Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are [2007]

Question

Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are [2007]

Smart Achievers · Accepted Answer

(3)

$∵$ $A r (∆ O P R) = A r (∆ P Q R) = A r (∆ O Q R)$

$∴$ $By simply geometry, R should be the centroid of ∆ P Q O$

$\Rightarrow co-ordinate of R = (\frac{3 + 6 + 0}{3}, \frac{4 + 0 + 0}{3}) = (3, \frac{4}{3})$

Solution

Q.
Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are [2007]

1 $(\frac{4}{3}, 3)$

2 $(3, \frac{2}{3})$

3 $(3, \frac{4}{3})$

4 $(\frac{4}{3}, \frac{2}{3})$

Ans.
(3)

$∵$ $A r (∆ O P R) = A r (∆ P Q R) = A r (∆ O Q R)$

$∴$ $By simply geometry, R should be the centroid of ∆ P Q O$

$\Rightarrow co-ordinate of R = (\frac{3 + 6 + 0}{3}, \frac{4 + 0 + 0}{3}) = (3, \frac{4}{3})$

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Solution

Q. Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are [2007]

1 (43,3)

2 (3,23)

3 (3,43)

4 (43,23)

Ans. (3) ∵ Ar(∆OPR)=Ar(∆PQR)=Ar(∆OQR) ∴ By simply geometry, R should be the centroid of ∆PQO ⇒co-ordinate of R=(3+6+03,4+0+03)=(3,43)