A variable line L passing through the origin | vedclass

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

Question

(C) Let the equation of the line passing through the origin be $y=mx$,which cuts the parallel lines $x-y+10=0$ and $x-y+20=0$ at points $A$ and $B$ re

Vedclass · Accepted Answer

$3x-3y+40=0$

Vedclass · Answer

(C) Let the equation of the line passing through the origin be $y=mx$,which cuts the parallel lines $x-y+10=0$ and $x-y+20=0$ at points $A$ and $B$ respectively. For point $A$: $x-mx+10=0$ $\Rightarrow x(1-m)=-10$ $\Rightarrow x=\frac{10}{m-1}, y=\frac{10m}{m-1}$. Thus,$OA = \sqrt{x^2+y^2} = \sqrt{\frac{100(1+m^2)}{(m-1)^2}} = \frac{10\sqrt{1+m^2}}{|m-1|}$. Similarly,for point $B$: $x-mx+20=0 \Rightarrow x=\frac{20}{m-1}, y=\frac{20m}{m-1}$. Thus,$OB = \frac{20\sqrt{1+m^2}}{|m-1|}$. Let $P(h, k)$ be a point on $y=mx$,so $m=\frac{k}{h}$ and $OP = \sqrt{h^2+k^2}$. Since $OA, OP, OB$ are in harmonic progression,$\frac{2}{OP} = \frac{1}{OA} + \frac{1}{OB}$. Substituting the values: $\frac{2}{\sqrt{h^2+k^2}} = \frac{|m-1|}{10\sqrt{1+m^2}} + \frac{|m-1|}{20\sqrt{1+m^2}} = \frac{|m-1|}{\sqrt{1+m^2}} \left(\frac{1}{10} + \frac{1}{20}\right) = \frac{|m-1|}{\sqrt{1+m^2}} \cdot \frac{3}{20}$. Since $m = \frac{k}{h}$,$\sqrt{1+m^2} = \sqrt{1+\frac{k^2}{h^2}} = \frac{\sqrt{h^2+k^2}}{|h|}$. So,$\frac{2}{\sqrt{h^2+k^2}} = \frac{|\frac{k}{h}-1|}{\frac{\sqrt{h^2+k^2}}{|h|}} \cdot \frac{3}{20} = \frac{|k-h|}{\sqrt{h^2+k^2}} \cdot \frac{3}{20}$. $40 = 3|k-h| \Rightarrow 3(k-h) = \pm 40$. Given the geometry,the locus is $3x-3y+40=0$.

$A$ variable line $L$ passing through the origin cuts two parallel lines $x-y+10=0$ and $x-y+20=0$ at two points $A$ and $B$ respectively. If $P$ is a point on line $L$ such that $OA, OP, OB$ are in harmonic progression,then the locus of $P$ is

Similar Questions

$A$ straight line passing through a fixed point $(2, 3)$ intersects the coordinate axes at distinct points $P$ and $Q$. If $O$ is the origin and the rectangle $OPRQ$ is completed,then the locus of $R$ is:

Consider a square $ABCD$ of side $12$ and let $M, N$ be the midpoints of $AB, CD$ respectively. Take a point $P$ on $MN$ and let $AP=r, PC=s$. Then,the area of the triangle whose sides are $r, s, 12$ is

Suppose that the three points $A, B$ and $C$ in the plane are such that their $x$-coordinates as well as $y$-coordinates are in $GP$ with the same common ratio. Then,the points $A, B$ and $C$

If $ABC$ is an isosceles triangle and the coordinates of the base points are $B(1, 3)$ and $C(-2, 7)$, then the coordinates of $A$ can be:

The locus of the incentre of the triangle formed by the lines $xy-4x-4y+16=0$ and $x+y=5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module