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In the triangle with vertices at $A(6,3), B(-6,3)$ and $C(-6,-3)$, the median through $A$ meets
$B C$ at $P$, the line $A C$ meets the $x$-axis at $Q$, while $R$ and $S$ respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-I to List-II is
$\begin{array}{llll} & \text { List-I } & & \text { List-II } \\ \text { (i) } & P & \text { (A) } & (0,0) \\ \text { (ii) } & Q & \text { (B) } & (6,0) \\ \text { (iii) } & R & \text { (C) } & (-2,1) \\ \text { (iv) } & S & \text { (D) }(-6,0) \\ & & \text { (E) }(-6,-3) \\ & & \text { (F) }(-6,3)\end{array}$
(i)
(ii)
(iii)
(iv)
Options:
$B C$ at $P$, the line $A C$ meets the $x$-axis at $Q$, while $R$ and $S$ respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-I to List-II is
$\begin{array}{llll} & \text { List-I } & & \text { List-II } \\ \text { (i) } & P & \text { (A) } & (0,0) \\ \text { (ii) } & Q & \text { (B) } & (6,0) \\ \text { (iii) } & R & \text { (C) } & (-2,1) \\ \text { (iv) } & S & \text { (D) }(-6,0) \\ & & \text { (E) }(-6,-3) \\ & & \text { (F) }(-6,3)\end{array}$
(i)
(ii)
(iii)
(iv)
Solution:
2220 Upvotes
Verified Answer
The correct answer is:
$\begin{array}{llll}\text { D } & \text { A } & \text { F } & \text { C }\end{array}$
$A(6,3), B(-6,3), C(-6,-3)$ forms a right angled triangle.
In which $\angle B=90^{\circ}$
Equation of $A C$ is $y-3=\frac{-3-3}{-6-6}(x-6)$
$\begin{aligned} & \Rightarrow \quad y-3=\frac{-6}{-12}(x-6) \\ & \Rightarrow 2 y-6=x-6 \\ & \Rightarrow \quad x=2 y\end{aligned}$
$\begin{aligned} & P=\text { Mid point of } B C=\left(\frac{-6-6}{2}, \frac{3-3}{2}\right)=(-6,0) \\ & Q=\text { The point where } A C \text { meets } x \text {-axis is }(0,0) \\ & R=\text { Orthocentre of } \triangle A B C=(-6,3) \\ & S=\text { Centroid of } \triangle A B C=(-2,1)\end{aligned}$
In which $\angle B=90^{\circ}$
Equation of $A C$ is $y-3=\frac{-3-3}{-6-6}(x-6)$
$\begin{aligned} & \Rightarrow \quad y-3=\frac{-6}{-12}(x-6) \\ & \Rightarrow 2 y-6=x-6 \\ & \Rightarrow \quad x=2 y\end{aligned}$
$\begin{aligned} & P=\text { Mid point of } B C=\left(\frac{-6-6}{2}, \frac{3-3}{2}\right)=(-6,0) \\ & Q=\text { The point where } A C \text { meets } x \text {-axis is }(0,0) \\ & R=\text { Orthocentre of } \triangle A B C=(-6,3) \\ & S=\text { Centroid of } \triangle A B C=(-2,1)\end{aligned}$
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