Question

Find the equation of the line AB in the following figure, given $OP={\displaystyle \frac{3}{2}}$.$$$$

Answer 1

Hint:We find the equation of the line AB in slope point form with slope $m$ and a point on the line $(x_1,y_1)$ as $y-y_1=m(x-x_1)$. We find co-ordinate of the point P lying on the line using$OP={\displaystyle \frac{3}{2}}$. We find the slope of AB taking tangent of the angle AB makes with positive $x-$axis whose measurement we find using the angle ${30}^{\circ }$ that AB subtends with $y-$axis in the figure.$$$$

Complete step by step answer:
We see in the given figure that the starlight line AB is inclined on $y-$axis in the coordinate plane. The line AB intersects the $y-$axis at point P and also subtends an angle of ${30}^{\circ }$We take a point R above the point P which also lies on the $y-$ axis. We denote the point of intersection of the line AB with the $x-$axis as Q and the angle of inclination the line AB makes with the positive direction of $x-$axis as $\theta .$ We have the constructed figure as,$$$$

We know the $x-$coordinate of any point is the distance of the point M from the $y-$axis and the $y-$coordinate is called ordinate is the distance of the point M from the $x-$axis .Then we can write the co-ordinates of M as$M(x,y)$. Here the$y-$coordinate of the point is equal to OP which given in the question as $OP={\displaystyle \frac{3}{2}}$ and the $x-$coordinate is 0 as P lies on $y-$axis. So the coordinate of the P is $P(0,{\displaystyle \frac{3}{2}}).$$$$$
We have $\mathrm{\angle }RPB=\mathrm{\angle }OPQ={30}^{\circ }$ because the both angles are vertical opposite to each other. We have the right angle$\mathrm{\angle }POQ={90}^{\circ }$. We use the property that sum of internal angles in a triangle is ${180}^{\circ }$ to find$\mathrm{\angle }PQO=\theta$. So we have
$$\begin{array}{rl}& \mathrm{\angle }POQ+\mathrm{\angle }PQO+\mathrm{\angle }OPQ={180}^{\circ }\\ & \Rightarrow {90}^{\circ }+\theta +{30}^{\circ }={180}^{\circ }\\ & \Rightarrow \theta ={60}^{\circ }\end{array}$$
We know that the slope of a line is tangent of the angle it makes with positive $x-$axis in anticlockwise direction. Here the lien AB makes angles $\theta$ with positive $x-$axis. So the slope $m$ of AB is
$$\tan \theta =\tan \left({60}^{\circ }\right)=\sqrt{3}$$
We know the equation of the line in slope point form with slope $m$ and a point on the line $(x_1,y_1)$ as
$$y-y_1=m(x-x_1)$$
The slope of the line AB is $m=\sqrt{3}$ and the point $P(0,{\displaystyle \frac{3}{2}})$ lies on it. So the equation of the line AB is
$$\begin{array}{rl}& y-{\displaystyle \frac{3}{2}}=\sqrt{3}(x-0)\\ & \Rightarrow 2y-3=2\sqrt{3}x\\ & \Rightarrow 2\sqrt{3}x+2y-3=0\end{array}$$
Note:
We can also find the equation of the any line with slope $m$ and $y-$intercept $c$ is given by $y=mx+c.$ Here in this problem$c={\displaystyle \frac{3}{2}}$. The equation of the lines with two points $(x_1,y_1),(x_2,y_2)$ is given by $y-y_1={\displaystyle \frac{y_2-y_1}{x_2-x_1}}(x-x_1)$.