A square of side a lies above the x-axis and | vedclass

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

Question

(B) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$B$,and $C$. Since the side $OA$ makes an angle $\alpha$ with the $x$

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$B$,and $C$. Since the side $OA$ makes an angle $\alpha$ with the $x$-axis,its coordinates are $A(a \cos \alpha, a \sin \alpha)$. The diagonal $OB$ makes an angle of $\alpha + \frac{\pi}{4}$ with the $x$-axis. The slope of $OB$ is $	an(\alpha + \frac{\pi}{4})$. The diagonal $AC$ is perpendicular to $OB$. The slope of $AC$ is $-\cot(\alpha + \frac{\pi}{4}) = -\frac{1}{	an(\alpha + \frac{\pi}{4})} = -\frac{1 - 	an \alpha 	an(\frac{\pi}{4})}{	an \alpha + 	an(\frac{\pi}{4})} = -\frac{1 - 	an \alpha}{1 + 	an \alpha} = \frac{	an \alpha - 1}{	an \alpha + 1} = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$. The equation of the line $AC$ passing through $A(a \cos \alpha, a \sin \alpha)$ is: $y - a \sin \alpha = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} (x - a \cos \alpha)$ $y(\sin \alpha + \cos \alpha) - a \sin \alpha(\sin \alpha + \cos \alpha) = x(\sin \alpha - \cos \alpha) - a \cos \alpha(\sin \alpha - \cos \alpha)$ $y(\sin \alpha + \cos \alpha) - x(\sin \alpha - \cos \alpha) = a \sin^2 \alpha + a \sin \alpha \cos \alpha - a \sin \alpha \cos \alpha + a \cos^2 \alpha$ $y(\sin \alpha + \cos \alpha) - x(\sin \alpha - \cos \alpha) = a(\sin^2 \alpha + \cos^2 \alpha) = a$. Thus,the equation is $y(\cos \alpha + \sin \alpha) - x(\sin \alpha - \cos \alpha) = a$.

$A$ square of side $a$ lies above the $x$-axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha, (0 < \alpha < \frac{\pi}{4})$ with the positive direction of the $x$-axis. The equation of its diagonal not passing through the origin is

Similar Questions

Two non-parallel sides of a rhombus are parallel to the lines $x+y-1=0$ and $7x-y-5=0$. If $(1,3)$ is the centre of the rhombus and one of its vertices $A(\alpha, \beta)$ lies on $15x-5y=6$,then one of the possible values of $(\alpha+\beta)$ is

If $\alpha$ is the angle made by the perpendicular drawn from the origin to the line $3x - 4y + 5 = 0$ with the positive $X$-axis in the positive direction,and $ax + by = 1$ is the equation of a line passing through the point $(1, -1)$ with $\tan \alpha$ as its slope,then $a + ab + b =$

$A$ pair of straight lines drawn through the origin form an isosceles right-angled triangle with the line $2x + 3y = 6$. Find the equations of the lines and the area of the triangle thus formed.

$L_1 \equiv ax-3y+5=0$ and $L_2 \equiv 4x-6y+8=0$ are two parallel lines. If $p, q$ are the intercepts made by $L_1=0$ and $m, n$ are the intercepts made by $L_2=0$ on the $X$ and $Y$ coordinate axes respectively,then the equation of the line passing through the points $(p, q)$ and $(m, n)$ is

The line $3x + 2y = 24$ meets the $y$-axis at $A$ and the $x$-axis at $B$. The perpendicular bisector of $AB$ meets the line through $(0, -1)$ parallel to the $x$-axis at $C$. The area of the triangle $ABC$ is ............... $sq. \, units$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module