Mathematics STRAIGHT LINES -General equation of a line , Distance of a Point From a Line and Distance between two parallel lines
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### Topics Covered

star  General Equation of a Line reduced into various forms of the equation of a line
star Slope-intercept form
star Intercept form
star Normal form
star Distance of a Point From a Line
star Distance between two parallel lines

### General Equation of a Line reduced into various forms of the equation of a line

In earlier classes, we have studied general equation of first degree in two variables, Ax + By + C = 0, where A, B and C are real constants such that A and B are not zero simultaneously.

Graph of the equation color(red)(Ax + By + C = 0" is always a straight line.")

Therefore, any equation of the form color(blue)(Ax + By + C = 0), where A  and B are not zero simultaneously is called color(blue)("general linear equation or general equation of a line.")

### Slope-intercept form

\color{green} ✍️ if color(red)(B ne 0) , them Ax +By+ C = 0 can be written as

color(blue)(y = - A/B x - C/B) or \ \ \ \ \ y = mx+c ......................(1)

where m = - A/B and c = - C/B

We know that Equation (1) is the slope-intercept form of the equation of a line

whose slope is -A/B , and y-intercept is -C/B

\color{green} ✍️ if color(red)(B = 0), then color(blue)(x = -C/A) which is a vertical line whose slope is undefined and x-intercept is - C/A

### Intercept form

\color{green} ✍️ If color(red)(C ≠ 0,) then Ax + By + C = 0 can be written as

color(blue)(x/(-C/A) +y/(-C/B) = 1) or x/a+y/b = 1

where a = -C/A and b = -C/B

We know that equation (1) is intercept form of the equation of a line whose

x-intercept is -C/A and y-intercept is - C/B

\color{green} ✍️ If color(red)(C = 0,) then Ax + By + C = 0 can be written as color(blue)(Ax + By = 0,) which is a line passing through the origin and, therefore, has zero intercepts on the axes.

### Normal form

\color{green} ✍️ Let color(blue)(x cos ω + y sin ω = p) be the normal form of the line represented by the equation color(blue)(Ax + By + C = 0) or Ax + By = – C.

Thus, both the equations are same and therefore A/(cos omega) = B/(sin omega) = - C/p

which gives cosomega = - (Ap)/C and sin omega = - (Bp)/C

Now sin^2 omega +cos^2 omega = (-(Ap)/C)^2+(-(Bp)/C)^2 = 1

or p^2 = C^2/(A^2+B^2)  or p = pm C/sqrt(A^2+B^2)

Therefore cosomega = pm A/sqrt(A^2+B^2)  and sinomega = pm B/sqrt(A^2+B^2)

\color{green} ✍️ Thus, the normal form of the equation Ax + By + C = 0 is x cos ω + y sin ω = p,

where color(red)(cosomega = pm A/sqrt(A^2+B^2) , \ \ sinomega = pm B/sqrt(A^2+B^2))  and color(red)(p = pm C/sqrt(A^2+B^2))

Proper choice of signs is made so that color(blue)(p" should be positive.")
Q 3176256176

Equation of a line is 3x – 4y + 10 = 0. Find its (i) slope, (ii) x - and
y-intercepts.

Solution:

(i) Given equation 3x – 4y + 10 = 0 can be written as

 y = 3/4 x + 5/2 .........(1)

Comparing (1) with y = mx + c, we have slope of the given line as m =3/4

3x-4y =-10 or  x/(-10/3) + y/(5/2) = 1 .......(2)

Comparing (2) with x/a + y/b =1, we have x-intercept as a =-10/3 and

y-intercept as b = 5/2
Q 3116356270

Reduce the equation  sqrt 3 x + y - 8 =0 into normal form. Find the values of p and ω.

Solution:

Given equation is

sqrt 3 x + y -8 =0 .......(1)

Dividing (1) by  sqrt ( (sqrt 3)^2 +(1)^2 ) = 2, we get

 (sqrt 3) /2 x + 1/2 y =4 or cos 30^ox + sin 30^o y = 4 ......(2)

Comparing (2) with x cos ω + y sin ω = p, we get p = 4 and ω = 30°.
Q 3176356276

Find the angle between the lines y − sqrt 3 x - 5 =0 and sqrt 3 y -x + 6 = 0 .

Solution:

Given lines are

 y - sqrt 3 x -5 =0 or  y = sqrt 3 x+ 5 ..... (1)

and sqrt 3 y -x+ 6 =0 or  y =1/(sqrt 3) x-2 sqrt 3 .....(2)

Slope of line (1) is m_1 = 3 and slope of line (2) is m_2 = 1/(sqrt 3)

The acute angle (say) θ between two lines is given by

tan theta = | (m_2 -m_1 )/(1+ m_1 m_2) | ..... (3)

Putting the values of m_1 and m_2 in (3), we get

tan theta = | (1/(sqrt 3) - sqrt 3)/(1+ sqrt 3 xx 1/(sqrt 3 )) | = | (1-3)/(2 sqrt 3) | = 1/(sqrt 3)

which gives θ = 30°. Hence, angle between two lines is either 30° or 180° – 30° = 150°.
Q 3156456374

Show that two lines a_1 x + b_1 y + c_1 = 0 and a_2 x + b_2 y + c_2 = 0 ,
where b_1, b_2 ≠ 0 are:

(i) Parallel if  a_1/b_1 = a_2/b_2, and

(ii) Perpendicular if a_1a_2 + b_1b_2 = 0 .

Solution:

Given lines can be written as

y = a_1/b_1 x - c_1/b_1 .........(1)

and  y = - a_2/b_2 x - c_2/b_2 .......(2)

Slopes of the lines (1) and (2) are m_1 = -a_1/b_1 and m_2 = - a_2/b_2 , respectively. Now

(i) Lines are parallel, if m_1 = m_2, which gives

-a_1/b_1 = -a_2/b_2 or a_1/b_1 = a_2/b_2

(ii) Lines are perpendicular, if m_1 * m-2 = – 1, which gives

a_1/b_1 * a_2/b_2 = -1 or a_1 a_2 + b_1 b_2 = 0
Q 3146656573

Find the equation of a line perpendicular to the line x − 2y + 3 = 0 and
passing through the point (1, – 2).

Solution:

Given line x − 2 y + 3 = 0 can be written as

 y =1/2 x + 3/2 ......(1)

Slope of the line (1) is m_1 = 1/2 . Therefore, slope of the line perpendicular to line (1) is

m_2 = - 1/(m_1) = -2

Equation of the line with slope – 2 and passing through the point (1, – 2) is

y − ( − 2) = −2(x −1) or y= − 2x ,

which is the required equation.

### Distance of a Point From a Line

\color{green} ✍️ The distance of a point from a line is the length of the perpendicular drawn from the point to the line.

Let color(blue)(L : Ax + By + C = 0 be a line, whose distance from the point P (x_1, y_1) is d. Draw a perpendicular PM from the point P to the line L (Fig10.19).

If the line meets the x-and y-axes at the points Q and R, respectively. Then, coordinates of the points are Q (-C/A , 0) and R (0 , - C/R).

Thus, the area of the triangle PQR is given by area color(blue)((Delta PQR) = 1/2 P M . QR)

which gives color(red)(P M = (2 text{area} (Delta PQR))/(QR)) .............(1)

Also, area (Delta PQR ) = 1/2 | x_1 (0+C/B) +(-C/A) (-C/B - y_1) +0 (y_1-0) |

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1/2 | x_1 C/B +y_1 C/A+C^2/(AB) |

or color(blue)(2 area \ \Delta PQR) = | C/(AB) | . | A_(x_1) +B_(y_1) +C |

color(blue)(QR) = sqrt((0+C/A)^2+(C/B -0)^2) = | C/(AB)| sqrt(A^2+B^2)

Substituting the values of area (ΔPQR) and QR in (1), we get

color(red)(P M = (| A_(x_1) +B_(y_1) + C|)/sqrt(A^2+B^2))

or color(blue)(d = (| A_(x_1) + B_(y_1) +C|)/sqrt(A^2+B^2))

\color{green} ✍️ Thus, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x_1, y_1) is given by

color(red)(d = (| A_(x_1) +B_(y_1) + C|)/sqrt(A^2+B^2))

### Distance between two parallel lines

\color{green} ✍️ We know that slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form

color(blue)(y = mx+c_1) ..............(1)

color(blue)(y = mx+c_2) ..............(2)

Line (1) will intersect x-axis at the point color(blue)(A (-c_1/m , 0)) as shown in Fig 10.20

Distance between two lines is equal to the length of the perpendicular from point A  to line (2).

Therefore, distance between the lines (1) and (2) is

(| (-m) (-c_1/m) +(-c_2)|)/sqrt(1+m^2)  or color(blue)(d = (| c_1-c_2|)/sqrt(1+m^2))

\color{green} ✍️ Thus, the distance d between two parallel lines color(red)(y = mx + c_1) and color(red)(y = mx + c_2) is given by

color(red)(d = ( | c_1-c_2|)/sqrt(1+m^2))

\color{green} ✍️ If lines are given in general form, i.e., color(blue)(Ax + By + C_1 = 0) and color(blue)(Ax + By + C_2 = 0,)

then above formula will take the form color(blue)(d = (| C_1-C_2|)/sqrt(A^2+B^2))

Q 3166756675

Find the distance of the point (3, – 5) from the line 3x – 4y –26 = 0.

Solution:

Given line is 3x – 4y –26 = 0 ... (1)
Comparing (1) with general equation of line Ax + By + C = 0, we get
A = 3, B = – 4 and C = – 26.
Given point is (x_1, y_1) = (3, –5) . The distance of the given point from given line is

d = ( | A x_1 +B y_1 +C | )/(sqrt (A^2+B^2) ) = ( | 3* 3 + )(-4) (-5) -26)/( sqrt (3^2 + (-4)^2 ) ) = 3/5