Find the equation of the straight line perpendicular to the line 2x+3y=0 and passing through the point of intersection of the lines x+3y-1=0 and x-2y+4=0.
Answers
Answer:
3x-2y+8=0
The photo is the answer
The required equation is 2x − 3y +7 = 0
Given:
2x+3y=0
and x+3y-1 = 0 and x - 2y+4 = 0
To find:
The equation of the straight line perpendicular to the line 2x+3y=0 and passing through the point of intersection of the lines x+3y-1=0 and x-2y+4=0
Solution:
Given lines
2x + 3y = 0 ------(1)
x + 3y - 1 = 0 ------(2)
x - 2y + 4 = 0 -----(3)
Note: The formula for equation of a straight line which is perpendicular to the line ax+by+c = 0 is bx–ay+λ = 0
From the above formula,
The equation of the line which is perpendicular to the line 2x+3y=0 is
2x −3y = λ ----(4)
Given that line (1) is passing through the point of intersection (2) and (3)
Solve (2) and (3) to get point of intersection
Subtract (3) from (2)
⇒ x + 3y - 1 - x + 2y - 4 = 0
⇒ 5y - 5 = 0
⇒ 5y = 5
⇒ y = 1
Substitute y = 1 in (3)
x + 3(1) - 1 = 0
x + 2 = 0
x = -2
The point of intersection of (2) and (3) is (-2, 1)
As we know line (4) will passing through (-2, 1)
Substitute (-2, 1) in (4)
⇒ 2(-2) − 3(1) = λ
⇒ − 4 − 3 = λ
⇒ λ = − 7
at λ = − 7 (4) ⇒ 2x −3y = −7
⇒ 2x −3y +7 = 0
Therefore,
The required equation is 2x − 3y +7 = 0
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