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Animated Solution for Mathematics - Straight Lines: If the equation of the locus of a point equidistant from the point $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$ , then the value of 'c' is

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the geometry. When a point

P (x, y)

is equidistant from two fixed points

A (a_{1}, b_{1})

and

B (a_{2}, b_{2})

, it is not just floating randomly; it is constrained to the perpendicular bisector of the segment

A B

. Can you see the symmetry?

Hint 2

The Arsenal: You have the distance formula

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

. Since

P A = P B

, you must set the square roots equal to each other. The most powerful tool in your kit here is squaring both sides to eliminate the radicals immediately.

Hint 3

The Execution: Start by writing

(x - a_{1})^{2} + (y - b_{1})^{2} = (x - a_{2})^{2} + (y - b_{2})^{2}

. Expand these squares carefully, and watch how the quadratic terms

x^{2}

and

y^{2}

vanish, leaving you with a linear equation that you can then rearrange to match the format

(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0

Visualized Solution

The Fixed Points

Let the fixed points be $A (a_{1}, b_{1})$ and $B (a_{2}, b_{2})$ .

The Moving Point and Locus

Let the moving point be $P (x, y)$ .
The condition is that $P$ is always equidistant from $A$ and $B$ .
Therefore, $P A = P B$ .
The locus of $P$ forms the perpendicular bisector of segment $A B$ .

Equating Distances

Distance from $P$ to $A$ : $P A$
Distance from $P$ to $B$ : $P B$
Condition: $P A = P B$

Applying the Distance Formula

Using the distance formula: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
$P A = (x - a_{1})^{2} + (y - b_{1})^{2}$
$P B = (x - a_{2})^{2} + (y - b_{2})^{2}$
$(x - a_{1})^{2} + (y - b_{1})^{2} = (x - a_{2})^{2} + (y - b_{2})^{2}$

Squaring Both Sides

To eliminate the square roots, square both sides:
$(x - a_{1})^{2} + (y - b_{1})^{2} = (x - a_{2})^{2} + (y - b_{2})^{2}$

Expanding the $x$ Terms

Expand $(x - a_{1})^{2}$ and $(x - a_{2})^{2}$ :
LHS: $x^{2} - 2 a_{1} x + a_{1}^{2} + (y - b_{1})^{2}$
RHS: $x^{2} - 2 a_{2} x + a_{2}^{2} + (y - b_{2})^{2}$

Expanding the $y$ Terms

Expand $(y - b_{1})^{2}$ and $(y - b_{2})^{2}$ :
LHS: $x^{2} - 2 a_{1} x + a_{1}^{2} + y^{2} - 2 b_{1} y + b_{1}^{2}$
RHS: $x^{2} - 2 a_{2} x + a_{2}^{2} + y^{2} - 2 b_{2} y + b_{2}^{2}$

Canceling Common Terms

Notice $x^{2}$ and $y^{2}$ appear on both sides.
Subtract $x^{2}$ and $y^{2}$ from both sides:
$- 2 a_{1} x + a_{1}^{2} - 2 b_{1} y + b_{1}^{2} = - 2 a_{2} x + a_{2}^{2} - 2 b_{2} y + b_{2}^{2}$

Grouping the Variables

Bring all terms involving $x$ and $y$ to the left side:
$2 a_{2} x - 2 a_{1} x + 2 b_{2} y - 2 b_{1} y + a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2} = 0$
Factor out $x$ and $y$ :
$2 (a_{2} - a_{1}) x + 2 (b_{2} - b_{1}) y + (a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2}) = 0$

Matching the Given Format

The given equation is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$ .
Our equation has $(a_{2} - a_{1})$ and $(b_{2} - b_{1})$ .
Divide the entire equation by $- 2$ :
$\frac{2 ( a _{2} - a _{1} )}{- 2} x + \frac{2 ( b _{2} - b _{1} )}{- 2} y + \frac{a _{1}^{2} + b _{1}^{2} - a _{2}^{2} - b _{2}^{2}}{- 2} = 0$
$(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + \frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2}) = 0$

Finding the Value of $c$

Compare our derived equation with the given equation:
$(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + \frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2}) = 0$
$(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$
Therefore, $c = \frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})$
This matches Option 2.

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The Sigma Insight: Distance and Section Formulas

Solution Diagram

Analyzing the Setup

Imagine you are standing on a vast, pristine coordinate plane. We place two anchors, point

A (a_{1}, b_{1})

and point

B (a_{2}, b_{2})

Now, we introduce a traveler, point

P (x, y)

. This traveler is bound by a singular, elegant law: it must always remain equidistant from

A

and

B

P

moves, it traces a path. Geometrically, this path is the perpendicular bisector of the segment

A B

The Algebraic Foundation

We begin with the definition of our condition:

P A = P B

. Using the distance formula, we set the distances equal:

(x - a_{1})^{2} + (y - b_{1})^{2} = (x - a_{2})^{2} + (y - b_{2})^{2}

To simplify, we square both sides to eliminate the radicals. This yields the following equation:

(x - a_{1})^{2} + (y - b_{1})^{2} = (x - a_{2})^{2} + (y - b_{2})^{2}

The Great Cancellation

Next, we expand the squares on both sides of the equation:

x^{2} - 2 a_{1} x + a_{1}^{2} + y^{2} - 2 b_{1} y + b_{1}^{2} = x^{2} - 2 a_{2} x + a_{2}^{2} + y^{2} - 2 b_{2} y + b_{2}^{2}

Observe that the

x^{2}

and

y^{2}

terms appear on both sides. By subtracting these terms from both sides, they vanish entirely.

This leaves us with a linear relationship, confirming that the locus is a straight line.

The Final Alignment

We rearrange the remaining terms to group the variables:

2 (a_{2} - a_{1}) x + 2 (b_{2} - b_{1}) y + (a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2}) = 0

To match the standard form

(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0

, we divide the entire equation by

- 2

. This transformation results in:

(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + \frac{a _{1}^{2} + b _{1}^{2} - a _{2}^{2} - b _{2}^{2}}{- 2} = 0

Celebrating the Result

By simplifying the constant term, we identify the value of

c

c = \frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})

We have successfully translated a spatial concept into a rigorous mathematical proof. This result demonstrates the perfect harmony between algebraic manipulation and geometric intuition.

Common Traps & FAQs

Wait, why does the locus turn out to be a straight line? Shouldn't it be a curve?

That is a fantastic intuition check! In many locus problems, we get parabolas or circles. However, because the distance formula involves squaring the coordinates, the

x^{2}

and

y^{2}

terms appear on both sides of the equation. When you subtract them, they cancel out perfectly. A linear equation is the hallmark of a straight line, which is exactly what a perpendicular bisector is!

Do I really need to square both sides? Can't I just solve it with the square roots?

You could, but that would be like trying to run a marathon in flip-flops! Squaring both sides is the standard algebraic 'shortcut' to remove the radicals. It simplifies the expression into a polynomial, which is much easier to manipulate and compare to the target format. Always look for the path that minimizes algebraic friction.

I got the equation, but my signs for 'c' are flipped. What happened?

This is the most common trap in this problem! When you compare your derived equation to the given format

(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0

, you likely had terms like

(a_{2} - a_{1}) x

. To match the given format, you must divide the entire equation by

- 2

. If you forget that negative sign, your constant

c

will have the wrong polarity. Always double-check your coefficients after division!

Animated Solution for Mathematics - Straight Lines: If the equation of the locus of a point equidistant from the point $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$ , then the value of 'c' is

Select Answer:

Visualized Solution

The Fixed Points

The Moving Point and Locus

Equating Distances

Applying the Distance Formula

Squaring Both Sides

Expanding the $x$ Terms

Expanding the $y$ Terms

Canceling Common Terms

Grouping the Variables

Matching the Given Format

Finding the Value of $c$

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

The Algebraic Foundation

The Great Cancellation

The Final Alignment

Celebrating the Result

Similar Questions

For $a > b > c > 0$ , the distance between $(1, 1)$ and the point of intersection of the lines $a x + b y + c = 0$ and $b x + a y + c = 0$ is less than $22$ . Then

A straight line segment of length $l$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio $1 : 2$ .

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Select Answer:

Visualized Solution

The Fixed Points

The Moving Point and Locus

Equating Distances

Applying the Distance Formula

Squaring Both Sides

Expanding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x Terms

Expanding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y Terms

Canceling Common Terms

Grouping the Variables

Matching the Given Format

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

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

The Algebraic Foundation

The Great Cancellation

The Final Alignment

Celebrating the Result

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a>b>c>0, the distance between (1,1) and the point of intersection of the lines ax+by+c=0 and bx+ay+c=0 is less than 22​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A straight line segment of length l moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio 1:2.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line 2x+y=k passes through the point which divides the line segment joining the points (1,1) and (2,4) in the ratio 3:2, then k equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (1,2),(4,6),(5,7) and (a,b) are the vertices of a parallelogram PQRS, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A triangle with vertices (4,0),(−1,−1),(3,5) is

Expanding the $x$ Terms

Expanding the $y$ Terms

Finding the Value of $c$

For $a > b > c > 0$ , the distance between $(1, 1)$ and the point of intersection of the lines $a x + b y + c = 0$ and $b x + a y + c = 0$ is less than $22$ . Then

A straight line segment of length $l$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio $1 : 2$ .

If the line $2 x + y = k$ passes through the point which divides the line segment joining the points $(1, 1)$ and $(2, 4)$ in the ratio $3 : 2$ , then $k$ equals

If $(1, 2), (4, 6), (5, 7)$ and $(a, b)$ are the vertices of a parallelogram $P QR S$ , then

A triangle with vertices $(4, 0), (- 1, - 1), (3, 5)$ is