JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 15)
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 :
parallel to y-axis
with slope $${2 \over 3}$$
parallel to x-axis
with slope $${3 \over 2}$$
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
$${{\alpha + 1 + 3} \over 3} = h\,$$ 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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