Area of the parallelogram formed by the lines | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) Area $ = \frac{{{p_1}{p_2}}}{{\sin 	heta }} = {p_1}{p_2}{m{cosec}}	heta $ ,

where ${p_1} = \frac{{{d_1} - {c_1}}}{{\sqrt {(a_1^2 + b_1^2)}

Vedclass · Accepted Answer

$\frac{{({d_1} - {c_1})({d_2} - {c_2})}}{{{a_1}{b_2} - {a_2}{b_1}}}$

Vedclass · Answer

(d) Area $ = \frac{{{p_1}{p_2}}}{{\sin 	heta }} = {p_1}{p_2}{m{cosec}}	heta $ ,

where ${p_1} = \frac{{{d_1} - {c_1}}}{{\sqrt {(a_1^2 + b_1^2)} }},{p_2} = \frac{{{d_2} - {c_2}}}{{\sqrt {(a_2^2 + b_2^2)} }}$

Also $	an 	heta = \frac{{{a_1}{b_2} - {a_2}{b_1}}}{{{a_1}{a_2} + {b_1}{b_2}}}$. Since ${m{cose}}{{m{c}}^2}	heta = 1 + {\cot ^2}	heta $

or ${m{cose}}{{m{c}}^2}	heta = \frac{{{{({a_1}{a_2} + {b_1}{b_2})}^2} + {{({a_1}{b_2} - {a_2}{b_1})}^2}}}{{{{({a_1}{b_2} - {a_2}{b_1})}^2}}}$

$ = \frac{{(a_1^2 + b_1^2)(a_2^2 + b_2^2)}}{{{{({a_1}{b_2} - {a_2}{b_1})}^2}}}$

Putting for ${p_1},{p_2}$ and ${m{cosec}}	heta $, we get

Area =$\frac{{({d_1} - {c_1})({d_2} - {c_2})}}{{{a_1}{b_2} - {a_2}{b_1}}}$.

Area of the parallelogram formed by the lines ${a_1}x + {b_1}y + {c_1} = 0$,${a_1}x + {b_1}y + {d_1} = 0$and ${a_2}x + {b_2}y + {c_2} = 0$, ${a_2}x + {b_2}y + {d_2} = 0$is

Similar Questions

In an isosceles triangle $ABC$, the coordinates of the points $B$ and $C$ on the base $BC$ are respectively $(1, 2)$ and $(2, 1)$. If the equation of the line $AB$ is $y = 2x$, then the equation of the line $AC$ is

The locus of the mid-points of the perpendiculars drawn from points on the line, $\mathrm{x}=2 \mathrm{y}$ to the line $\mathrm{x}=\mathrm{y}$ is

The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta ,2\sin \theta )$ internally in the ratio $2 : 3$ for all $\theta $, is a

$A(-1, 1)$, $B(5, 3)$ are opposite vertices of a square in $xy$-plane. The equation of the other diagonal (not passing through $(A, B)$ of the square is given by

Two sides of a parallelogram are along the lines, $x + y = 3$ and $x -y + 3 = 0$. If its diagonals intersect at $(2, 4)$, then one of its vertex is