Question

If a point (x,y) is equidistant from the Q(9,8) and S(17,8), then
(i) x+y=13

(ii) x-13=0

(iii) y-13=0

(iv)x-y=13​

Answer 1

Answer:

(ii) x - 13 = 0

Step-by-step explanation:

m = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

y - $y_{1}$ = m( x - $x_{1}$ )

Coordinates of midpoint:

( $\frac{x_{1} +x_{2} }{2}$ , $\frac{y_{1} +_{2} }{2}$ )

~~~~~~~~~~~~~~~~~~

[(9 + 17)/2 , (8 + 8)/2] = (13, 8)

$m_{QS}$ = $\frac{8-8}{17-9}$ = 0

Equation of line passing through Q and S is y = 8

Equation of the perpendicular bisector is x = 13 or x - 13 = 0

Answer 2

Given,

The coordinates of point Q$\[ = \left( {9,8} \right)\]$

The coordinates of point S$\[ = \left( {17,8} \right)\]$

Solution,

Formula used, The coordinates of the midpoint of a line segment$\[ = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{x_1} + {x_2}}}{2}} \right)\]$

Apply the formula of the mid-point of a line segment.

Therefore,

$\[\begin{array}{l} \Rightarrow \left( {\frac{{9 + 17}}{2},\frac{{8 + 8}}{2}} \right)\\ \Rightarrow \left( {\frac{{26}}{2},\frac{{16}}{2}} \right)\\ \Rightarrow \left( {13,8} \right)\end{array}\]$

The slope of the line is

$\[\begin{array}{l} \Rightarrow \frac{{8 - 8}}{{17 - 9}}\\ \Rightarrow 0\end{array}\]$

So, the equation of the perpendicular bisector is

$\[\begin{array}{l} \Rightarrow x = 13\\ \Rightarrow x - 13 = 0\end{array}\]$

Hence, the correct option is (ii) i.e. $\[x - 13 = 0\].$