Question

Question 6
Find the coordinates of the point Q on the X-axis which lies on the perpendicular bisector of the line segment joining the points A(-5,-2) and B(4,-2). Name the type of triangle formed by the point Q, A and B.

Answer 1

Firstly, we plot the points of the line segment on the paper and join them.

We know that, the perpendicular bisector of the line segment AB bisect the segment AB, i.e. perpendicular bisector of the line segment AB passes through the mid-point of AB.
$\therefore \text{Mid-point of AB}=(\frac{-5+4}{2},\frac{-2-2}{2})\Rightarrow R=(-\frac{1}{2},-2)$
$\left[\begin{array}{c}\because \text{Mid-point of a line segment passes}\\ \text{through the points}(x_1,y_1)and(x_2,y_2)\\ =(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\end{array}\right]$

$\text{Now, we draw a straight line on paper passes through}\text{the mid-point R. we see that}\text{perpendicular bisector cuts the X-axis at the point}Q(-\frac{1}{2},0)\text{Hence, the required coordinates of}Q=(-\frac{1}{2},0)$

Alternate Method
(i) To find the coordinates of the point of Q on the X-axis. We find the equation of perpendicular bisector of the line segment AB.
Now, slope of line segment AB,
$Let{m}_1=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-(-2)}{4-(-5)}=\frac{-2+2}{4+5}=\frac{0}{9}\Rightarrow m_1=0$
Let the slope of perpendicular bisector of line segment is m₂.
Since, perpendicular bisector is perpendicular to the line segment AB.
By perpendicularity condition of two lines,
$m_1.m_2=-1\Rightarrow m_2=\frac{-1}{m_1}=\frac{-1}{0}\Rightarrow m_2=\infty$
Also, we know that, the perpendicular bisector is always passes through the mid-point of the line segment.
$\therefore Mid-point=(\frac{-5+4}{2},\frac{-2-2}{2})=(\frac{-1}{2},-2)\because mid-point=\left[(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\right]$
To find the equation of perpendicular bisector of line segment, we find the slope and a point through which perpendicular bisector is pass.
Now, equation of perpendicular bisector having slope infinity and passing through the point $(\frac{-1}{2},-2)$ is,
$(y+2)\equiv \infty (x+\frac{1}{2})\because (y-y_1)=m_2(x-x_1)\Rightarrow \frac{y+2}{x+\frac{1}{2}}\equiv \infty =\frac{1}{0}\Rightarrow x+\frac{1}{2}=0\therefore x=\frac{-1}{2}$
So, the coordinate of the point Q is $(\frac{-1}{2},0)$ on the X-axis which lies on the perpendicular bisector of the line segment joining the point AB.
To know the type of triangle formed by the points Q, A and B. We find the length of all three sides and see whatever condition of triangle is satisfy by these sides.
Now, using distance formula between two points.
$AB=\sqrt{(4+5{)}^2+(-2+2{)}^2}=\sqrt{(9{)}^2+0}\left[\begin{array}{c}\because \text{Distance between the points}(x_1,y_1)and(x_2,y_2),d\\ =\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\end{array}\right]BQ=\sqrt{{(\frac{-1}{2}-4)}^2+(0+2{)}^2}=\sqrt{{\left(\frac{-9}{2}\right)}^2+(2{)}^2}=\sqrt{\frac{81}{4}+4}=\sqrt{\frac{97}{4}}AndQA=\sqrt{(-5+\frac{1}{2}+(-2-0{)}^2)}=\sqrt{{\left(\frac{-9}{2}\right)}^2+(2{)}^2}=\sqrt{\frac{81}{4}+4}=\sqrt{\frac{97}{4}}=\sqrt{\frac{97}{4}}$

We see that, BQ=QA$\ne$ AB which shows that the triangle formed by the points Q,A and B is an isosceles.