A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R (3, – 2) are fixed points, then the locus of the centroid of $$\Delta $$PQR is a line :
JEE · Math · previous-year question
- A.parallel to y-axis
- B.with slope $${2 \over 3}$$correct
- C.parallel to x-axis
- D.with slope $${3 \over 2}$$
Answer
B. with slope $${2 \over 3}$$
Explanation
Let the centroid of $$\Delta $$PQR is (h, k) & P is ($$\alpha $$, $$\beta $$), then $${{\alpha + 1 + 3} \over 3} = h\,$$ and $${{\beta + 4 - 2} \over 3} = k$$ $$\alpha = \left( {3h - 4} \right)$$ $$\beta = \left( {3k - 4} \right)$$ Point P($$\alpha $$, $$\beta $$) lies on the line 2x $$-$$ 3y + 4 = 0 $$ \therefore $$ 2(3h $$-$$ 4) $$-$$ 3 (3k $$-$$ 2) + 4 = 0 $$ \Rightarrow $$ locus is 6x $$-$$ 9y + 2 = 0
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