Question

suppose a,b are two points on 2x+_y+3=0 and p (1,2)is such that pa=pb then the mid point of about is​

Answer 1

Step-by-step explanation:

2x+-y+3=0

2x-y+3=0

.

2x-y=0-3

2x-y=3

x-y=3/2

Answer 2

Given = 2x-y+3=0

point = (1,2)

To find : The coordinates of mid point

Solution :

point P be (1,2) and A and B are the two points  and PA = PB

since PA = PB

PQ is perpendicular to AB

and Q is the foot of the perpendicular

Q is also the mid point of AB

and AQ = AB

thus ,

let the points of Q be (x,y)

then using the mid point formula

$\frac{x-1}{2}=\frac{y-2}{-1}= \frac{-(2-2+3)}{4+1}\\\\\frac{x-1}{2}=\frac{y-2}{-1}=\frac{-3}{5}\\\\\frac{x-1}{2}=\frac{-3}{5}\\\\x-1 = \frac{-6}{5}\\\\x = \frac{-1}{5}\\\\y-2 = \frac{3}{5}\\\\y = \frac{13}{5}$

hence , the coordinates of Q are $(\frac{-1}{5},\frac{13}{5})$

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