The straight lines x+2y-9=0, 3x+5y-5=0, and ax+by-1=0 are concurrent if the straight line 35x-22y+1=0 passes through the point:
- (-a,-b)
- (a,-b)
- (-a,b)
- (a,b)
The Correct Option is D
Approach Solution - 1
Three lines are concurrent if they intersect at a single point. For the lines \(x + 2y - 9 = 0\), \(3x + 5y - 5 = 0\), and \(ax + by - 1 = 0\) to be concurrent, the determinant of their coefficients must be zero:
\(\begin{vmatrix} 1 & 2 & -9 \\ 3 & 5 & -5 \\ a & b & -1 \end{vmatrix} = 0\)
Expanding the determinant:
\(1(5(-1) - (-5)b) - 2(3(-1) - (-5)a) - 9(3b - 5a) = 0\)
\(-5 + 5b + 6 - 10a - 27b + 45a = 0\)
\(35a - 22b + 1 = 0\)
We are given that the line \(35x - 22y + 1 = 0\) passes through one of the points \((-a, -b)\), \((a, -b)\), \((-a, b)\), or \((a, b)\).
Let's test each point:
1. If the line passes through \((a, b)\):
\(35a - 22b + 1 = 0\). This matches the equation we derived from the concurrency condition.
2. If the line passes through \((a, -b)\):
\(35a - 22(-b) + 1 = 0 \implies 35a + 22b + 1 = 0\). This does not match.
3. If the line passes through \((-a, b)\):
\(35(-a) - 22b + 1 = 0 \implies -35a - 22b + 1 = 0\). This does not match.
4. If the line passes through \((-a, -b)\):
\(35(-a) - 22(-b) + 1 = 0 \implies -35a + 22b + 1 = 0\). This does not match.
The only point that satisfies the equation \(35a - 22b + 1 = 0\) is \((a, b)\).
Approach Solution -2
Given: Three lines are:
- $L_1: x + 2y - 9 = 0$
- $L_2: 3x + 5y - 5 = 0$
- $L_3: ax + by - 1 = 0$
These three lines are concurrent if they all pass through a common point.
That means the determinant formed by their coefficients must be zero:
$\begin{vmatrix} 1 & 2 & -9 \\ 3 & 5 & -5 \\ a & b & -1 \end{vmatrix} = 0$
We will expand this determinant using the first row:
$= 1 \cdot \begin{vmatrix} 5 & -5 \\ b & -1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 3 & -5 \\ a & -1 \end{vmatrix} + (-9) \cdot \begin{vmatrix} 3 & 5 \\ a & b \end{vmatrix}$
Now calculate each 2×2 determinant:
- $\begin{vmatrix} 5 & -5 \\ b & -1 \end{vmatrix} = 5 \cdot (-1) - (-5) \cdot b = -5 + 5b$
- $\begin{vmatrix} 3 & -5 \\ a & -1 \end{vmatrix} = 3 \cdot (-1) - (-5) \cdot a = -3 + 5a$
- $\begin{vmatrix} 3 & 5 \\ a & b \end{vmatrix} = 3b - 5a$
Substitute back:
$= 1(-5 + 5b) - 2(-3 + 5a) - 9(3b - 5a)$
$= -5 + 5b + 6 - 10a - 27b + 45a$
$= (5b - 27b) + (-10a + 45a) + (-5 + 6)$
$= -22b + 35a + 1$
For the lines to be concurrent:
$-22b + 35a + 1 = 0 \quad \Rightarrow \quad 35a - 22b + 1 = 0$
So the point $(a, b)$ must satisfy the line equation:
$35x - 22y + 1 = 0$
Hence, the correct answer is: (a, b) lies on the line $35x - 22y + 1 = 0$.
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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