Question

Question 8
The point A(2,7) lies on the perpendicular bisector of the lie segment joining the points P(6,5) and Q(0,-4).

Answer 1

False
If A(2,7) lies on perpendicular bisector of P(6,5) and Q(0,-4), then AP = AQ
$\therefore AP=\sqrt{(6-2{)}^2+(5-7{)}^2}=\sqrt{(4{)}^2+(-2{)}^2}=\sqrt{16+4}=\sqrt{20}AndAQ=\sqrt{(0-2{)}^2+(-4-7{)}^2}=\sqrt{(-2{)}^2+(-11{)}^2}=\sqrt{4+121}=\sqrt{125}$
So, A does not lies on the perpendicular bisector of PQ.
Alternate Method
If the point A(2,7) lies on the perpendicular bisector of the line segment, then the point A satisfy the equation of perpendicular bisector.
Now, we find the equation of perpendicular bisector. For this, we find the slope of perpendicular bisector.
$\therefore \text{Slope of perpendicular bisector}=\frac{-1}{Slopeofthesegmentjoining}=\frac{-1}{\frac{-4-5}{0-6}}=-\frac{2}{3}[\because Slope=\frac{y_2-y_1}{x_2-x_1}]$
[since, perpendicular bisector is perpendicular to the line segment, so its slopes have the condition, $m_1.m_2=-1$]
Since, the perpendicular bisector passes through the mid-point of the line segment joining the points (6,5) and (0,-4).
$\therefore \text{Mid-point of PQ}=(\frac{6+0}{2},\frac{5-4}{2})=(3,\frac{1}{2})$ So, the equation of perpendicular bisector having slope $\frac{-2}{3}$ and passes through the mid-point $(3,\frac{1}{2})$ is
$(y-3)=\frac{-2}{3}(x-\frac{-1}{2})\because \left[\text{equation of line is}\right(y-y_1)=m(x-x_1\left)\right]\Rightarrow y-3=\frac{-2}{3}x+\frac{-1}{3}\Rightarrow 3y-9+2x+1=0\Rightarrow 3y+2x-8=0....\left(i\right)$ Now, check whether the point A(2,7) lie on the Eq.(i) or not. $3\times 7-8+2\times 2=17\ne 0$. Hence, the point A(2,7) does not lie on the perpendicular bisector of the line segment.