MCQ
If points $(1, 2), (-5, 6)$ and $(a, -2)$ are collinear, then $a =$
- A$-3$
- ✓$7$
- C$2$
- D$-2$
✓
Answer
Correct option: B.
$7$
The area of a triangle whose vertices are $(1, 2), (-5, 6)$ and $(a, -2)$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[1(6+2)+(-5)(-2-2)+\text{a}(2-6)]$
$=\frac{1}{2}[1\times8+(-5)(-4)+\text{a}(-4)]$
$=\frac{1}{2}[8+20-4\text{a}]$
$\Rightarrow\ \frac{1}{2}(28-4\text{a})$
$=(14-2\text{a})\text{ sq.units}$
$\because$ The points are collinear.
$\therefore$ Area $= 0$
$\Rightarrow\ 14-2\text{a}=0$
$\Rightarrow\ 2\text{a}=14$
$\Rightarrow\ \text{a}=\frac{14}{2}=7$
Hence $, a = 7$
$=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[1(6+2)+(-5)(-2-2)+\text{a}(2-6)]$
$=\frac{1}{2}[1\times8+(-5)(-4)+\text{a}(-4)]$
$=\frac{1}{2}[8+20-4\text{a}]$
$\Rightarrow\ \frac{1}{2}(28-4\text{a})$
$=(14-2\text{a})\text{ sq.units}$
$\because$ The points are collinear.
$\therefore$ Area $= 0$
$\Rightarrow\ 14-2\text{a}=0$
$\Rightarrow\ 2\text{a}=14$
$\Rightarrow\ \text{a}=\frac{14}{2}=7$
Hence $, a = 7$
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