Question

If a straight line perpendicular to 2x-3y+7=0forms a triangle with coordinate axes whose area is 3 sq units then equation of straight line is

Answer 1

Step-by-step explanation:

Given If a straight line perpendicular to 2x-3y+7=0forms a triangle with coordinate axes whose area is 3 sq units then equation of straight line is    

• Given 2x – 3y + 7 = 0
• We can write this as 3y = 2x + 7
•                      Or y = 2/3 x + 7/3
• So slope of the line will be m 1 = 2/3 (since y = mx + c)
• Now slope of line perpendicular to given line will be m2
• If two lines are perpendicular then the product of their slope will be = to – 1
• Therefore m1.m2 = - 1
•                 2/3 . m2 = - 1
•                     Or m2 = - 3/2
• So equation of the perpendicular line will be y = - 3/2 x + c
•                                                   Now 2y = - 3x + 2c
•                                                   Or 2c = 3x + 2y ---------1
•                                                   Or 1 = 3x / 2c + 2y / 2c
•            Now we can write this as x / 2c / 3 + y/c = 1
• We will now have a triangle formed with the coordinate axis.
•                   Area of triangle = ½ x base x height
•                                              = ½ x 2/3 c x c
•        Now area of triangle = 3 sq units
•                       So 3 = c^2 / 3
•                   Or c^2 = 9
•                   C = + - 3
• Substituting the value in equation 1 we get
•             3x + 2y + 6 = 0 and 3x + 2y – 6 = 0

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https://brainly.in/question/14790298