Question

The equation of the perpendicular bisector of line segment joining points A(4,5) and B(-2,3) is​

Answer 1

Step-by-step explanation:

Since, the slope of the line AB= x 2 −x 1 y 2 −y 1 = 6−2−5−3 = 4−8 =−2 But slope of perpendicular bisector will be=− m1 = 21 Therefore, the equation of the liney+1= 21 (x−4)2y+2=x−4x−2y=6

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Answer 2

To find:

The equation of the perpendicular bisector of line segment joining the points A (4, 5) and B (- 2, 3)

Step-by-step explanation:

The equation of the line segment AB is

$\quad \dfrac{y-5}{5-3}=\dfrac{x-4}{4-(-2)}$

$\Rightarrow \dfrac{y-5}{2}=\dfrac{x-4}{6}$

$\Rightarrow y-5=\dfrac{x-4}{3}$

⇒ 3y - 15 = x - 4

⇒ x - 3y + 11 = 0

Let, the line perpendicular to the line segment AB be

3x + y = k ... ... (i)

The coordinates of the mid-point of AB are

$\quad (\dfrac{4-2}{2},\:\dfrac{5+3}{2})$

i.e., $(\dfrac{2}{2},\:\dfrac{8}{2})$

i.e., (1, 4)

Here (i) no. line passes through the ppint (1, 4). Then

3 + 4 = k

⇒ k = 7

From (i), we get

3x + y = 7

Final answer: 3x + y = 7

The equation of the perpendicular bisector of line segment joining the points A (4, 5) and B (- 2, 3) is 3x + y = 7.

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