Question

Find the perpendicular distance from the point

(-1, 2) from the line x + 3y - 4 = 0.

Answer 1

Answer:

it is easy for you

find the values of x and y by shifting values from equation

use distance formula

put the value

you will get answer

Answer 2

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

Let assume that the l represents the equation of line x + 3y - 4 = 0 and let the coordinates of the point (- 1, 2) be denoted by P.

Now, we have to find distance, d between line l and the point P.

We know,

The perpendicular distance (d) of the line ax + by + c = 0 from the point (p, q) is given by

$\boxed{\sf{  \:\rm \: d \: = \: \bigg | \frac{ap + bq + c}{ \sqrt{ {a}^{2} + {b}^{2} } } \bigg| \: \: }}$

So, here

$\rm \: a = 1$

$\rm \: b = 3$

$\rm \: c = - 4$

$\rm \: p = - 1$

$\rm \: q = 2$

So, on substituting the values, we get

$\rm \: d \: = \: \bigg | \dfrac{1( - 1) + 3(2) - 4}{ \sqrt{ {1}^{2} + {3}^{2} } } \bigg|$

$\rm \: d \: = \: \bigg | \dfrac{ - 1 + 6 - 4}{ \sqrt{1 + 9} } \bigg|$

$\rm \: d \: = \: \bigg | \dfrac{1}{ \sqrt{10} } \bigg|$

$\rm\implies \:d \: = \: \dfrac{1}{ \sqrt{10} } \: units$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

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

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

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.