The origin and the points where the line $L_1$ intersect the $x$ -axis and $y$ -axis are vertices of right angled triangle $T$ whose area is $8$. Also the line $L_1$ is perpendicular to line $L_2$ : $4x -y = 3$, then perimeter of triangle $T$ is -
The origin and the points where the line $L_1$ intersect the $x$ -axis and $y$ -axis are vertices of right angled triangle $T$ whose area is $8$. Also the line $L_1$ is perpendicular to line $L_2$ : $4x -y = 3$, then perimeter of triangle $T$ is -
- A
$10 + \sqrt {68}$
- B
$8 + \sqrt {32}$
- C
$17 + \sqrt {257}$
- D
$4 \sqrt {2}+ 4$
Similar Questions
Two mutually perpendicular straight lines through the origin from an isosceles triangle with the line $2x + y = 5$ . Then the area of the triangle is :
Two mutually perpendicular straight lines through the origin from an isosceles triangle with the line $2x + y = 5$ . Then the area of the triangle is :
Let the points $\left(\frac{11}{2}, \alpha\right)$ lie on or inside the triangle with sides $x + y =11, x +2 y =16$ and $2 x +3 y =29$. Then the product of the smallest and the largest values of $\alpha$ is equal to :
- [JEE MAIN 2025]
There are two candles of same length and same size. Both of them burn at uniform rate. The first one burns in $5\,hr$ and the second one burns in $3\,h$. Both the candles are lit together. After how many minutes the length of the first candle is $3$ times that of the other?
There are two candles of same length and same size. Both of them burn at uniform rate. The first one burns in $5\,hr$ and the second one burns in $3\,h$. Both the candles are lit together. After how many minutes the length of the first candle is $3$ times that of the other?
- [KVPY 2016]
The $x -$ co-ordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3 |x| + 2 = 0$ and the $y -$ co-ordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$ then the possible vertices of the square is/are :
The $x -$ co-ordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3 |x| + 2 = 0$ and the $y -$ co-ordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$ then the possible vertices of the square is/are :
The equation of base $BC$ of an equilateral triangle is $3x + 4y = 1$ and vertex is $(-3,2),$ then the area of triangle is-
The equation of base $BC$ of an equilateral triangle is $3x + 4y = 1$ and vertex is $(-3,2),$ then the area of triangle is-