Question

Question 4
Point P(0,2) is the point of intersection of Y-axis and perpendicular bisector of line segment joining the points A(-1,1) and B(3,3).

Answer 1

False
We know that the point lying on perpendicular bisector of the line segment joining two points is equidistant from these two points.
$\therefore PA=\sqrt{[-1-(0){]}^2+(1-2{)}^2}=\sqrt{(-1{)}^2+(-1{)}^2}=\sqrt{2}PB=\sqrt{[3-0{]}^2+(3-2{)}^2}=\sqrt{(3{)}^2+(1{)}^2}=\sqrt{10}\because PA\ne PB$
So, the point P does not lie on the perpendicular bisector of AB.
Alternate Method
Slope of the line segment joining the points A(-1,1) and B(3,3)
$m_1=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}[\because m=\frac{y_2-y_1}{x_2-x_1}]$
Since, the perpendicular bisector is perpendicular to the line segment, so its slope,
$m_2=-\frac{1}{m_1}=-2$ [by perpendicularity condition, m₁m₂ = - 1]
Also, the perpendicular bisector will be passing through the mid-point of the line segment joining the points A(-1,1) and B(3,3).
$\therefore \text{Mid-point}=(\frac{-1+3}{2},\frac{1+3}{2})=(1,2)$
[Since, mid-point of the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is
$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
Now, equation of perpendicular bisector having slope (-2) and passing through the point (1,2) is
$(y-2)=(-2)(x-1)\Rightarrow y-2=-2x+2\Rightarrow 2x+y=4..............\left(i\right)$
[Since, the equation of line is $(y-y_1)=m(x-x_1)$]
If the perpendicular bisector cuts the Y-axis, then put x = 0 in Eq.$\left(i\right)$, we get
$2\times 0+y=4$
$\Rightarrow y=4$
Hence, point P(0,4) is the point of intersection of Y-axis and the perpendicular bisector of line segment joining points A(-1,1) and B(3,3).