Question

If $(a,b)$ is the co-ordinates of the point obtained in previous question, then the equation of line which is at the distance $|b-2a-1|$ units from origin and make equal intercept on co-ordinate axes in first quadrant, is

• A. $x+y+4\sqrt{6}=0$
• B. $x+y+2\sqrt{6}=0$
• C. $x+y-4\sqrt{6}=0$
• D. $x+y-2\sqrt{6}=0$

Answer 1

The correct option is A $x+y+4\sqrt{6}=0$

Given line

$AB:3x+4y=5$

$AC:4x-3y=15$

On comparing above equation with $y=mx+c$ where m is slope of line then we get

$m_{AB}=-\frac{3}{4}$ and $m_{AC}=\frac{4}{3}$

Let the equation of BC from point $(1,2)$ and slope m

$BC:y-2=m(x-1)---\left(1\right)$

Here ABC is isosceles triangle So

Angle between line $AB$ and $BC$ is equal to angle between line $AC$ and $BC$

By angle between two lines formula $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\left|\frac{-\frac{3}{4}-m}{1-\frac{3}{4}m}\right|=\left|\frac{\frac{4}{3}-m}{1+\frac{4}{3}m}\right|$

$\left|\frac{-3-4m}{4-3m}\right|=\left|\frac{4-3m}{3+4m}\right|$

$\frac{4m+3}{4-3m}=\frac{4-3m}{4m+3}$

$(4m+3{)}^2=(4-3m{)}^2$

$16m^2+9+24m=16+9m^2-24m$

$7m^2+48m-7=0$

On sovling we get

$m=\frac{1}{7},-7$

Equation of line BC

$7(y-2)=x-1$ and $y-2=-7(x-1)$

$7y-14=x-1$ and $y-2=-7x+7$

$x-7y+13=0$ and $7x+y-9=0$

On comparing above equations with $ax+by+x=0$ and $dx+ey+f=0$ respectively we get

$c=13,f=-9$

$c+f=13-9=4$

Now Point $P(2,c+f-1)\equiv (2,3)$

The line inclined ${60}^0$ at Y-axis $x=0$in clockwise direction

Hence Comparing Y-axis equation then the line make a right angle triangle

while intersecting X-axis So it inclined angle of ${30}^0$ with x-axis

Hence slope of line be $m=\tan {30}^0=\frac{1}{\sqrt{3}}$

Equation of line from point $P(2,3)$ and slope $m=\frac{1}{\sqrt{3}}$

$y-3=\frac{1}{\sqrt{3}}(x-2)$

$y=\frac{1}{\sqrt{3}}(x-2)+3----\left(1\right)$

Let the coordinates of point whose distance is $c+f$ from P is $Q(x,c+f+1)\equiv (x,5)$

From equation (1)

$5-3=\frac{1}{\sqrt{3}}(x-2)$

$2\sqrt{3}=x-2$

$x=2+2\sqrt{3}$

Hence point $Q(2+2\sqrt{3},5)$

Here point $(a,b)\equiv Q(2+2\sqrt{3},5)$

$a=2+2\sqrt{3},b=5$

$|b-2a-1|=|5-2(2+2\sqrt{3})-1|$

$|b-2a-1|=|5-4-4\sqrt{3}-1|$

$|b-2a-1|=|-4\sqrt{3}|$

$|b-2a-1|=4\sqrt{3}$

Here equation of line make equal intercept on co-ordinate axes in first quadrant is

$\frac{x}{a}+\frac{y}{a}=1$

$x+y=a\Rightarrow x+y-a=0----\left(2\right)$

Distance of above line from origin is $4\sqrt{3}$

Hence $4\sqrt{3}=\frac{0+0-a}{\sqrt{1^2+1^2}}$

$4\sqrt{3}=\frac{-a}{\sqrt{2}}$

$4\sqrt{3}=\frac{-a}{\sqrt{2}}$

$a=-4\sqrt{6}$

Putting a in eq (2)

$x+y+4\sqrt{6}=0$