Question

Find the equation of the straight line perpendicular to the line 5x-3y+1=0 and passing through the point (4, -3).

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Answer 1

Answer:

Step-by-step explanation:

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.