Question

Point B is collinear with A and C and partitions AC in a 3:4 ratio. B is located at (4,1) and C is located at (12,5). Find the coordinates of endpoint A. Show steps

Answer 1

Answer:

$A(-\frac{20}{3} ,-\frac{13}{3})$

Step-by-step explanation:

Point B partitions AC in a 3:4 ratio.

B is located at (4,1) and C is located at (12,5).

Let $(x_1,y_1)$ be the coordinates of A.

Then $(\frac{mx_1+nx_2}{m+n}, \frac{my_1+ny_2}{m+n})=(4,1)$

But m:n=3:4 implies m=3 and n=4 and $(x_2=12,y_2=5)$

$(\frac{3x_1+4*12}{3+4}, \frac{3y_1+4*5}{3+4})=(4,1)$

$(\frac{3x_1+48}{7}, \frac{3y_1+20}{7})=(4,1)$

$\frac{3x_1+48}{7}=4, \frac{3y_1+20}{7}=1$

$3x_1+48=28, 3y_1+20=7$

$3x_1=-20, 3y_1=-13$

$x_1=-\frac{20}{3} , y_1=-\frac{13}{3}$