Dear Student,
Midpoint of points P(2,0) and Q(0,29) is given by, (x1+x22,y1+y22)Midpoint of P(2,0) and Q(0,29) = (22,292) = (1,19)Given (1,p3) is the midpoint of points P(2,0) and Q(0,29)So, p3 = 19p =13We have to show that ,Line 5x+3y+2=0 ....(i), passes through the point (−1,3×13)Substituting the values of x = −1 and y =1 in equ (i), we get5(−1)+3(1)+2 = 0−5+3+2=0−5+5 = 0LHS = RHS.So, the given line 5x+3y+2=0 passes through the point (−1,1)
That is (-1,3p)
Another method
2)
let ( x1 , y1 ) = A( 2 , 0 ) ,
( x2 , y2 ) = B ( 0 , 2/9 ) ;
mid point of joining of A and B = ( x1+ x2 /2 , y1 + y2 /2 )
( 1 , p/ 3 ) = ( 0 + 2 /2 , 0 + 2/9 / 2 )
= ( 1 , 1/9 )
p/3 = 1/9
[ ∵ If ( a , b ) = ( c , d ) then a = c and b = d ]
p = 1/3 --- ( 1 )
according to the problem given ,
put ( -1 , 3p ) in the equation 5x + 3y + 2 =0
5 ( -1 ) + 3 × ( 3p ) + 2 = 0
-5 + 3 × 3 ( 1/3 ) + 2 =0 [ from ( 1 ) ]
-5 + 3 + 2 =0
0 = 0 [ true ]
Therefore ,
5x + 3y + 2 =0 line passes through the
point ( -1 , 3p ) .
Another method 3)
Coordinates of the mid point X on the line joining of the points A and B are:
X= [( x1+x2)/2 ,( y1+y2)/2 ] ( midpoint Formula)
Since (1,p/3) is the midpoint of the line segment joining the points (2,0) & (0,2/9)
Here,
x= 1, y= p/3 , x1= 2, y1= 0 , x2= 0 , y2= 2/9
y= (y1+y2)/2
p/3=( 0+2/9)/2
p/3= (2/9)/2
2p =( 2/9)× 3
p = 3/9
p = ⅓
Given : (-1,3p)
x= -1 , y= 3p
y= 3× ⅓
y = 1
5x+3y+2=0. (Given)
Put the value of x & y
5 (-1) + 3(1)+2= 0
-5+3+2= 0
-2+2= 0
Hence, The line 5x + 3y + 2 = 0 passes through the point (–1, 1) as 5(–1) + 3(1) + 2 = 0.