Question

The point $P$ is equidistant from $A\left( 1,3 \right)$,$B\left( -3,5 \right)$ and $C\left( 5,-1 \right)$. Find length of $PA$.
A.5
B. $5\sqrt{5}$
C.25
D. $5\sqrt{10}$

Answer 1

Hint: In the given question, the point $P$ is equidistant from $A\left( 1,3 \right)$,$B\left( -3,5 \right)$ and $C\left( 5,-1 \right)$. We will find the perpendicular bisector of $AB$ and $AC$. The intersection point of $AB$ and $AC$ is the coordinate of $P$. By using the distance formula, we will find the length of $PA$.

Formula Used:
The slope of a line joining by two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.
The relation between the slopes of two perpendicular lines is ${m_1}{m_2} = - 1$.
The formula of midpoints of points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.
The equation of a line which slope is $m$ and passing through the point $\left( {{x_1},{y_1}} \right)$ is $y - {y_1} = m\left( {x - {x_1}} \right)$.
The distance between the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $.

Complete step by step solution:
The coordinate of points $A,B,$ and $C$are $A\left( {1,3} \right)$,$B\left( { - 3,5} \right)$ and $C\left( {5, - 1} \right)$.
The slope of a line joining by two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.
The slope of line $AB$ is $\dfrac{{5 - 3}}{{ - 3 - 1}}$
                                         $ = \dfrac{2}{{ - 4}}$
                                       $ = - \dfrac{1}{2}$
Let ${m_1}$ be the slope of the perpendicular on $AB$.
Therefore,
${m_1} \times \left( { - \dfrac{1}{2}} \right) = - 1$
Calculate the value of ${m_1}$.
${m_1} = 2$
The formula of midpoints of points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.
The midpoint of $A\left( {1,3} \right)$ and $B\left( { - 3,5} \right)$ is $\left( {\dfrac{{1 - 3}}{2},\dfrac{{3 + 5}}{2}} \right) = \left( { - 1,4} \right)$
The equation of line which slope is $m$ and passing through the point $\left( {{x_1},{y_1}} \right)$ is $y - {y_1} = m\left( {x - {x_1}} \right)$.
The equation of line perpendicular to the line $AB$ is
$y - 4 = 2\left( {x + 1} \right)$
Simplify the above equation,
$y - 4 = 2x + 2$
$2x - y + 6 = 0$ …….(i)
The slope of a line joining by two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.
The slope of line $AC$ is $\dfrac{{ - 1 - 1}}{{5 - 3}}$
                                         $ = \dfrac{{ - 2}}{2}$
                                       $ = - 1$
Let ${m_2}$ be the slope perpendicular on $AC$.
Therefore,
${m_2} \times \left( { - 1} \right) = - 1$
Calculate the value of ${m_2}$.
${m_2} = 1$
The formula of midpoints of points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.
The midpoint of $A\left( {1,3} \right)$ and$C\left( {5, - 1} \right)$ is $\left( {\dfrac{{1 + 5}}{2},\dfrac{{3 - 1}}{2}} \right) = \left( {3,1} \right)$
The equation of line which slope is $m$ and passing through the point $\left( {{x_1},{y_1}} \right)$ is $y - {y_1} = m\left( {x - {x_1}} \right)$.
The equation of line perpendicular to the line $AB$ is
$y - 1 = 1\left( {x - 3} \right)$
Simplify the above equation,
$y - 1 = x - 3$
$x - y - 2 = 0$ …(ii)
Now we will solve (i) and (ii).
Subtract equation (ii) from (i) we get.
     $2x - y + 6 = 0$
     $x - y - 2 = 0$
$( - )$ $( + )$$( + )$
$\overline {x\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 8 = 0} $
Therefore $x = - 8$.
Putting $x = - 8$ in the equation (ii)
$ - 8 - y - 2 = 0$
$ \Rightarrow - y - 10 = 0$
$ \Rightarrow y = - 10$
The coordinate of $P$ is $\left( { - 8, - 10} \right)$.
The distance between the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $.
The length of $PA$ is $\sqrt {{{\left( {1 + 8} \right)}^2} + {{\left( {3 + 10} \right)}^2}} $
                                   $ = \sqrt {{9^2} + {{13}^2}} $
                                  $ = 5\sqrt {10} $ units

Option ‘D’ is correct

Note: Another formula to find the equation of perpendicular bisector of the line segment joining by the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $2x\left( {{x_1} - {x_2}} \right) + 2y\left( {{y_1} - {y_2}} \right) = \left( {{x_1}^2 + {y_1}^2} \right) - \left( {{x_2}^2 + {y_2}^2} \right)$. We can use this formula to find the equations of $AB$ and $AC$.