Question

1. Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than to B.
2. Given points A(3, -5) and B(19, -1), find the coordinates of point C such that CB/AC = 1/7

Answer 1

1. C(x, y) = (7.3, –3.9)

2. C(x, y) = (17, –1.5)

Solution:

Question 1:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction $\frac{3}{8}$.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)$

Here, $x_1=3, y_1=-5, x_2=19, y_2=-1$ and m = 3, n = 8

$C(x, y)=\left(\frac{3\times19+8\times3}{3+8} , \frac{3\times(-1)+8\times(-5)}{3+8}\right)$

           $=\left(\frac{57+24}{11} , \frac{-3-40}{11}\right)$

           $=\left(\frac{81}{11} , \frac{-43}{11}\right)$

C(x, y) = (7.3, –3.9)

Question 2:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction $\frac{3}{8}$.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)$

Here, $x_1=3, y_1=-5, x_2=19, y_2=-1$ and m = 7, n = 1

$C(x, y)=\left(\frac{7\times19+1\times3}{7+1} , \frac{7\times(-1)+1\times(-5)}{7+1}\right)$

           $=\left(\frac{133+3}{8} , \frac{-7-5}{8}\right)$

           $=\left(\frac{136}{8} , \frac{-12}{8}\right)$

C(x, y) = (17, –1.5)