The point A (2, 7) lies on the perpendicular

Solution:

False

If $A(2,7)$ lies on perpendicular bisector of $P(6,5)$ and $Q(0,-4)$, then $A P=A Q$

$\therefore$ $A P=\sqrt{(6-2)^{2}+(5-7)^{2}}$

$=\sqrt{(4)^{2}+(-2)^{2}}$

$=\sqrt{16+4}=\sqrt{20}$

and $A=\sqrt{(0-2)^{2}+(-4-7)^{2}}$

$=\sqrt{(-2)^{2}+(-11)^{2}}$

$=\sqrt{4+121}=\sqrt{125}$

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.

$\therefore$ Slope of perpendicular bisector $=\frac{-1}{\text { Slope of line segment joining }}$

the points $(5,-3)$ and $(0,-4)$

$=\frac{-1}{\frac{-4-(-3)}{0-5}}=5$ $\left[\because\right.$ slope $\left.=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right]$

[since, perpendicular bisector is perpendicular to the line segment, so its slopes have the condition, $m_{1} \cdot m_{2}=-1$ ]

 

Since, the perpendicular bisector passes through the mid-point of the line segment joining the points $(5,-3)$ and $(0,-4)$.

$\therefore$ Mid-point of $P Q=\left(\frac{5+0}{2}, \frac{-3-4}{2}\right)=\left(\frac{5}{2}, \frac{-7}{2}\right)$

So, the equation of perpendicular bisector having slope $\frac{1}{3}$ and passes through the mid-point $\left(\frac{5}{2}, \frac{-7}{2}\right)$ is,

$\left(y+\frac{7}{2}\right)=5\left(x-\frac{5}{2}\right)$

$\left[\because\right.$ equation of line is $\left.\left(y-y_{1}\right)=m\left(x-x_{1}\right)\right]$

$\Rightarrow \quad 2 y+7=10 x-25$

$\Rightarrow \quad 10 x-2 y-32=0$

$\Rightarrow \quad 10 x-2 y=32$

$\Rightarrow \quad 5 x-y=16$ ...(1)

Now check whether the Doint $A(2,7)$ lie on the Eq. (i) or not.

$5 \times 2-7=10-7=3 \neq 16$

Hence, the point A (2,7) does not lie on the perpendicular bisector of the line segment.