`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

`star` Slope-intercept form

`star` Intercept form

`star` Normal form

`star` Distance of a Point From a Line

`star` Distance between two parallel lines

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.")`

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.")`

`\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`

`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`

`\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.

`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.

`\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.")`

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.

y-intercepts.

(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.

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` .

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` .

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` .

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).

passing through the point (1, – 2).

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.

`\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))`

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))`

`\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))`

`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.

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`