Question

Find the equation of the perpendicular drawn from the point (1, −2) on the line 4x − 3у − 5 = 0. Also find the co-ordinates of the foot of the perpendicular.

Answer 1

The given equation of the line is:

4x − 3у − 5 = 0

To find its slope (m₁) we rewrite it in slope-intercept form (y = mx + c):

−3у = −4x + 5

`y = 4/3 x - 5/3`

Comparing this with y = mx + c, the slope `m_1 = 4/3`.

Let the slope of the perpendicular line m₂ be. Since the lines are perpendicular, the product of their slopes is −1:

m₁ × m₂ = −1

`4/3 xx m_2 = -1`

∴ `m_2 = -3/4`

The perpendicular line passes through the point (1, −2),

Using the point-slope form:

y − y₁​ = m(x − x₁​)

`y - (-2) = - 3/4 (x - 1)`

`y + 2 = - 3/4 (x - 1)`

4(y + 2) = −3(x − 1)      ....[Multiplied by 4 to clear the fraction]

4y + 8 = −3x + 3

∴ 3x + 4y + 5 = 0

Solving the equations:

(1) 4x − 3у = 5

(2) 3x + 4y = −5

Multiply equation (1) by 4 and equation (2) by 3:

(3) 16x − 12у = 20

(4) 9x + 12y = −15

Adding (3) and (4) equations:

25x = 5

`x = 5/25`

∴ `x = 1/5`

Substitute `x = 1/5` back into equation (1):

`4(1/5) - 3y = 5`

`4/5 - 5 = 3y`

`(4 - 25)/5 = 3y`

`- (21)/5 = 3y`

∴ `y = - 7/5`

Hence, the equation of the perpendicular line is 3x + 4y + 5 = 0, and the coordinates of the foot of the perpendicular are `1/5, - 7/5`.