Please Wait... While Loading Full Video### STRAIGHT LINES -Various Forms of the Equation of a Line

`star` Horizontal and vertical lines

`star` Point-slope form

`star` Two-point form

`star` Slope-intercept form

`star` Intercept - form

`star` Normal form

`star` Point-slope form

`star` Two-point form

`star` Slope-intercept form

`star` Intercept - form

`star` Normal form

`\color{green} ✍️ ` If a horizontal line `L` is at a distance a from the `x`-axis then ordinate of every point lying on the line is either `a` or `– a` [Fig 10.11 (a)].

`\color{green} ✍️ ` Therefore, equation of the line `L` is either `y = a` or `y = – a.` Choice of sign will depend upon the position of the line according as the line is above or below the `y`-axis.

`\color{green} ✍️ ` Similarly, the equation of a vertical line at a distance `b` from the `x`-axis is either `x = b` or `x = – b` [Fig 10.11(b)].

`\color{green} ✍️ ` Therefore, equation of the line `L` is either `y = a` or `y = – a.` Choice of sign will depend upon the position of the line according as the line is above or below the `y`-axis.

`\color{green} ✍️ ` Similarly, the equation of a vertical line at a distance `b` from the `x`-axis is either `x = b` or `x = – b` [Fig 10.11(b)].

Q 3116445379

Find the equations of the lines parallel to axes and passing through

(– 2, 3).

(– 2, 3).

Position of the lines is shown in the Fig 10.12. The y-coordinate of every point on

the line parallel to x-axis is 3, therefore, equation of the line parallel to x-axis and passing through

(– 2, 3) is y = 3. Similarly, equation of the line parallel to y-axis and passing through (– 2, 3)

is x = – 2.

Suppose that `P_0 (x_0, y_0)` is a fixed point on a non-vertical line `L`, whose slope is `m.`

Let `P (x, y)` be an arbitrary point on `L` (Fig 10.13).

Then, by the definition, the slope of `L` is given by

`color(red)(m=( y- y_0)/(x-x_0))` i.e., `color(blue)(( y- y_0) =m(x-x_0))` ..........................(1)

Since the point `P_0 (x_0 , y_0)` along with all points `(x, y)` on `L` satisfies `(1)` and no other point in the plane satisfies` (1).`

Equation `(1)` is indeed the equation for the given line `L.`

Thus, the point `(x, y)` lies on the line with slope `m` through the fixed point `(x_0, y_0),` if and only if, its coordinates satisfy the equation

`color(red)(y – y_0 = m (x – x_0))`

Let `P (x, y)` be an arbitrary point on `L` (Fig 10.13).

Then, by the definition, the slope of `L` is given by

`color(red)(m=( y- y_0)/(x-x_0))` i.e., `color(blue)(( y- y_0) =m(x-x_0))` ..........................(1)

Since the point `P_0 (x_0 , y_0)` along with all points `(x, y)` on `L` satisfies `(1)` and no other point in the plane satisfies` (1).`

Equation `(1)` is indeed the equation for the given line `L.`

Thus, the point `(x, y)` lies on the line with slope `m` through the fixed point `(x_0, y_0),` if and only if, its coordinates satisfy the equation

`color(red)(y – y_0 = m (x – x_0))`

Q 3156545474

Find the equation of the line through (– 2, 3) with slope – 4.

Here m = – 4 and given point `(x_0 , y_0)` is (– 2, 3).

By slope-intercept form formula

(1) above, equation of the given

line is

y – 3 = – 4 (x + 2) or

4x + y + 5 = 0, which is the

required equation.

Let the line `L` passes through two given points `P_1 (x_1, y_1)` and `P_2 (x_2, y_2).`

Let `P (x, y)` be a general point on `L` (Fig 10.14).

The three points `color(blue)(P_1, P_2 \ \ "and" \ \P" are collinear")`, therefore, we have

`color(red)("slope of " \ \ P_1P =" slope of" \ \ P_1P_2)`

`(y- y_1)/(x-x_1)= (y_2- y_1)/(x_2-x_1)` or `(y- y_1)= (y_2- y_1)/(x_2-x_1)* (x-x_1)`

Thus, equation of the line passing through the points `(x_1, y_1)` and `(x_2, y_2)` is given by

`color(blue)((y- y_1)= (y_2- y_1)/(x_2-x_1)* (x-x_1))`

Let `P (x, y)` be a general point on `L` (Fig 10.14).

The three points `color(blue)(P_1, P_2 \ \ "and" \ \P" are collinear")`, therefore, we have

`color(red)("slope of " \ \ P_1P =" slope of" \ \ P_1P_2)`

`(y- y_1)/(x-x_1)= (y_2- y_1)/(x_2-x_1)` or `(y- y_1)= (y_2- y_1)/(x_2-x_1)* (x-x_1)`

Thus, equation of the line passing through the points `(x_1, y_1)` and `(x_2, y_2)` is given by

`color(blue)((y- y_1)= (y_2- y_1)/(x_2-x_1)* (x-x_1))`

Q 3116645570

Write the equation of the line through the points (1, –1) and (3, 5).

Here `x_1 = 1, y_1 = – 1, x_2 = 3` and `y_2 = 5`. Using two-point form (2) above

for the equation of the line, we have

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

or `− 3x + y + 4 = 0` , which is the required equation.

Sometimes a line is known to us with its slope and an intercept on one of the axes. We will now find equations of such lines.

`color(red)("Case I")` Suppose a line `L` with slope `m` cuts the `y`-axis at a distance `c` from the origin (Fig10.15).

The distance `c` is called the `y` intercept of the line `L.` Obviously, coordinates of the point where the line meet the `y`-axis are `(0, c).`

Thus, `L` has slope `m` and passes through a fixed point `(0, c).`

Therefore, `color(blue)("by point-slope form")`,

The equation of `L` is `color(blue)(y − c = m( x − 0 ))` or `color(red)(y = mx + c)`

Thus, the point `(x, y)` on the line with slope `m` and `y`-intercept c lies on the line if and only if

`color(red)(y = mx + c)`

`color(red)"Note : "` that the value of `c` will be positive or negative according as the intercept is made

on the positive or negative side of the `y`-axis, respectively.

`color("Case II")` Suppose line L with slope m makes x-intercept `d.`

Then equation of `L` is `color(red)(y = m(x − d ))`

`color(red)("Case I")` Suppose a line `L` with slope `m` cuts the `y`-axis at a distance `c` from the origin (Fig10.15).

The distance `c` is called the `y` intercept of the line `L.` Obviously, coordinates of the point where the line meet the `y`-axis are `(0, c).`

Thus, `L` has slope `m` and passes through a fixed point `(0, c).`

Therefore, `color(blue)("by point-slope form")`,

The equation of `L` is `color(blue)(y − c = m( x − 0 ))` or `color(red)(y = mx + c)`

Thus, the point `(x, y)` on the line with slope `m` and `y`-intercept c lies on the line if and only if

`color(red)(y = mx + c)`

`color(red)"Note : "` that the value of `c` will be positive or negative according as the intercept is made

on the positive or negative side of the `y`-axis, respectively.

`color("Case II")` Suppose line L with slope m makes x-intercept `d.`

Then equation of `L` is `color(red)(y = m(x − d ))`

Q 3166645575

Write the equation of the lines for which` tan θ = 1/2` , where ` θ` is the

inclination of the line and (i) y-intercept is ` -3/2` (ii) x-intercept is 4.

inclination of the line and (i) y-intercept is ` -3/2` (ii) x-intercept is 4.

(i) Here, slope of the line is `m = tan θ =1/2` and `y ` intercept c ` = -3/2`

Therefore, by slope-intercept form (3) above, the equation of the line is

` y =1/2 x - 3/2` or `2y -x +3 = 0`, which is the required equation.

(ii) Here, we have `m = tanθ =1/2` and `d = 4`

Therefore, by slope-intercept form (4) above, the equation of the line is

`y= 1/2 (x-4)` or `2y -x +4 = 0`,

which is the required equation.

Suppose a line `L` makes `x`-intercept `a` and `y`-intercept `b` on the axes.

Obviously `L` meets `x`-axis at the point `(a, 0)` and `y`-axis at the point `(0, b)` (Fig .10.16).

By two-point form of the equation of the line, we have

`y-0=(b-0)/(0-a) *(x-a)` or `ay=-bx+ab`

`color(blue)(x/a+y/b=1)`

Thus, equation of the line making intercepts `a` and `b` on `x`-and `y`-axis, respectively, is `color(red)(x/a+y/b=1)`

Obviously `L` meets `x`-axis at the point `(a, 0)` and `y`-axis at the point `(0, b)` (Fig .10.16).

By two-point form of the equation of the line, we have

`y-0=(b-0)/(0-a) *(x-a)` or `ay=-bx+ab`

`color(blue)(x/a+y/b=1)`

Thus, equation of the line making intercepts `a` and `b` on `x`-and `y`-axis, respectively, is `color(red)(x/a+y/b=1)`

Q 3136145972

Find the equation of the line, which makes intercepts –3 and 2 on the

x- and y-axes respectively.

x- and y-axes respectively.

Here a = –3 and b = 2. By intercept form (5) above, equation of the line is

`x/(-3) + y/2 =1` or `2x - 3y + 6 =0`

Suppose a non-vertical line is known to us with following data:

`(i)` Length of the perpendicular (normal) from origin to the line.

`(ii)` Angle which normal makes with the positive direction of `x`-axis.

Let `L` be the line, whose perpendicular distance from origin `O` be `OA = p` and the angle between the positive `x`-axis and `OA` be `∠XOA = ω.`

The possible positions of line `L` in the Cartesian plane are shown in the Fig 10.17.

Now, our purpose is to find slope of `L` and a point on it. Draw perpendicular `AM` on the `x`-axis in each case.

In each case, we have `OM = p cos ω` and `MA = p sin ω,` so that the coordinates of the point `A` are `(p cos ω, p sin ω).`

Further, line `L` is perpendicular to `OA.` Therefore

The slope of the line `color(blue)(L =- 1/(Slope of OA)=-1/(tan ω)= -(cos ω)/(sin ω)).`

Thus, the line `L` has slope `(− cosω )/(sinω)` and point `A(pcosω, p sinω)` on it.

Therefore, `color(blue)("by point-slope form")`, the equation of the line `L` is

`color(blue)(x cos ω + y sin ω = p.)`

Hence, the equation of the line having normal distance `p` from the origin and angle `ω` which the normal makes with the positive direction of `x`-axis is given by `color(red)(x cos ω + y sin ω = p)`

`(i)` Length of the perpendicular (normal) from origin to the line.

`(ii)` Angle which normal makes with the positive direction of `x`-axis.

Let `L` be the line, whose perpendicular distance from origin `O` be `OA = p` and the angle between the positive `x`-axis and `OA` be `∠XOA = ω.`

The possible positions of line `L` in the Cartesian plane are shown in the Fig 10.17.

Now, our purpose is to find slope of `L` and a point on it. Draw perpendicular `AM` on the `x`-axis in each case.

In each case, we have `OM = p cos ω` and `MA = p sin ω,` so that the coordinates of the point `A` are `(p cos ω, p sin ω).`

Further, line `L` is perpendicular to `OA.` Therefore

The slope of the line `color(blue)(L =- 1/(Slope of OA)=-1/(tan ω)= -(cos ω)/(sin ω)).`

Thus, the line `L` has slope `(− cosω )/(sinω)` and point `A(pcosω, p sinω)` on it.

Therefore, `color(blue)("by point-slope form")`, the equation of the line `L` is

`color(blue)(x cos ω + y sin ω = p.)`

Hence, the equation of the line having normal distance `p` from the origin and angle `ω` which the normal makes with the positive direction of `x`-axis is given by `color(red)(x cos ω + y sin ω = p)`

Q 3116156070

Find the equation of the line whose perpendicular distance from the

origin is 4 units and the angle which the normal makes with positive direction of x-axis

is 15°.

origin is 4 units and the angle which the normal makes with positive direction of x-axis

is 15°.

Here, we are given p = 4 and

ω = 150 (Fig10.18).

Now `cos 15^o = (sqrt 3 +1 )/(2 sqrt 2)`

and `sin 15^o = (sqrt 3-1)/(2 sqrt 2)` (why ?)

By the normal form (6) above, the equation of the

line is

`x cos 15^o + y sin 15^o = 4` or ` (sqrt 3 +1)/(2 sqrt 2 ) x + (sqrt 3 -1)/(2 sqrt 2) y = 4` or ` (sqrt 3 +1) x + (sqrt 3 -1) y = 8 sqrt 2`

This is the required equation.

Q 3126256171

The Fahrenheit temperature F and absolute temperature K satisfy a

linear equation. Given that K = 273 when F = 32 and that K = 373 when F = 212.

Express K in terms of F and find the value of F, when K = 0.

linear equation. Given that K = 273 when F = 32 and that K = 373 when F = 212.

Express K in terms of F and find the value of F, when K = 0.

Assuming F along x-axis and K along y-axis, we have two points (32, 273)

and (212, 373) in XY-plane. By two-point form, the point (F, K) satisfies the equation

`K - 273 = (373 -273)/( 212-32) (F -32)` or `K- 273 = 100/180 (F- 32)`

or `K= 5/9 (F-32 ) +273` ..........(1)

which is the required relation.

When K = 0, Equation (1) gives

`0 = 5/9 (F-32 ) + 273 ` or `F-32 = - (273 xx 9)/5 = - 491.4` or `F= -459 .4`

`"Alternate"` method We know that simplest form of the equation of a line is y = mx + c.

Again assuming F along x-axis and K along y-axis, we can take equation in the form

K = mF + c ... (1)

Equation (1) is satisfied by (32, 273) and (212, 373). Therefore

273 = 32m + c ... (2)

and 373 = 212m + c ... (3)

Solving (2) and (3), we get

`m =5/9` and ` c = 2297/9`

Putting the values of m and c in (1), we get

`k = 5/9 F + 2297/9 .................................(4)

which is the required relation. When K = 0, (4) gives F = – 459.4.

`color{blue} "Note"` We know, that the equation y = mx + c, contains two constants, namely, m and c. For finding these two constants, we need two conditions satisfied by the equation of line. In all the examples above, we are given two conditions to determine the equation of the line.