MHT CET · Maths · Straight Lines
If \(\mathrm{G}(3,-5, \mathrm{r})\) is the centroid of \(\triangle \mathrm{ABC}\), where \(\mathrm{A} \equiv(7,-8,1)\), \(B \equiv(p, q, 5), C \equiv(q+1,5 p, 0)\) are vertices of the triangle \(A B C\), then the values of \(p, q, r\) are respectively
- A -2,3,2
- B -4,5,4
- C 6,5,4
- D 2,-2,3
Answer & Solution
Correct Answer
(A) -2,3,2
Step-by-step Solution
Detailed explanation
\(\mathrm{A}(7,-8,1) ; \mathrm{B}(\mathrm{p}, \mathrm{q}, 5)\) and \(\mathrm{C}(\mathrm{q}+1,5 \mathrm{p}, 0)\) are vertices of \(\triangle \mathrm{ABC}\) having centroid \(\mathrm{G}(3,-5, \mathrm{r})\)
\(\therefore 3=\frac{7+p+q+1}{3},-5=\frac{-8+q+5 p}{3}, r=\frac{1+5+0}{3}\)
\(\mathrm{r}=2\)
Solving (1) and (2), we get \(p=-2, q=3\)
\(\therefore 3=\frac{7+p+q+1}{3},-5=\frac{-8+q+5 p}{3}, r=\frac{1+5+0}{3}\)
\(\mathrm{r}=2\)
Solving (1) and (2), we get \(p=-2, q=3\)
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