Question

11. Find the equation of the straight line passing through the point of intersection of the lines 2x+ y -3=0 and 5x + y -6 = 0 and parallel to the line joining the points
(1, 2) and (2, 1).

Answer 1

straight line passing through the point of intersection of the lines 2x+ y -3=0 and 5x + y -6 = 0 and parallel to the line joining the points
(1, 2) and (2, 1).

step 1:- solve 2x + y - 3 = 0 and 5x + y -6 = 0
2x + y - 3 - (5x + y - 6) = 0
=> 2x + y - 3 -5x - y + 6 = 0
=> -3x + 3 = 0
=> x = 1
put x = 1 in 2x + y - 3 = 0
2(1) + y - 3 = 0
=> 2 + y - 3 = 0
=> y = 1
hence, point of intersection of 2x + y -3 = 0 and 5x + y - 6 = 0 is (1,1)

step 2 :- unknown straight line is parallel to the line joining the points (1,2) and (2,1).
so, slope of line joining the points (1,2) and (2,1) = slope of unknown line.
=> (1 - 2)/(2 - 1) = slope of unknown line
=> slope of unknown line = -1

step 3 :- use formula (y - y1) = m(x - x1) to find equation of unknown straight line.
here, (x1,y1) = (1,1) and m = -1
so, equation : (y - 1) = -1(x - 1)
=> y -1 + x - 1 = 0
=> x + y - 2 = 0

hence, equation of St. line is x + y - 2 = 0

Answer 2

Solution:

i ) Finding the intersecting point of

two straight lines

2x + y - 3 = 0

=> y = -2x + 3 ---( 1 )

and 5x + y - 6 = 0 ---( 2 )

substitute y = -2x + 3 in equation ( 2 ),

5x - 2x + 3 - 6 = 0

=> 3x - 3 = 0

=> 3x = 3

=> x = 3/3 = 1

Substitute x = 1 in equation ( 1 ), we get

y = -2 × 1 + 3

=> y = 1

Intersecting point = ( 1 , 1 )

ii ) Slope of a line joining A( 1,2) =(x1,y1)

and B( 2,1) = ( x2 , y2 )

slope of AB = m1 = ( y2 - y1 )/( x2 - x1 )

=> m1 = ( 1 - 2 )/( 2 - 1 )

=> m1 = -1

iii ) slope of a line parallel to AB is

m2 = m1 = -1

iv ) We equation of the straight line

passing through ( 1 , 1 ) and slope

m2 = -1 is

y - y1 = m2 ( x - x 1 )

=> y - 1 = ( -1 ) ( x - 1 )

=> y - 1 + ( x - 1 ) = 0

=> y -1 + x - 1 = 0

=> x + y - 2 = 0

Therefore ,

Required equation ,

x + y - 2 = 0

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