Question

the equation of line passing through (3,-3) and perpendicular to the line 2x+5y-2=0 is

Answer 1

SOLUTION

TO DETERMINE

The equation of line passing through (3,-3) and perpendicular to the line 2x + 5y - 2 = 0

EVALUATION

Here the given equation of the line is

2x + 5y - 2 = 0

Since the required line is perpendicular to the line 2x + 5y - 2 = 0

Let the required equation of the line is

5x - 2y + k = 0 - - - - - - (1)

Where k is a constant to be determined

Now the line given by Equation 1 passes through the point (3, - 3)

So we have

15 + 6 + k = 0

⇒ k + 21 = 0

⇒ k = - 21

Hence the required equation of the line is

5x - 2y - 21 = 0

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Answer 2

Given : line 2x+5y-2=0 is

point ( 3 , - 3)

To Find : the equation of line passing through (3,-3) and perpendicular to the line 2x+5y-2=0

Solution:

2x+5y-2=0

=> 5y = -2x + 2

=> y = -2x/5  + 2/5

Comparing with y = mx + c

m = -2/5

Product of slope of perpendicular lines = - 1

Hence slope of perpendicular line  = -1/(-2/5)  = 5/2

line passing through ( 3 , - 3)

Hence Equation of the line

y - (-3) = (5/2)(x - 3)

=> 2y + 6  = 5x  - 15

=> 2y  = 5x - 21

5x -2y  - 21 = 0

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