Question

Find the foot of the perpendicular drawn from (4, 1) upon the straight line 3x - 4y + 12 = 0.

Answer 1

$\bold{Answer :}$

The line is

3x - 4y + 12 = 0 ...(i)

Any line perpendicular to (i) can be written as

4x + 3y + c = 0 ...(ii)

where c is an arbitrary constant.

Given that (ii) passes through the point (4, 1), then

4 (4) + 3 (1) + c = 0

or, 16 + 3 + c = 0

or, 19 + c = 0

or, c = - 19

Thus, the required line be

4x + 3y - 19 = 0 ...(ii)

Now, we solve (i) and (iii) to get the required foot of the perpendicular drawn from (4, 1)

3x - 4y + 12 = 0 ...(i) × 4

4x + 3y - 19 = 0 ...(iii) × 3

==>

12x - 16y + 48 = 0

12x + 9y - 57 = 0

On subtraction, we get

- 25y + 105 = 0

or, y = 21/5

Putting y = 21/5 in (ii), we get

4x + 3 (21/5) - 19 = 0

or, 4x = 19 - 63/5

or, 4x = 32/5

or, x = 8/5

So, the required foot of the perpendicular be

(8/5, 21/5)

$\textbf{Have a good day :)}$

Answer 2

Answer:

refer to the image please