P (x, y) moves such that area of triangle formed by P, Q (a , 2 a) and R (- a,

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9.Straight Line

normal

$P (x, y)$ moves such that the area of the triangle formed by $P, Q (a , 2 a)$ and $R (- a, - 2 a)$ is equal to the area of the triangle formed by $P, S (a, 2 a)\,\,\, \&\,\, \,T (2 a, 3 a)$. The locus of $'P'$ is a straight line given by :

$3x - y = a$

$(A)$ or $(C)$ both

$y = 2ax$

$5x - 3y + a = 0$

Solution

$\Delta P Q R=\Delta P S T$

$P(x, y), Q(a, 2 a), R(-a,-2 a)$

$P(x, y), S(a, 2 a) \quad, T(2 a, 3 a)$

$\frac{1}{2}\left|\begin{array}{ccc}x & y & 1 \\ a & 2 a & 1 \\ -a & -2 a & 1\end{array}\right|=\frac{1}{2}\left|\begin{array}{cccc}x & y & 1 \\ a & 2 a & 1 \\ 2 a & 3 a & 1\end{array}\right|$

$\left(2 a x-2 a^{2}-a y\right)-\left(-2 a^{2}+29 x-4 y\right)=\left(29 x+3 a^{2}+2 a y\right)-\left(4 a^{2}-3 a x-a y\right)$

$2 a x+3 a^{2}+2 a y=4 a^{2}-3a x-a y$

$5 a x+3 a y=a^{2}$

$5 x+3 y=a$

Standard 11

Mathematics

If the coordinates of the vertices of the triangle $ABC$ be $(-1, 6)$, $(-3, -9)$, and $(5, -8)$ respectively, then the equation of the median through $C$ is

easy

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medium

(AIEEE-2012)

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normal

Co-ordinates of the orthocentre of the triangle whose vertices are $A(0, 0) , B(3, 4)$ and $C(4, 0)$ is

hard

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normal

(KVPY-2011)

$P (x, y)$ moves such that the area of the triangle formed by $P, Q (a , 2 a)$ and $R (- a, - 2 a)$ is equal to the area of the triangle formed by $P, S (a, 2 a)\,\,\, \&\,\, \,T (2 a, 3 a)$. The locus of $'P'$ is a straight line given by :

Solution

Similar Questions

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