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Question : 31 of 160
Marks:
+1,
-0
Solution:
If four points are collinear, then
⇒|| x2−x1 | y2−y1 | z2−z1 |
| x3−x1 | y3−y1 | z3−z1 |
| x4−x1 | y4−y1 | z4−z1 |
|=0 ⇒|| −3 | 2 | 1 |
| (x−1) | −6 | 2 |
| 0 | (y−1) | −2 |
|=0 ⇒−3[12−2(y−1)]−(x−1)[−4−(y−1)]=0 ⇒−3(12−2y+2−(x−1)(−4−y+1)=0 ⇒−42+6y+xy+3x−y−3=0 ⇒3x+5y+xy−45=0⇒(x+5)(y+3)=60
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