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Find the equation of the straight line passing through the point of intersection of the lines 4x-5y-1 = 0 and 2x+y+5=0 and perpendicular to the line 4x-
6y+11=0.
A.21x - 14y - 58 = 0
B.21x + 14y + 57 = 0
C.21x - 14y - 57 = 0
D.21x + 14y + 58 = 0
Answers
Given : Lines 4x-5y-1 = 0 and 2x+y+5=0
To find : equation of the straight line passing through the point of intersection of the lines and perpendicular to the line 4x- 6y+11=0
Solution:
Lines 4x-5y-1 = 0 EqA
2x+y+5=0 EqB
EqA + 5EqB
=> 4x + 10x -1 + 25 = 0
=> 14x = -24
=>x = -12/7
2(-12/7) + y + 5 = 0
=> y = -11/7
intersection Point ( -12/7 , - 11/7)
perpendicular to the line 4x- 6y+11=0.
here Slope = 4/6 = 2/3
Slope of perpendicular line = -3/2
y = -3x/2 + c
-11/7 = -3(-12/7)/2 + c
=> c = (-22 - 36)/14
=> c = -29/7
y = -3x/2 - 29/7
=> 14y = -21x - 58
=> 21x + 14y + 58 = 0
21x + 14y + 58 = 0 is Equation of line
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The equation of the straight line is D. 21x + 14y + 58 = 0
Step-by-step explanation:
From question, the perpendicular line is:
4x - 6y + 11 = 0
The slope of the perpendicular line is:
m₁ = -4/-6 = 2/3
Now, the condition of the slope of perpendicular lines is:
m₁ × m₂ = -1
⇒ 2/3 × m₂ = -1
⇒ m₂ = -3/2 = Slope of required line
Let the point of intersection of lines be (x, y).
The two lines are:
4x - 5y - 1 = 0 and 2x + y + 5 = 0
On solving the two equations, we get,
x = 12/7 and y = -11/7
Now,
y = mx + c
-11/7 = (-3/2 × -12/7) + c
∴ c = -29/7
Now, the equation of the required line is given as:
y = -3x/2 - 29/7
14y = -21x - 58
∴ 21x + 14y + 58 = 0