Question

A straight line through $P(2,c+f-1)$, inclined at an angle of ${60}^{\circ }$ with positive Y-axis in clockwise direction. The co-ordinates of one of the points on its at a distance $(c+f)$ units from point $P$ is $(c,f$ obtained from previous question).

• A. $(2+2\sqrt{3},5)$
• B. $(3+2\sqrt{3},3)$
• C. $(2+3\sqrt{2},4)$
• D. $(2+3\sqrt{2},3)$

Answer 1

Answer: (A)

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

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