Question

suppose p is (6, -1) and q is (2, 5). a find the equation of the perpendicular bisector of pq. b show that r(1, 0) lies on the perpendicular bisector. c show that r is equidistant from p and q.

Explanation:

Step1: Find the mid - point of PQ

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $P(6,-1)$ and $Q(2,5)$, the mid - point $M$ is $(\frac{6 + 2}{2},\frac{-1+5}{2})=(4,2)$.

Step2: Find the slope of PQ

The slope formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For $P(6,-1)$ and $Q(2,5)$, the slope of $PQ$ is $m_{PQ}=\frac{5+1}{2 - 6}=\frac{6}{-4}=-\frac{3}{2}$.

Step3: Find the slope of the perpendicular bisector

The slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$. So the slope of the perpendicular bisector of $PQ$, $m_{perp}=\frac{2}{3}$.

Step4: Find the equation of the perpendicular bisector

Using the point - slope form of a line $y - y_0=m(x - x_0)$ with $(x_0,y_0)=(4,2)$ and $m = \frac{2}{3}$, we have $y - 2=\frac{2}{3}(x - 4)$. Expanding gives $y-2=\frac{2}{3}x-\frac{8}{3}$, or $y=\frac{2}{3}x-\frac{8}{3}+2=\frac{2}{3}x-\frac{2}{3}$.

Step5: Check if R(1,0) lies on the perpendicular bisector

Substitute $x = 1$ and $y = 0$ into the equation $y=\frac{2}{3}x-\frac{2}{3}$. The right - hand side is $\frac{2}{3}\times1-\frac{2}{3}=0$, which is equal to the left - hand side. So $R(1,0)$ lies on the perpendicular bisector.

Step6: Find the distance between R and P

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For $R(1,0)$ and $P(6,-1)$, $d_{RP}=\sqrt{(6 - 1)^2+(-1-0)^2}=\sqrt{25 + 1}=\sqrt{26}$.

Step7: Find the distance between R and Q

For $R(1,0)$ and $Q(2,5)$, $d_{RQ}=\sqrt{(2 - 1)^2+(5 - 0)^2}=\sqrt{1+25}=\sqrt{26}$. Since $d_{RP}=d_{RQ}$, $R$ is equidistant from $P$ and $Q$.

Answer:

a. The equation of the perpendicular bisector is $y=\frac{2}{3}x-\frac{2}{3}$.
b. Substituting $x = 1$ and $y = 0$ into $y=\frac{2}{3}x-\frac{2}{3}$ gives $0=\frac{2}{3}\times1-\frac{2}{3}$, so $R(1,0)$ lies on the perpendicular bisector.
c. $d_{RP}=\sqrt{(6 - 1)^2+(-1-0)^2}=\sqrt{26}$ and $d_{RQ}=\sqrt{(2 - 1)^2+(5 - 0)^2}=\sqrt{26}$, so $R$ is equidistant from $P$ and $Q$.