Question

Find the equation of the right bisector of the line segment joining the points A (1, 0) and B (2, 3).

Answer 1

$\text{The given piont are A (1, 0) and B (2, 3)}\text{Let M be the midpoint of AB.}\text{Coordinates of M}=(\frac{1+2}{2},\frac{0+3}{2})=(\frac{3}{2},\frac{3}{2})$

$\text{And, slope of AB}=\frac{3-0}{2-1}=3\text{Let m be the slope of the perpendicular bisector of the line joining the points A (1, 0) and B (2, 3)}\therefore m\times SlopeofAB=-1\Rightarrow m\times 3=-1\Rightarrow m=-\frac{1}{3}\text{So, the equation of the line that passes}\text{through M}(\frac{3}{2},\frac{3}{2})\text{and has slope}-\frac{1}{3}isy-\frac{3}{2}=-\frac{1}{3}(x-\frac{3}{2})\Rightarrow x+3y.-6=0\text{Hence, the equation of the right bisector of the line segment joining the points A (1, 0) and B (2, 3) is x + 3y - 6 = 0}\text{P.Q. Find the equation of the line passing through (1, 2) and making angle of}{30}^{\circ }\text{with y=axis.}.\text{Equation of the line passing through}(x_1,y_1)\text{and making angle}\theta \text{with the x-axis is,}(y-y_1)=tan\theta (x-x_1)Here,(x_1,y_1)=(1,2),\text{angle with y-axis is}{30}^{\circ }\therefore \text{angle with x-axis is}\theta ={90}^{\circ }-{30}^{\circ }={60}^{\circ }(y-y_1)=tan\theta (x-x_1)(y-2)=\left(tan{60}^{\circ }\right)(x-1)y-2=\sqrt{3}x-\sqrt{3}\sqrt{3}x-y+2-\sqrt{3}=0$