Question:

If 4a²+9b²-c²+12ab=0, then the family of straight lines ax+by+c=0 is concurrent at

Updated On: Apr 11, 2025

(2,3)or(-2,-3)
(-2,3)or(2,-3)
(3,2)or(-3,2)
(-3,2)or(2,3)

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The Correct Option is A

Approach Solution - 1

Given:
Equation: \( 4a^2 + 9b^2 + 12ab - c^2 = 0 \)

Step 1: Rewrite the expression
Group the terms: \[ 4a^2 + 9b^2 + 12ab = (2a + 3b)^2 \]
So the given expression becomes: \[ (2a + 3b)^2 - c^2 = 0 \]

Step 2: Apply difference of squares:
\[ (2a + 3b)^2 - c^2 = (2a + 3b + c)(2a + 3b - c) = 0 \]
So either: \[ 2a + 3b + c = 0 \quad \text{or} \quad 2a + 3b - c = 0 \]

Step 3: Equation of line:
Let the general equation of a line be: \[ ax + by + c = 0 \]
We now test the points (2, 3) and (-2, -3) in this line.

Case 1: Let (2, 3) satisfy the line: \[ a(2) + b(3) + c = 2a + 3b + c = 0 \]
This matches one of the factors found above.

Case 2: Let (-2, -3) satisfy the line: \[ a(-2) + b(-3) + c = -2a - 3b + c = 0 \]
Which is equivalent to: \[ 2a + 3b - c = 0 \] Again, this matches the other factor found.

Conclusion:
The line \( ax + by + c = 0 \) passes through both points (2, 3) and (-2, -3).

Answer: Option (A): (2, 3) or (–2, –3)

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Approach Solution -2

The given equation is:

\[ 4a^2 + 9b^2 + 12ab - c^2 = 0 \]

Step 1: Simplify the expression:

\[ 4a^2 + 9b^2 + 12ab = (2a + 3b)^2 \]
So the equation becomes: \[ (2a + 3b)^2 - c^2 = 0 \]

Step 2: Apply the identity for difference of squares:

\[ (2a + 3b - c)(2a + 3b + c) = 0 \]

Step 3: Solve each case:

Case 1: \( 2a + 3b = c \)
Case 2: \( 2a + 3b = -c \)

Step 4: Rewrite the line equation \( ax + by + c = 0 \)

Using Case 1: \( c = -2a - 3b \), substitute in the line equation: \[ ax + by - 2a - 3b = 0 \Rightarrow a(x - 2) + b(y - 3) = 0 \]
So, this line passes through the point \( (2, 3) \)

Using Case 2: \( c = 2a + 3b \), substitute in the line equation: \[ ax + by + 2a + 3b = 0 \Rightarrow a(x + 2) + b(y + 3) = 0 \]
So, this line passes through the point \( (-2, -3) \)

Final Conclusion:
The line \( ax + by + c = 0 \) passes through either \( (2, 3) \) or \( (-2, -3) \).

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Questions Asked in WBJEE exam

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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

If 4a2+9b2-c2+12ab=0, then the family of straight lines ax+by+c=0 is concurrent at