Question

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


ello koi jinda h yaha XD​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us assume that l be the required equation of line which passes through the point P(1, - 2) and perpendicular to the line 4x - 3y - 5 = 0.

Let assume that the equation of line 4x - 3y - 5 = 0 is represented as l'.

We know, slope of line ax + by + c = 0 is represented by m and given by

$\red{\rm :\longmapsto\:\boxed{\tt{ m \: = - \: \frac{coefficient \: of \: x}{coefficient \: of \: y}}}}$

So, using this,

$\rm :\longmapsto\:Slope \: of \: l' \: = \: - \dfrac{4}{( - 3)} = \dfrac{4}{3}$

We know,

Two lines having slope m and M are perpendicular iff Mm = - 1

Let assume that slope of line l be m

So,

$\rm :\longmapsto\:m \times \dfrac{4}{3} = - 1$

$\bf\implies \:m = - \: \dfrac{3}{4}$

We know,

Equation of line having slope m and passes through the point (a, b) is given by

$\red{\rm :\longmapsto\:\boxed{\tt{ \: y - b = m(x - a) \: }}}$

So, equation of line l which passes through the point P(1, - 2) and having slope - 3/4 is

$\rm :\longmapsto\:y + 2 = - \dfrac{3}{4} (x - 1)$

$\rm :\longmapsto\:4y + 8 = - 3x + 3$

$\rm :\longmapsto\:3x + 4y + 5 = 0$

Now, To find the coordinates of Foot of Perpendicular

We have to solve

$\rm :\longmapsto\:3x + 4y + 5 = 0 - - - (1)$

and

$\rm :\longmapsto\:4x - 3y - 5 = 0 - - - (2)$

On multiply equation (1) by 3 and (2) by 4, we get

$\rm :\longmapsto\:9x + 12y + 15 = 0 - - - (3)$

and

$\rm :\longmapsto\:16x - 12y - 20 = 0 - - - (4)$

On adding equation (3) and (4), we get

$\rm :\longmapsto\:25x - 5 = 0$

$\rm :\longmapsto\:25x = 5$

$\rm \implies\:x = \dfrac{1}{5}$

On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:3 \times \dfrac{1}{5} + 4y + 5 = 0$

$\rm :\longmapsto\:4y = - 5 - \dfrac{3}{5}$

$\rm :\longmapsto\:4y = \dfrac{ - 25 - 3}{5}$

$\rm :\longmapsto\:4y = \dfrac{ - 28}{5}$

$\bf\implies \:y = - \dfrac{7}{5}$

Hence,

$\boxed{\tt{ Coordinates \: of \: foot \: of \: perpendicular = \bigg(\dfrac{1}{5} , \: \dfrac{ - 7}{5} \bigg) \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Explore more

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m is y - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y - axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle β with the positive X-axis is x cosβ + y sinβ = p.

Answer 2

$\large\underline{\sf{Solution-}}$

Let us assume that l be the required equation of line which passes through the point P(1, - 2) and perpendicular to the line 4x - 3y - 5 = 0.

Let assume that the equation of line 4x - 3y - 5 = 0 is represented as l'.

We know, slope of line ax + by + c = 0 is represented by m and given by

$\red{\rm :\longmapsto\:\boxed{\tt{ m \: = - \: \frac{coefficient \: of \: x}{coefficient \: of \: y}}}}$

So, using this,

$\rm :\longmapsto\:Slope \: of \: l' \: = \: - \dfrac{4}{( - 3)} = \dfrac{4}{3}$

We know,

Two lines having slope m and M are perpendicular iff Mm = - 1

Let assume that slope of line l be m

So,

$\rm :\longmapsto\:m \times \dfrac{4}{3} = - 1$

$\bf\implies \:m = - \: \dfrac{3}{4}$

We know,

Equation of line having slope m and passes through the point (a, b) is given by

$\red{\rm :\longmapsto\:\boxed{\tt{ \: y - b = m(x - a) \: }}}$

So, equation of line l which passes through the point P(1, - 2) and having slope - 3/4 is

$\rm :\longmapsto\:y + 2 = - \dfrac{3}{4} (x - 1)$

$\rm :\longmapsto\:4y + 8 = - 3x + 3$

$\rm :\longmapsto\:3x + 4y + 5 = 0$

Now, To find the coordinates of Foot of Perpendicular

We have to solve

$\rm :\longmapsto\:3x + 4y + 5 = 0 - - - (1)$

and

$\rm :\longmapsto\:4x - 3y - 5 = 0 - - - (2)$

On multiply equation (1) by 3 and (2) by 4, we get

$\rm :\longmapsto\:9x + 12y + 15 = 0 - - - (3)$

and

$\rm :\longmapsto\:16x - 12y - 20 = 0 - - - (4)$

On adding equation (3) and (4), we get

$\rm :\longmapsto\:25x - 5 = 0$

$\rm :\longmapsto\:25x = 5$

$\rm \implies\:x = \dfrac{1}{5}$

On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:3 \times \dfrac{1}{5} + 4y + 5 = 0$

$\rm :\longmapsto\:4y = - 5 - \dfrac{3}{5}$

$\rm :\longmapsto\:4y = \dfrac{ - 25 - 3}{5}$

$\rm :\longmapsto\:4y = \dfrac{ - 28}{5}$

$\bf\implies \:y = - \dfrac{7}{5}$

Hence,

$\boxed{\tt{ Coordinates \: of \: foot \: of \: perpendicular = \bigg(\dfrac{1}{5} , \: \dfrac{ - 7}{5} \bigg) \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Explore more

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m is y - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y - axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle β with the positive X-axis is x cosβ + y sinβ = p.