Question

The line segment $AB$ is divided into five congruent parts at $P,Q,R$ and $S$ such that $AP=PQ=QR=RS=SB$. If point $Q(12,14)$ and $S(4,18)$ are given find the coordinates of $A,P,R,B$.

Answer 1

Let the coordinates be $A(x_1,y_1),P(x_2,y_2),R(x_3,y_3)$ and $B(x_4,y_4)$.

Now, $QR=RS$, therefore, $R$ is the mid point of $QS$

Hence, using mid point formula, we get,

$\Rightarrow R(x_3,y_3)=(\frac{12+4}{2},\frac{14+18}{2})=(8,16)$

Hence, $R(x_3,y_3)=(8,16)$

Similarly, $RS=SB$, therefore, $S$ is the mid point of $RB$.

$\Rightarrow S(4,18)=(\frac{8+x_4}{2},\frac{16+y_4}{2})$

$\Rightarrow 4=\frac{8+x_4}{2}$ and $18=\frac{16+y_4}{2}$

$\Rightarrow x_4=8-8=0$ and $y_4=36-16=20$

Hence, $B(x_4,y_4)=(0,20)$.

Similarly, $PQ=QR$, therefore, $Q$ is the mid point of $PR$.

$\Rightarrow Q(12,14)=(\frac{8+x_2}{2},\frac{16+y_2}{2})$

$\Rightarrow 12=\frac{8+x_2}{2}$ and $14=\frac{16+y_2}{2}$

$\Rightarrow x_2=24-8=16$ and $y_2=28-16=12$

Hence, $P(x_2,y_2)=(16,12)$

Similarly, $AP=PQ$, therefore, $P$ is the mid point of $AQ$.

$\Rightarrow P(16,12)=(\frac{12+x_1}{2},\frac{14+y_1}{2})$

$\Rightarrow 16=\frac{12+x_1}{2}$ and $12=\frac{14+y_1}{2}$

$\Rightarrow x_1=32-12=20$ and $y_1=24-14=10$

Hence, $A(x_1,y_1)=(20,10)$

Hence, all the coordinates of the points have been obtained.