The coordinates of point R, are (1.6, 0)