One side of a square is inclined at an acute | vedclass

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

Question

$	herefore \,\,\,$ slope of line $AC =$ $\frac{{
ot 4\sin\, \alpha  - 
ot 4\cos\, \alpha }}{{
ot 4\cos\, \alpha  + 
ot 4\sin\, \al

Vedclass · Accepted Answer

$(cos\, \alpha - sin\, \alpha) x + (cos\, \alpha + sin\, \alpha) y = 4$

Vedclass · Answer

$	herefore \,\,\,$ slope of line $AC =$ $\frac{{
ot 4\sin\, \alpha  - 
ot 4\cos\, \alpha }}{{
ot 4\cos\, \alpha  + 
ot 4\sin\, \alpha }}$

$	herefore \,\,\,$ equation of line = $\frac{{\sin\, \alpha  - \sin\, \alpha }}{{\cos\, \alpha \sin\, \alpha }}$

$y - sin\, \alpha =$ $\frac{{\sin\, \alpha  - \sin\, \alpha }}{{\cos\, \alpha \sin\, \alpha }}$ $(x - 4cos\, \alpha)$

$y\, cos\, \alpha + y\, sin\, \alpha - 4 \, sin\, \alpha &middot; cos\, \alpha = x \,sin\, \alpha - x \,cos\, \alpha - 4 \,sin^2\, \alpha - sin\, \alpha cos\, \alpha + 4\, cos^2\, \alpha$

$x (cos\, \alpha - sin\, \alpha) + y (cos\, \alpha + sin\, \alpha) = 4\Rightarrow(D)$

Similar Questions

Area of the rhombus bounded by the four lines, $ax \pm by \pm c = 0$ is :

If the sum of the distances of a point from two perpendicular lines in a plane is $1$, then its locus is

Let $A B C$ be a triangle formed by the lines $7 x-6 y+3=0, x+2 y-31=0$ and $9 x-2 y-19=0$, Let the point $( h , k )$ be the image of the centroid of $\Delta A B C$ in the line $3 x+6 y-53=0$. Then $h^2+k^2+h k$ is equal to

In a triangle $ABC,$ side $AB$ has the equation $2 x + 3 y = 29$ and the side $AC$ has the equation , $x + 2 y = 16$ . If the mid - point of $BC$ is $(5, 6)$ then the equation of $BC$ is :

Let the equation of two sides of a triangle be $3x\,-\,2y\,+\,6\,=\,0$ and $4x\,+\,5y\,-\,20\,=\,0.$ If the orthocentre of this triangle is at $(1, 1),$ then the equation of its third side is