find the equation of the straigth line passing througg the point (3, -4) and perpendicular to the line 5x+3y-1=0
Answers
Answer:
The given line
5 x - 3 y + 1 = 0
can be rewritten as
3 y = 5 x + 1
or
y = (5/3) x + ⅓ —————————————-(1)
has a slope = 5/3
The slope of the straight line perpendicular to the given line = - 3/5
The equation of straight line with slope m and passing through the point (x¹, y¹) is
y - y¹ = m (x - x¹).
In our case (x¹, y¹) = (4, - 3) and the equation of the required line becomes
y - (-3) = (- 3/5) (x -4)
=> y + 3 = -3 x/5 + 12/5
=> 5 y + 15 = - 3 x + 12
=> 3 x + 5 y + 3 = 0
The required equation is:
3 x + 5 y + 3 = 0.
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Kalash Mangal
, Bsc,PGCE Physics and Computer Science
Answered 8 months ago · Author has 383 answers and 55.3K answer views
What is the equation of the line perpendicular to 5x-3y+1=0 and passing through (4,-3)?
Rearrange the given equation: 3y = 5x +1
Divide through by 3: y = (5/3)x + 1/3 Slope of this line = 5/3
Line perpendicular to this will have Slope -3/5 (Negative reciprocal of 5/3)
Using the general form of a straight line; Y = MX + C where M = Slope
The perpendicular line will have equation; Y = (-3/5)X + C; C = Y intercept
To find the Y-Intercept we substitute the given point into the above partial equation i.e. Y = -3 AND X = 4
-3 = (-3/5) * 4 + C
C = -3 + 12/5
C = - 3/5 ( -15/5 + 12/5)
Substitute this value of C in Y = (-3/5)X + C
Y = (-3/5)X - 3/5; Multiply through by 5 and rearrange
5Y = -3X - 3
5Y + 3X +3 = 0 is the required equation of line perpendicular to 5x-3y+1=0
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