Question

Find the equation of the straight line which passes through the midpoint of the line segment joining $(4,2)$ and $(3,1)$ whose angle of inclination is ${30}^o$.

• A. $2x-2\sqrt{3}y+(3\sqrt{3}-7)=0$
• B. $2x+2\sqrt{3}y+(3\sqrt{3}-7)=0$
• C. $2x-2\sqrt{3}y-(3\sqrt{3}-7)=0$
• D. None of these

Answer 1

Answer: (A)

$2x-2\sqrt{3}y+(3\sqrt{3}-7)=0$

Consider the given points.

$(4,2)$ and $(3,1)$

Mid-point $=(\frac{4+3}{2},\frac{2+1}{2})$

$=(\frac{7}{2},\frac{3}{2})$

Since, the angle $={30}^0$

$m=\tan {30}^0=\frac{1}{\sqrt{3}}$

We know that the equation of the line

$y-y_1=m(x-x_1)$

So,

$y-\frac{3}{2}=\tan {30}^0(x-\frac{7}{2})$

$y-\frac{3}{2}=\frac{1}{\sqrt{3}}(x-\frac{7}{2})$

$\left(\frac{2y-3}{2}\right)=\frac{1}{\sqrt{3}}\left(\frac{2x-7}{2}\right)$

$\sqrt{3}\left(\frac{2y-3}{2}\right)=\left(\frac{2x-7}{2}\right)$

$\sqrt{3}(2y-3)=(2x-7)$

$2\sqrt{3}y-3\sqrt{3}=2x-7$

$2x-2\sqrt{3}y+3\sqrt{3}-7=0$

Hence, this is the answer.