Question
Find the coordinate of the points which trisect the line segment joining the points A(2, 1, -3) and B(5, -8, 3).
✓
Answer
Let P(x1, y1, z1) and Q(x2, y2, z2) trisect line segment AB.
Since point P divides AB in the ratio 1 : 2 internally, we have
$\text{P}(\text{x}_1,\text{y}_1,\text{z}_1)\equiv\text{P}\Big(\frac{1(5)+2(2)}{2+1},\frac{1(-8)+2(1)}{2+1},\frac{1(3)+2(-3)}{2+1}\Big)$
$\equiv\text{P}(3,-2,-1)$
Since point Q divides AB in the ratio 2 : 1 internally, we have
$\text{P}(\text{x}_2,\text{y}_2,\text{z}_2)\equiv\text{Q}\Big(\frac{2(5)+1(2)}{2+1},\frac{2(-8)+1(1)}{2+1},\frac{2(3)+(-3)}{2+1}\Big)$
$\equiv\text{Q}(4,-5,1)$
Since point P divides AB in the ratio 1 : 2 internally, we have
$\text{P}(\text{x}_1,\text{y}_1,\text{z}_1)\equiv\text{P}\Big(\frac{1(5)+2(2)}{2+1},\frac{1(-8)+2(1)}{2+1},\frac{1(3)+2(-3)}{2+1}\Big)$
$\equiv\text{P}(3,-2,-1)$
Since point Q divides AB in the ratio 2 : 1 internally, we have
$\text{P}(\text{x}_2,\text{y}_2,\text{z}_2)\equiv\text{Q}\Big(\frac{2(5)+1(2)}{2+1},\frac{2(-8)+1(1)}{2+1},\frac{2(3)+(-3)}{2+1}\Big)$
$\equiv\text{Q}(4,-5,1)$
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