MCQ
If the points $A(2, 3), B(5 , k)$ and $C(6, 7)$ are collinear then :
- A$\text{k}=4$
- ✓$\text{k}=6$
- C$\text{k}=\frac{-3}{2}$
- D$\text{k}=\frac{11}{4}$
✓
Answer
Correct option: B.
$\text{k}=6$
The given points are $A (2, 3), B(5, k)$ and $C(6, 7)$ are collinear.
$\therefore(\text{x}_1=2,\text{y}_1=3),(\text{x}_2=5,\text{y}_2=\text{k)}$and $(\text{x}_3=6,\text{y}_3=7)$
The given points are collinear.
$\Rightarrow\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)=0$
$\Rightarrow2(\text{k}-7)+5(7-3)+6(3-\text{k)}=0$
$\Rightarrow5\text{k}-14+20+18-6\text{k}=0$
$\Rightarrow4\text{k}=24$
$\Rightarrow\text{k}=6$
$\therefore(\text{x}_1=2,\text{y}_1=3),(\text{x}_2=5,\text{y}_2=\text{k)}$and $(\text{x}_3=6,\text{y}_3=7)$
The given points are collinear.
$\Rightarrow\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)=0$
$\Rightarrow2(\text{k}-7)+5(7-3)+6(3-\text{k)}=0$
$\Rightarrow5\text{k}-14+20+18-6\text{k}=0$
$\Rightarrow4\text{k}=24$
$\Rightarrow\text{k}=6$
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