Answer to Question #90264 in Analytic Geometry for Juan .D

Question

The coordinates of points A and B are A (16,4) and B (34,40). What are the coordinates of a point that partitions segment AB in a ratio of 1:5?

Answer 1

Let's find the length of AB:

(1) $AB=\sqrt{\smash[b]{(x_2-x_1)}^{\smash{2}}+{(y_2-y_1)}^{\smash{2}}}$

$AB=\sqrt{\smash[b]{(34-16)}^{\smash{2}}+{(40-4)}^{\smash{2}}}$

$AB=\sqrt{\smash[b]324+1296}=\sqrt{\smash[b]1620}=18*\sqrt{\smash[b]5}$

Let point L (x_L;y_L) will be a point that partition segment AB in a ratio of 1:5. It means that:

$AL=AB*\frac{1}{6}$ and $BL=AB*\frac{5}{6}$

Lengths of segments AL and BL are respectively equal:

(2) $AL=\sqrt{\smash[b]{(x_A-x_L)}^{\smash{2}}+{(y_A-y_L)}^{\smash{2}}}=18*\sqrt{\smash[b]{(5)}}/6=3*\sqrt{\smash[b]{(5)}}$

(3) $BL=\sqrt{\smash[b]{(x_B-x_L)}^{\smash{2}}+{(y_B-y_L)}^{\smash{2}}}=18*\sqrt{\smash[b]{(5)}}*5/6=15*\sqrt{\smash[b]{(5)}}$

We now coordinates of points A (16;4) and B (34;40):

$\begin{cases} \sqrt{\smash[b]{(16-x_L)}^{\smash{2}}+{(4-y_L)}^{\smash{2}}}=3*\sqrt{\smash[b]{(5)}} \\ \sqrt{\smash[b]{(34-x_L)}^{\smash{2}}+{(40-y_L)}^{\smash{2}}}=15*\sqrt{\smash[b]{(5)}} \end{cases}$

$\begin{cases} (16-x_L)^{\smash{2}}+(4-y_L)^{\smash{2}}=45 \\ (34-x_L)^{\smash{2}}+(40-y_L)^{\smash{2}}=1125 \end{cases}$

$\begin{cases} 256-32*x_L+x_L^2+16-8*y_L+y_L^2=45 \\ 1156-68*x_L+x_L^2+1600-80*y_L+y_L^2=1125 \end{cases}$

Let's subtract the first equation from the second:

$900-36*x_L+1584-72*y_L=1080$

$39-x_L-2*y_L=0$

(4) $x_L=39-2*y_L$

Substitute value of x_L in equation (2):

$\sqrt{\smash[b]{(16-39+2*y_L)}^{\smash{2}}+{(4-y_L)}^{\smash{2}}}=3*\sqrt{\smash[b]{(5)}}$

$(2*y_L-23)^{\smash{2}}+(4-y_L)^{\smash{2}}=45$

$4*y_L^2-92*y_L+529+16-8*y_L+y_L^2-45=0$

$500-100*y_L+5*y_L^2=0$

$100-20*y_L+*y_L^2=0$

$(10-y_L)^2=0\implies y_L=10$

Substitute value of y_L in equation (4):

$x_L=39-2*10=19$

Coordinates of point that partitions segment AB in a ratio of 1:5 is (19;10)