Question

If x + 3y = 16 is the perpendicular bisector of AB and A(5, 7), then B =​

Answer 1

$\textbf{Given:}$

$\text{x+3y-16=0 is the perpendicular bisector of AB and A(5,7)}$

$\textbf{To find:}\;\;B$

$\text{Since AB is perpendicular to x+3y-16=0,}$

$\text{the equation of AB can be written as 3x-y+k=0}$

$\text{It passes through (5,7)}$

$\implies\,3(5)-7+k=0$

$\implies\,8+k=0$

$\implies\,k=-8$

$\textbf{The equation of AB is 3x-y-8=0}$

$\text{Now,}$

$\text{The midpoint of AB is the point of intersection of the lines x+3y-16=0 and 3x-y-8=0}$

$x+3y-16=0$

$9x-3y-24=0$

$\text{Adding,}\;10x-40=0$

$\implies\,x=4$

$\text{and y=4}$

$\text{Let the coordinates of B be (h,k)}$

$\text{Clearly (4,4) is the midpoint of AB}$

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(4,4)$

$(\frac{5+h}{2},\frac{7+k}{2})=(4,4)$

$\frac{5+h}{2}=4\;\text{and}\;\frac{7+k}{2}=4$

$5+h=8\;\text{and}\;7+k=8$

$h=3\;\text{and}\;k=1$

$\therefore\textbf{The coordinates of B is (3,1)}$

Find more:

1.Find the coordinates of the foot of perpendicular drawn from the point(-2,3) on the line 3x-2y+5=0

https://brainly.in/question/11747216

2.Find the coordinates of the foot of the perpendicular drawn from the point (1 -2) to the line y=2x+1

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Answer 2

Given :

The equation of line x + 3 y = 16

The line is perpendicular to line AB

The co-ordinate of point A = 5 , 7

To Find :

The co-ordinate of point B

Solution :

Let The  co-ordinate of point B = $x_1 , y_1$

The line x + 3 y = 16 is perpendicular to line AB

The mid point of perpendicular of line AB= $\dfrac{5+x_1}{2}$ ,  $\dfrac{7+y_1}{2}$

As line x + 3 y = 16 is meet at point $\dfrac{5+x_1}{2}$ ,  $\dfrac{7+y_1}{2}$

So, Equation of line written as

  $\dfrac{5+x_1}{2}$ + 3 ( $\dfrac{7+y_1}{2}$ ) = 16

Or, 5 + $x_1$ + 21 + 3 $y_1$ = 32

Or,  $x_1$ + 3 $y_1$ = 6                      .............1

Again

The slope of line $x_1$ + 3 $y_1$ = 6

i.e        3 $y_1$ = 6 - $x_1$

Or,          y = - $\dfrac{x_1}{3}$ + $\dfrac{6}{3}$

         $m_1$ = $\dfrac{-1}{3}$

Again

Slope of line AB

   $m_2$ = $\dfrac{y_1-7}{x_1-5}$

∵ When slopes are perpendicular , product of slope = - 1

∴  $m_1 m_2$ = -1

Or, ( $\dfrac{-1}{3}$ ) × ( $\dfrac{y_1-7}{x_1-5}$ ) = - 1

Or,   - 1 ( $y_1$ - 7 ) = - 1 ( 3$x_1$ - 15 )

Or,     $y_1$ - 7 = 3$x_1$ - 15

i.e     $y_1$ =  3$x_1$ - 8                    ............2

Solving eq 1and eq 2

$x_1$ + 3 ( 3$x_1$ - 8 ) = 6  

Or,  $x_1$ + 9$x_1$ - 24 = 6  

Or,  10 $x_1$  = 24 + 6

Or,      $x_1$  = $\dfrac{30}{10}$

i.e     $x_1$  = 3

Put the value of $x_1$  in eq 2  , we get

 $y_1$ =  3 × 3 - 8    

     = 9 - 8

     = 1

i.e  Point  $x_1$ , $y_1$ = 3 , 1

So, The co-ordinate of point B = ( $x_1$ , $y_1$ ) = 3 , 1

Hence,  The co-ordinate of point B is 3 , 1     Answer