Question

Find the foot of the perpendicular drawn from (3,0) upon the straight line 5x + 12y - 41=O

Answer 1

Given:

5x+12y-41 = 0

To find:

The foot of the perpendicular.

Solution:

Let the coordinate of the foot of the perpendicular is (a,b)

we have the equation

5x+12y-41 = 0

the foot of the perpendicular also satisfies the given equation:

• 5x+12y-41 = 0

x = a, y = b

• 5a+12b-41 = 0------------(I)

As the lines are perpendicular so:

m1m2 = -1

m1 will be the slope of the line 5x+12y-41 = 0 which is given by:

• 5x+12y-41 = 0
• 5x+12y = 41
• 12y = -5x + 41
• y = -5/12x + 41/12

so from the equation y = mx+c  we have the slope as

• m1 = -5/12

and

• m2 = -1/m1
• m2 = -1/-5/12
• m2 = 12/5

Slope of the line with (a,b) and (3,0)

• m2 = y2-y1/x2-x1
• 12/5 = 0-b/3-a
• 12(3-a) = -5b
• 36 - 12a = -5b
• 12a - 5b = 36 -------------(II)

we have the equations as

• (5a+12b = 41 )5
• (12a - 5b = 36)12

25a+60b = 205

144a -60b = 432

169a          = 637

• a = 637/169
• a = 49/13

putting the value of a

• 12a - 5b = 36
• 12.49/13 - 5b = 36
• -5b = 36-588/13
• -5b = 468-588/13
• -5b = -120/13
• b = 24/13

Hence the foot of the perpendicular is (49/13, 24/13)

Answer 2

Here is your answer in the attachment

hope it helps you.....