The point $P \,(3, 6)$ is first reflected on the line $y = x$ and then the image point $Q$ is again reflected on the line $y = - x$ to get the image point $Q'$. Then the circumcentre of the $\Delta PQQ'$ is
- $(6, 3)$
- $(6, - 3)$
- $(3, -6)$
- $(0, 0)$
The Correct Option is D
Solution and Explanation
The coordinates of $Q$ and $Q^{\prime}$ will be $(6,3)$ and $(-3,-6)$, respectively.
Now, Slope of $P Q=\frac{6-3}{3-6}=-1$
and Slope of $Q Q^{\prime}=\frac{-6-3}{-3-6}=1$
$\therefore$ Slope of $P Q \times$ Slope of $Q Q^{\prime}=-1 \times 1=-1$
$\therefore \triangle P Q Q^{\prime}$ is right angled triangle at $Q$.
$\therefore$ Circumcentre will be the mid-point of hypotenuse
$P Q^{\prime}=\left(\frac{-3+3}{2}, \frac{-6+6}{2}\right)$
$=(0,0)$
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Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c