Origin and points where line L₁ intersect x -axis and y -axis are vertices of right angled tr

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9.Straight Line

normal

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 -

$10 + \sqrt {68}$

$8 + \sqrt {32}$

$17 + \sqrt {257}$

$4 \sqrt {2}+ 4$

Solution

Let $L_1$ $\equiv$ $x + 4y -c = 0$
$\therefore$ Points $(c, 0),$ $\left( {0,\frac{c}{4}} \right)$ $(0, 0)$
area $= 8 \Rightarrow c^2 = 64 \Rightarrow c = 8$ or $-8$
$\therefore$ perimeter $=8 + 2 + \sqrt {68} = 10 + \sqrt {68}$

Standard 11

Mathematics

The area of a parallelogram formed by the lines $ax \pm by \pm c = 0$, is

easy

(IIT-1973)

Let $A B C D$ be a tetrahedron such that the edges $AB , AC$ and $AD$ are mutually perpendicular. Let the areas of the triangles $ABC , ACD$ and $ADB$ be $5,6$ and $7$ square units respectively. Then the area (in square units) of the $\triangle BCD$ is equal to :

hard

(JEE MAIN-2025)

In the triangle $ABC$ with vertices $A$$(2,3), B(4,-1)$ and $C(1,2),$ find the equation and length of altitude from the vertex $A$.

medium

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 :