Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the

Q: Wait, why is $M$ always at $y=3$? Does it not move?

That is a fantastic observation! Because $A$ is fixed at $(0, 6)$ and $B$ moves along the $x$-axis at $(2t, 0)$, the $y$-coordinate of the midpoint $M$ is always $\frac{6+0}{2} = 3$. So, while $M$ moves horizontally as $t$ changes, its vertical position is locked. This simplifies our geometry significantly!

Q: How do we handle the parameter $t$ when finding the locus?

Great question. When finding a locus, we want an equation involving only $x$ and $y$. We express the coordinates of $P$ as $(h, k)$ in terms of $t$. We then have two equations: $h = \frac{t}{2}$ and $k = 3 - \frac{t^2}{6}$. We isolate $t$ in the first equation ($t = 2h$) and substitute it into the second. This 'eliminates' the parameter $t$, leaving us with the path $P$ traces.

Q: Why does the perpendicular bisector equation look so messy?

It might look intimidating, but it is just a line! The equation $y - 3 = \frac{t}{3}(x - t)$ is just a standard linear equation where the slope and intercept depend on $t$. Don't let the variables scare you; treat $t$ as a constant while you manipulate the equation, and only substitute it at the very end. You've got this!

Question

Let $A$ be a fixed point $(0, 6)$ and $B$ be a moving point $(2t, 0)$. Let $M$ be the mid-point of $AB$ and the perpendicular bisector of $AB$ meets the $y$-axis at $C$. The locus of the mid-point $P$ of $MC$ is :

Options:

(A) $3x^2 - 2y - 6 = 0$

(B) $3x^2 + 2y - 6 = 0$

(C) $2x^2 + 3y - 9 = 0$

(D) $2x^2 - 3y + 9 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You see a fixed point $A$ at $(0, 6)$ on the $y$-axis, and a point $B$ at $(2t, 0)$ sliding along the $x$-axis.

As $t$ varies, the line segment $AB$ sweeps across the first quadrant, constantly changing its orientation. This is the stage for our problem, where we observe a dynamic system to determine the locus of a specific point $P$.

### The Midpoint $M$ and the Perpendicular Bisector

First, let us find the heart of this segment, the midpoint $M$. Using the midpoint formula, we find:

$$M = \left( \frac{0 + 2t}{2}, \frac{6 + 0}{2} \right) = (t, 3)$$

Notice the elegance here: $M$ is always at a height of $3$, regardless of where $B$ is. It is dancing along the horizontal line $y = 3$.

Now, we determine the perpendicular bisector of $AB$. The slope of $AB$ is:

$$m_{AB} = \frac{0 - 6}{2t - 0} = -\frac{3}{t}$$

For the perpendicular bisector, we require a slope $m_{\perp}$ such that $m_{AB} \cdot m_{\perp} = -1$. Thus, $m_{\perp} = \frac{t}{3}$.

Using the point-slope form at $M(t, 3)$, the equation of our bisector is:

$$y - 3 = \frac{t}{3}(x - t)$$

### The Intercept $C$ and the Target $P$

This bisector hits the $y$-axis at point $C$. Setting $x = 0$ in our equation, we get $y - 3 = \frac{t}{3}(0 - t)$, which simplifies to:

$$y = 3 - \frac{t^2}{3}$$

Thus, the coordinates of $C$ are $(0, 3 - \frac{t^2}{3})$. Now, we define $P$ as the midpoint of $MC$.

Using the midpoint formula on $M(t, 3)$ and $C(0, 3 - \frac{t^2}{3})$, we find the coordinates of $P(h, k)$:

$$h = \frac{t + 0}{2} = \frac{t}{2}$$

$$k = \frac{3 + (3 - \frac{t^2}{3})}{2} = 3 - \frac{t^2}{6}$$

### The Final Revelation

We are at the finish line. We have the parametric equations $h = \frac{t}{2}$ and $k = 3 - \frac{t^2}{6}$. To find the locus, we must eliminate the parameter $t$.

From the first equation, we find $t = 2h$. Substituting this into the second equation yields:

$$k = 3 - \frac{(2h)^2}{6} = 3 - \frac{4h^2}{6} = 3 - \frac{2h^2}{3}$$

Multiplying by $3$ gives $3k = 9 - 2h^2$, or $2h^2 + 3k - 9 = 0$. Replacing $h$ and $k$ with $x$ and $y$, we arrive at the final equation:

**$$2x^2 + 3y - 9 = 0$$**

This is the path $P$ traces—a beautiful, downward-opening parabola. You have successfully mapped the motion of a point through the language of algebra.

Animated Solution for Mathematics - Straight Lines: Let $A$ be a fixed point $(0, 6)$ and $B$ be a moving point $(2 t, 0)$ . Let $M$ be the mid-point of $A B$ and the perpendicular bisector of $A B$ meets the $y$ -axis at $C$ . The locus of the mid-point $P$ of $M C$ is :

Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Midpoint $M$

Calculating Coordinates of $M$

Finding the Slope of $A B$

Slope of the Perpendicular Bisector

Equation of the Perpendicular Bisector

Finding Point $C$ on the $y$ -axis

Defining the Target Point $P$

Coordinates of $P$ in terms of $t$

Eliminating the Parameter $t$

Substituting $t$ into the $k$ Equation

Simplifying to the Final Form

Final Locus Equation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Midpoint $M$ and the Perpendicular Bisector

The Intercept $C$ and the Target $P$

The Final Revelation

Similar Questions

The locus of mid points of the perpendiculars drawn from points on the line $x = 2 y$ to the line $x = y$ is :

The locus of the mid-point of the perpendiculars drawn from points on the line, $x = 2 y$ to the line $x = y$ is:

A straight line through a fixed point $(2, 3)$ intersects the coordinate axes at distinct points $P$ and $Q$ . If $O$ is the origin and the rectangle $O P R Q$ is completed, then the locus of $R$ is :

Locus of mid point of the portion between the axes of $x cos α + y sin α = p$ where $p$ is constant is

The end $A, B$ of a straight line segment of constant length $c$ slide upon the fixed rectangular axes $O X, O Y$ respectively. If the rectangle $O A P B$ be completed, then show that the locus of the foot of the perpendicular drawn from $P$ to $A B$ is $x^{2/3} + y^{2/3} = c^{2/3}$ .

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

A line cuts the x-axis at $A (7, 0)$ and the y-axis at $B (0, - 5)$ . A variable line $P Q$ is drawn perpendicular to $A B$ cutting the x-axis in $P$ and the y-axis in $Q$ . If $A Q$ and $B P$ intersect at $R$ , find the locus of $R$ .

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

In a $△ A B C$ , suppose $y = x$ is the equation of the bisector of the angle $B$ and the equation of the side $A C$ is $2 x - y = 2$ . If $2 A B = B C$ and the point $A$ and $B$ are respectively $(4, 6)$ and $(α, β)$ , then $α + 2 β$ is equal to

A rectangle $P QR S$ has its side $P Q$ parallel to the line $y = m x$ and vertices $P, Q$ and $S$ on the lines $y = a, x = b$ and $x = - b$ , respectively. Find the locus of the vertex $R$ .

Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Calculating Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Finding the Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Slope of the Perpendicular Bisector

Equation of the Perpendicular Bisector

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C on the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-axis

Defining the Target Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Eliminating the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t into the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k Equation