If a line passing through the point P(1, 1) m | vedclass

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

Question

(C) The equation of the line passing through $P(1, 1)$ with angle $	heta$ is $\frac{x - 1}{\cos	heta} = \frac{y - 1}{\sin	heta} = r$.The coordinate

Vedclass · Accepted Answer

$1 + \sqrt{3}$

Vedclass · Answer

(C) The equation of the line passing through $P(1, 1)$ with angle $	heta$ is $\frac{x - 1}{\cos	heta} = \frac{y - 1}{\sin	heta} = r$.The coordinates of $A$ (where $y=0$) are $(1 - \cot	heta, 0)$ and $B$ (where $x=0$) are $(0, 1 - 	an	heta)$.Given $AB^2 = OP^2$,we have $(1 - \cot	heta)^2 + (1 - 	an	heta)^2 = 1^2 + 1^2 = 2$.Expanding this: $(1 - 2\cot	heta + \cot^2	heta) + (1 - 2	an	heta + 	an^2	heta) = 2$.$2 - 2(\cot	heta + 	an	heta) + (\cot^2	heta + 	an^2	heta) = 2$.Let $u = 	an	heta + \cot	heta$. Since $\cot^2	heta + 	an^2	heta = u^2 - 2$,we get:$u^2 - 2 - 2u = 0 \Rightarrow u^2 - 2u - 2 = 0$.Solving for $u$: $u = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.Since $|	an	heta + \cot	heta| \ge 2$,the value $1 - \sqrt{3}$ is rejected as it is approximately $-0.732$.Thus,$	an	heta + \cot	heta = 1 + \sqrt{3}$.

If a line passing through the point $P(1, 1)$ makes an angle $\theta$ with the positive direction of the $x$-axis and intersects the coordinate axes at $A$ and $B$ such that $AB = OP$,where $O$ is the origin,then what is the value of $\tan\theta + \cot\theta$?

Similar Questions

$A$ straight line is drawn through the point $A(1,2)$ such that its point of intersection with the straight line $x+y=4$ is at a distance $\frac{\sqrt{6}}{3}$ from the given point $A$. Find the angle which the line makes with the positive direction of $X$-axis.

One side of a rectangle lies along the line $4x + 7y + 5 = 0.$ Two of its vertices are $(-3, 1)$ and $(1, 1).$ Then the equations of the other three sides are

If the equation of the base of an equilateral triangle is $2x - y = 1$ and the vertex is $(-1, 2)$,then the length of the side of the triangle is

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

The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equations to the sides $AB$ and $AC$ are respectively $px + qy = 1$ and $qx + py = 1$. Then the equation to the median through $A$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module