Question

what is the perpendicular from any point on line 2x + 11y =1 upon the line 24x + 7y =20 ?????

Answer 1

Given:

• The equation of line from whose points the perpendicular is to be drawn (L1) is 2x+11y=1.
• The equation of the line upon which the perpendicular is to be drawn (L2) is 24x+7y=20

To find:

The equation of perpendicular from any point on the line L1 upon the line L2.

Solution:

• The slope op the line L2 (m1) is equal to (-x_coefficient/y_coefficient) = -24/7.
• The slope of any line perpendicular to L2 (m) will be equal to -1/m1 = 7/24
• Any point on the line L1 can be parametrically expressed as (t,(1-2t)/11), where t is any real number.

• So, the equation of the perpendicular can be written as (y-(1-2t)/11) = m (x-t)
• Simplifying the equation we get, 77x-264y=125t-24, where t is a real number.

Answer:

The equation of the perpendicular is 77x-264y=125t-24, where t is a real number.