Question
Determine if the points $(1, 5), (2, 3)$ and $(-2, -11)$ are collinear.
✓
Answer
The points of trisection means that the points which divide the line into three equal parts. From the figure, it is clear that C, and D are these two points. Let $C (x_1, y_1)$ and $D (x_2, y_2)$ are the points of trisection of the line segment joining the given points i.e., $BC = CD = DA$
Let $BC = CD = DA = k$, Point C divides BC and CA as: $BC = kCA = CD + DA = k + k = 2k$
Hence the ratio between BC and CA is: $\frac{B C}{C A}=\frac{k}{2 k}=\frac{1}{2}$
Therefore, point C divides BA internally in the ratio 1:2 then by section formula we have that if a point P(x, y) divides two points $P (x_1, y_1)$ and $Q (x_2, y_2)$ in the ratio m:n then, the point (x, y) is given by $(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}, \frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}\right)$
Therefore C(x, y) divides B(–2, –3) and A(4,–1) in the ratio 1:2, then
$C(x, y)=\left(\frac{(1 \times 4)+(2 \times-2)}{1+2}, \frac{(1 \times-1)+(2 \times-3)}{1+2}\right)$
$C(x, y)=\left(\frac{4-4}{1+2}, \frac{-1-6}{1+2}\right)$
$C(x, y)=\left(0, \frac{-7}{3}\right)$
Point D divides the BD and DA as:$DA = kBD = BC + CD = k + k = 2k$
Hence the ratio between BD and DA is: $\frac{B D}{D A}=\frac{2 k}{k}=\frac{2}{1}$
The point D divides the line BA in the ratio 2:1
So now applying section formula again we get,
$D(x, y)=\left(\frac{(2 \times 4)+(1 \times-2)}{2+1}, \frac{(2 \times-1)+(1 \times-3)}{2+1}\right)$
$D(x, y)=\left(\frac{8-2}{3}, \frac{-2-3}{3}\right)$
$D(x, y)=\left(\frac{6}{3}, \frac{-5}{3}\right)$
$D(x, y)=\left(2, \frac{-5}{3}\right)$
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