Math, asked by sametithirumaniramu4, 1 month ago

find the equation of the straigth line passing througg the point (3, -4) and perpendicular to the line 5x+3y-1=0​

Answers

Answered by pranav17087
0

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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