Find the equation of the line passing through the point of intersection of the lines
x- 3y + 1 = 0 and 2x + 5y - 9 = 0 and whose distance from the origin is √5.
Answers
Given lines :- x-3y+1 = 0 ___(1)
2x+5y -9 = 0 ____''(2)
family of lines passing through two given Equation :-y + LY1 =0
Since, x-3y+1 +L ( 2x+5y-9 ) = 0
(1+2L)x + (5L-3)y +1-9L______(3)
Distance from (0,0 ) is √5 = |c|/√a²+b²
Since ,
=> 1-9L/ √(1+2L)²+(√5L-3)² = √5
squaring on both side,
=> (1-9L)²/(1+2L)²+(5L-3)² =( √5)²
=> 1²+(9L)²-18L / 1²+(2L)²+4L + (5L)²+(3)²-2*5L*3 = 5
=> 1 + 81L² -18L/ 29L² - 26L+ 10 =5
=> 1+81L²-18L = 5(29L²-26L+10)
=> 1+81L² -18L = 145L² - 130L + 50
=> 145L²-81L² -130L+18L +50-1 =0
=> 64L² - 112L +49 =0
We get L = 7/8
Now, put L = 7/8 in the (3) equation
=>(1+2*7/8)x+(5*7/8-3)y+1-9*7/8 = 0
=>(1+14/8)x+ (35/8-3 )y +8-63/8=0
=> (8+14/8)x+(35-24/8)y -55/8=0
=> 22/8x + 11/8y -55/8 =0
=> 11/8(2x + y -5) = 0
=> 2x+y-5 is required equation .
_________________________
2nd method is in given attachment .
Answer:
hii dii
Step-by-step explanation:
please mark me brainlest and follow me I will follow u back