If (1,5) is the midpoint of the segment of a line between the lines 5x −y −4 = 0 and 3x +4y −4=0,then the equation of the line will be:
Show Hint
- 83x+35y −92 = 0
- 83x−35y +92 = 0
- 83x−35y −92 = 0
- 83x+35y +92 = 0
The Correct Option is B
Solution and Explanation
1. The midpoint of a line segment is the average of the coordinates of the two endpoints.
The given midpoint of the segment is \((1, 5)\).
2. The line passing through \((1, 5)\) is the locus of all points equidistant from the given lines
\[ 5x - y - 4 = 0 \quad \text{and} \quad 3x + 4y - 4 = 0. \]
We use the formula for the distance of a point from a line:
\[ \text{Distance from } (x, y) \text{ to } ax + by + c = 0 \text{ is } \frac{|ax + by + c|}{\sqrt{a^2 + b^2}}. \]
Equate the distances of any point \((x, y)\) from the two lines:
\[ \frac{|5x - y - 4|}{\sqrt{5^2 + (-1)^2}} = \frac{|3x + 4y - 4|}{\sqrt{3^2 + 4^2}}. \]
Simplify:
\[ \frac{|5x - y - 4|}{\sqrt{26}} = \frac{|3x + 4y - 4|}{5}. \]
Cross-multiply:
\[ 5|5x - y - 4| = \sqrt{26}|3x + 4y - 4|. \]
Square both sides to eliminate the absolute values:
\[ 25(5x - y - 4)^2 = 26(3x + 4y - 4)^2. \]
Expand both sides and simplify to find the equation of the locus:
\[ 83x - 35y + 92 = 0. \]
Conclusion: The equation of the line is:
\[ \boxed{83x - 35y + 92 = 0}. \]
Top Questions on Straight lines
- If \( \triangle ABC \) is an isosceles triangle and the coordinates of the base points are \( B(1, 3) \) and \( C(-2, 7) \), the coordinates of \( A \) can be:
- WBJEE - 2024
- Mathematics
- Straight lines
- A square with each side equal to \( a \) lies above the \( x \)-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle \( \alpha \) (\( 0 < \alpha < \frac{\pi}{4} \)) with the positive direction of the \( x \)-axis. The equation of the diagonals of the square is:
- WBJEE - 2024
- Mathematics
- Straight lines
- In \(\triangle ABC\), coordinates of \(A\) are \((1, 2)\), and the equations of the medians through \(B\) and \(C\) are \(x + y = 5\) and \(x = 4\), respectively. Then the midpoint of \(BC\) is:
- WBJEE - 2024
- Mathematics
- Straight lines
- A line of fixed length a + b, moves so that its ends are always on two fixed perpendicular straight lines. The locus of a point which divides the line into two parts of length a and b is
- WBJEE - 2024
- Mathematics
- Straight lines
- If lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are mutually perpendicular, then $k$ is equal to:
- KCET - 2024
- Mathematics
- Straight lines
Questions Asked in WBJEE exam
- Choose the correct statement:
- WBJEE - 2024
- Trigonometry
- The points of extremum of \[ \int_{0}^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \, dt \] are:
- WBJEE - 2024
- Integration
- If \( a_i, b_i, c_i \in \mathbb{R} \) (\( i = 1, 2, 3 \)) and \( x \in \mathbb{R} \), and
\[ \begin{vmatrix} a_1 + b_1x & a_1x + b_1 & c_1 \\ a_2 + b_2x & a_2x + b_2 & c_2 \\ a_3 + b_3x & a_3x + b_3 & c_3 \end{vmatrix} = 0, \]
then:- WBJEE - 2024
- Matrices and Determinants
- The function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = e^x + e^{-x} \) is:
- WBJEE - 2024
- Relations and Functions
- If the quadratic equation \( ax^2 + bx + c = 0 \) (\( a > 0 \)) has two roots \( \alpha \) and \( \beta \) such that \( \alpha < -2 \) and \( \beta > 2 \), then:
- WBJEE - 2024
- Quadratic Equation