Question

If 2x+3y+4=0is perpendicular bisector of the segment joining the points A (1, 2) and Bα,βand α+β=81kthen k =___​

Answer 1

Answer:

-1/13

Step-by-step explanation:

the mid point of line segment joining A and B will be given by

(a+1)/2,(b+2)/2

this point lies on the line 2x + 3y + 4 = 0

2(a+1)/2 + 3(b+2)//2 + 4 = 0

(a+1)+3b/2 + 3 + 4 = 0

multiplying by 2

2a+2+3b+6+8= 0

2a+3b= -16.......(1)

again

slope of line AB = (b-2)/a-1)

the given line

2x+3y+4=0

3y= -2x-4

y = (-2/3) + (-4/3(

the slope of above line is -2/3

since the lines are perpendicular

(-2/3)*(b-2)/(a-1) = -1

2(b-2) = 3(a-1)

2b -4 = 3a-3

3a-2b= -1.........(2)

from (1)x2 and (2)x3

4a + 6b = -32

9a - 6b = -3

this gives

13a = -35

a = -35/13

this gives b = (3a+1)/2

= (3*(-35/13)+1)/2

=(-105+13)/26

= -92/26

b = -46/13

now

a+b = 81k

-35/13 + (-46/13) = 81k

-81/13 = 81k

k = -81/(13*81)

k= -1/13