`star` Introduction

`star` Slope or gradient of a Line

`star` Slope of a line when coordinates of any two points on the line are given

`star` Conditions for parallelism and perpendicularity of lines in terms of their slopes

`star` Angle between two lines

`star` Collinearity of three points

`star` Slope or gradient of a Line

`star` Slope of a line when coordinates of any two points on the line are given

`star` Conditions for parallelism and perpendicularity of lines in terms of their slopes

`star` Angle between two lines

`star` Collinearity of three points

`\color{green} ✍️` Let us have a brief recall of coordinate geometry done in earlier classes.

`\color{green} ✍️` To recapitulate, the location of the points `(6, – 4)` and `(3, 0)` in the `XY-`plane is shown in Fig 1.

We may note that the point `(6, – 4)` is at `6` units distance from the `y-`axis measured along the positive `x-`axis and at `4` units distance from the `x-`axis measured along the negative `y-`axis.

`\color{green} ✍️` Similarly, the point `(3, 0)` is at `3` units distance from the `y`-axis measured along the positive `x`-axis and has zero distance from the `x-` axis.

We also studied there following important formulae:

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \ \ \ 1.` `color{green}("Distance between the points")` `P \ (x_1, y_1)` and `Q \ (x_2, y_2)` is

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{blue}(PQ=sqrt((x_2-x_1)^2+(y_2-y_1)))`

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \ \ \ \ 2.` The coordinates of a point dividing the line segment joining the points `(x_1, y_1)` and `(x_2, y_2)` internally, in the ratio `m: n` are `color{blue} ( (((mx_2+nx_1)/(m+n) , (my_2+ny_1)/(m+n)) )`

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \ \ \ \ \ 3. ` In particular, if `m = n,` the coordinates of the mid-point of the line segment joining the points `(x_1, y_1)` and `(x_2, y_2)` are `color{blue} (((x_2+x_1)/2 , (y_2+y_1)/2))`

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` `\ \ \ \ \ \ \ 4. ` `color{green}("Area of the triangle")` whose vertices are `(x_1, y_1), (x_2, y_2)` and `(x_3, y_3)` is

`color{blue} (=1/2| x_1( y_2 − y_3) + x_2( y_3 − y_1) + x_3( y_1 − y_2)|) `

`"Remark "` If the area of the triangle `ABC` is zero, then three points `A, B` and `C` lie on a line, i.e., they are collinear.

`\color{green} ✍️` To recapitulate, the location of the points `(6, – 4)` and `(3, 0)` in the `XY-`plane is shown in Fig 1.

We may note that the point `(6, – 4)` is at `6` units distance from the `y-`axis measured along the positive `x-`axis and at `4` units distance from the `x-`axis measured along the negative `y-`axis.

`\color{green} ✍️` Similarly, the point `(3, 0)` is at `3` units distance from the `y`-axis measured along the positive `x`-axis and has zero distance from the `x-` axis.

We also studied there following important formulae:

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \ \ \ 1.` `color{green}("Distance between the points")` `P \ (x_1, y_1)` and `Q \ (x_2, y_2)` is

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{blue}(PQ=sqrt((x_2-x_1)^2+(y_2-y_1)))`

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \ \ \ \ 2.` The coordinates of a point dividing the line segment joining the points `(x_1, y_1)` and `(x_2, y_2)` internally, in the ratio `m: n` are `color{blue} ( (((mx_2+nx_1)/(m+n) , (my_2+ny_1)/(m+n)) )`

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \ \ \ \ \ 3. ` In particular, if `m = n,` the coordinates of the mid-point of the line segment joining the points `(x_1, y_1)` and `(x_2, y_2)` are `color{blue} (((x_2+x_1)/2 , (y_2+y_1)/2))`

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` `\ \ \ \ \ \ \ 4. ` `color{green}("Area of the triangle")` whose vertices are `(x_1, y_1), (x_2, y_2)` and `(x_3, y_3)` is

`color{blue} (=1/2| x_1( y_2 − y_3) + x_2( y_3 − y_1) + x_3( y_1 − y_2)|) `

`"Remark "` If the area of the triangle `ABC` is zero, then three points `A, B` and `C` lie on a line, i.e., they are collinear.

`\color{green} ✍️` A line in a coordinate plane forms two angles with the x-axis, which are supplementary.

`\color{green} ✍️` The angle (say) `θ` made by the line `l` with positive direction of `x-`axis and measured `color(blue)("anti clockwise is called the inclination of the line.")` Obviously `color(red)(0° ≤ θ ≤ 180°)` (Fig 2).

`\color{green} ✍️` We observe that lines parallel to `x-`axis, or coinciding with `x`-axis, have inclination of `0°.` The inclination of a vertical line (parallel to or coinciding with `y-`axis) is `90°.`

`\color{purple}ul(✓✓) \color{purple} (" DEFINITION ALERT")`

`\color{green} ✍️` If `θ` is the inclination of a line `l`, then `color{green}(tan θ)` is called the `color(blue)"slope or gradient"` of the line `l`.

`\color{green} ✍️` The slope of a line is denoted by `color(blue)(m.)`

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` Thus, `color(red)(m = tan θ, θ ≠ 90°)`

` \color{red} \ox \color{red} \mathbf(COMMON \ CONFUSION) ` The slope of a line whose inclination is `90°` is not defined.

`\color{green} ✍️` It may be observed that the slope of `x-`axis is zero and slope of `y-`axis is not defined.

`\color{green} ✍️` The angle (say) `θ` made by the line `l` with positive direction of `x-`axis and measured `color(blue)("anti clockwise is called the inclination of the line.")` Obviously `color(red)(0° ≤ θ ≤ 180°)` (Fig 2).

`\color{green} ✍️` We observe that lines parallel to `x-`axis, or coinciding with `x`-axis, have inclination of `0°.` The inclination of a vertical line (parallel to or coinciding with `y-`axis) is `90°.`

`\color{purple}ul(✓✓) \color{purple} (" DEFINITION ALERT")`

`\color{green} ✍️` If `θ` is the inclination of a line `l`, then `color{green}(tan θ)` is called the `color(blue)"slope or gradient"` of the line `l`.

`\color{green} ✍️` The slope of a line is denoted by `color(blue)(m.)`

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` Thus, `color(red)(m = tan θ, θ ≠ 90°)`

` \color{red} \ox \color{red} \mathbf(COMMON \ CONFUSION) ` The slope of a line whose inclination is `90°` is not defined.

`\color{green} ✍️` It may be observed that the slope of `x-`axis is zero and slope of `y-`axis is not defined.

`\color{green} ✍️` Let `P(x_1, y_1)` and `Q(x_2, y_2)` be two points on non-vertical line `l` whose inclination is `color(blue)(θ.)`

Obviously, `x_1 ≠ x_2,` otherwise the line will become perpendicular to x-axis and its slope will not be defined. The inclination of the line `l` may be acute or obtuse.

Let us take these two cases. Draw perpendicular `QR` to x-axis and `PM` perpendicular to `RQ` as shown .

`color{blue}("Case 1 :") color{green}("When angle θ is acute")`

In Fig 3 (i), `∠MPQ = θ.` ....... (1)

Therefore, slope of line `color(purple)(l = m = tan θ.)`..........(2)

But in `ΔMPQ,` we have `tanθ =(MQ)/(MP)=(y_2-y_1)/(x_2-x_1)`

From equations (1) and (2), we have

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \color(red)(m=(y_2-y_1)/(x_2-x_1))`

`color{blue}("Case 2 :")` `color{green}("When angle θ is obtuse:")`

In Fig 3 (ii), we have `∠MPQ = 180° – θ.`

Therefore, `θ = 180° – ∠MPQ.`

Now, slope of the line `color(purple)(l =m = tan θ = tan ( 180° – ∠MPQ)) `

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = – tan ∠MPQ=-(MQ)/(MP)=- (y_2-y_1)/(x_1-x_2)=(y_2-y_1)/(x_2-x_1)`

Consequently, we see that in both the cases the slope m of the line through the points `(x_1, y_1)` and `(x_2, y_2)` is given by

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \ color(red)(m=(y_2-y_1)/(x_2-x_1))`

Obviously, `x_1 ≠ x_2,` otherwise the line will become perpendicular to x-axis and its slope will not be defined. The inclination of the line `l` may be acute or obtuse.

Let us take these two cases. Draw perpendicular `QR` to x-axis and `PM` perpendicular to `RQ` as shown .

`color{blue}("Case 1 :") color{green}("When angle θ is acute")`

In Fig 3 (i), `∠MPQ = θ.` ....... (1)

Therefore, slope of line `color(purple)(l = m = tan θ.)`..........(2)

But in `ΔMPQ,` we have `tanθ =(MQ)/(MP)=(y_2-y_1)/(x_2-x_1)`

From equations (1) and (2), we have

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \color(red)(m=(y_2-y_1)/(x_2-x_1))`

`color{blue}("Case 2 :")` `color{green}("When angle θ is obtuse:")`

In Fig 3 (ii), we have `∠MPQ = 180° – θ.`

Therefore, `θ = 180° – ∠MPQ.`

Now, slope of the line `color(purple)(l =m = tan θ = tan ( 180° – ∠MPQ)) `

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = – tan ∠MPQ=-(MQ)/(MP)=- (y_2-y_1)/(x_1-x_2)=(y_2-y_1)/(x_2-x_1)`

Consequently, we see that in both the cases the slope m of the line through the points `(x_1, y_1)` and `(x_2, y_2)` is given by

`\color { maroon} ® \color{maroon} ul (" REMEMBER")` ` \ \ color(red)(m=(y_2-y_1)/(x_2-x_1))`

`\color{green} ✍️` In a coordinate plane, suppose that non-vertical lines `l_1` and `l_2` have slopes `m_1` and `m_2,` respectively. Let their inclinations be `α `and `β,` respectively.

`\color{green} ✍️` `color{blue}("If the line")` `color{blue}(l_1)` `color{blue}("is parallel to")` `color{blue}(l_2)` (Fig 10 .4), then their inclinations are equal, i.e., `color(red)(α = β),` and hence, `color(red)(tan α = tan β)`

Therefore `m_1 = m_2,` i.e., their slopes are equal.

Conversely, if the slope of two lines `l_1` and `l_2` is same, i.e.,

`m_1 = m_2.`

Then `tan α = tan β.`

By the property of tangent function `(`between `0°` and `180°), α = β.` Therefore, the lines are parallel.

Hence, two non vertical lines `l_1` and `l_2 ` are parallel if and only if their slopes are equal.

`\color{green} ✍️` `color{blue}("If the lines" \ \ l_1" and" \ \ l_2" are perpendicular")` (Fig10.5), then `color(red)(β = α + 90°.)`

Therefore, `tan β = tan (α + 90°)`

`= – cot α = - 1/ tanα`

`m_2 =-1/m_1` or `color(red)(m_1*m_2=-1)`

Conversely, if `m_1 m_2 = – 1,` i.e., `tan α tan β = – 1.`

Then `tan α = – cot β = tan (β + 90°)` or `tan (β – 90°)`

Therefore, `α` and `β` differ by `90°.`

Thus, lines `l_1` and `l_2` are perpendicular to each other.

Hence, two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other,

`m_2 =-1/m_1` or `color(red)(m_1*m_2=-1)`

`\color{green} ✍️` `color{blue}("If the line")` `color{blue}(l_1)` `color{blue}("is parallel to")` `color{blue}(l_2)` (Fig 10 .4), then their inclinations are equal, i.e., `color(red)(α = β),` and hence, `color(red)(tan α = tan β)`

Therefore `m_1 = m_2,` i.e., their slopes are equal.

Conversely, if the slope of two lines `l_1` and `l_2` is same, i.e.,

`m_1 = m_2.`

Then `tan α = tan β.`

By the property of tangent function `(`between `0°` and `180°), α = β.` Therefore, the lines are parallel.

Hence, two non vertical lines `l_1` and `l_2 ` are parallel if and only if their slopes are equal.

`\color{green} ✍️` `color{blue}("If the lines" \ \ l_1" and" \ \ l_2" are perpendicular")` (Fig10.5), then `color(red)(β = α + 90°.)`

Therefore, `tan β = tan (α + 90°)`

`= – cot α = - 1/ tanα`

`m_2 =-1/m_1` or `color(red)(m_1*m_2=-1)`

Conversely, if `m_1 m_2 = – 1,` i.e., `tan α tan β = – 1.`

Then `tan α = – cot β = tan (β + 90°)` or `tan (β – 90°)`

Therefore, `α` and `β` differ by `90°.`

Thus, lines `l_1` and `l_2` are perpendicular to each other.

Hence, two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other,

`m_2 =-1/m_1` or `color(red)(m_1*m_2=-1)`

Q 3146145073

Find the slope of the lines:

(a) Passing through the points (3, – 2) and (–1, 4),

(b) Passing through the points (3, – 2) and (7, – 2),

(c) Passing through the points (3, – 2) and (3, 4),

(d) Making inclination of 60° with the positive direction of x-axis.

(a) Passing through the points (3, – 2) and (–1, 4),

(b) Passing through the points (3, – 2) and (7, – 2),

(c) Passing through the points (3, – 2) and (3, 4),

(d) Making inclination of 60° with the positive direction of x-axis.

(a) The slope of the line through (3, – 2) and (– 1, 4) is

` m = (4- (-2) )/(-1-3) = 6/(-4) = -3/2`

(b) The slope of the line through the points (3, – 2) and (7, – 2) is

`m = (-2 - (-2) )/(7-3) = 0/4= 0`

(c) The slope of the line through the points (3, – 2) and (3, 4) is

`m = (4- (-2) )/( 3-3) = 6/0` , which is not defined.

(d) Here inclination of the line α = 60°. Therefore, slope of the line is

`m = tan 60^o = sqrt 3`

`\color{green} ✍️` Here we will discuss the angle between two lines in terms of their slopes.

Let `L_1` and `L_2` be two non-vertical lines with slopes `m_1` and `m_2`, respectively. If `α_1` and `α_2` are the inclinations of lines `L_1` and `L_2`, respectively. Then

`color(blue)(m_1 = tan α_1)` and `color(blue)(m_2 = tan α_2)` .

We know that when two lines intersect each other, they make two pairs of vertically opposite angles such that sum of any two adjacent angles is `180°`.

Let `θ` and `φ` be the adjacent angles between the lines `L_1` and `L_2` (Fig10.6). Then

` color(blue)(θ = α_2 – α_1)` and `α_1, α_2 ≠ 90°`

Therefore `color(red)(tan θ = tan (α_2 – α_1) = (tan α_2 - tan α_1 )/(1+ tan α_1 tan α_2) = (m_2 - m_1)/(1+ m_1 m_2))` (as `1 + m_1 m_2 ≠ 0`)

and `color(blue)(φ = 180° – θ)` so that

`color(red)(tan φ = tan (180^o - θ ) = - tan θ = -(m_2 - m_1 )/( 1+ m_1 m_2)) ` , as `1 + m_1 m_2 ≠ 0`

Now, there arise two cases:

`color(red)(text(Case I)) :` If ` color(green)((m_2 - m_1 )/( 1+ m_1 m_2))` is `color(blue)("positive"),` then `tan θ` will be positive and `tan φ ` will be negative, which means `θ` will be acute and `φ` will be obtuse.

`color(red)(text(Case II)) :` If ` color(green)((m_2 - m_1 )/(1+ m_1 m_2))` is `color(blue)("negative")`, then `tan θ` will be negative and `tan φ` will be positive, which means tha t` θ` will be obtuse and `φ` will be acute.

Thus, the acute angle (say `θ`) between lines `L_1` and `L_2 `with slopes `m_1` and `m_2`, respectively, is given by

`color(red)(tan theta = | (m_2- m_1 )/(1+ m_1 m_2) |)` , as `1 + m_1 m_2 ≠ 0` .... (1)

The obtuse angle (say `φ`) can be found by using `φ =180^o – θ`.

Let `L_1` and `L_2` be two non-vertical lines with slopes `m_1` and `m_2`, respectively. If `α_1` and `α_2` are the inclinations of lines `L_1` and `L_2`, respectively. Then

`color(blue)(m_1 = tan α_1)` and `color(blue)(m_2 = tan α_2)` .

We know that when two lines intersect each other, they make two pairs of vertically opposite angles such that sum of any two adjacent angles is `180°`.

Let `θ` and `φ` be the adjacent angles between the lines `L_1` and `L_2` (Fig10.6). Then

` color(blue)(θ = α_2 – α_1)` and `α_1, α_2 ≠ 90°`

Therefore `color(red)(tan θ = tan (α_2 – α_1) = (tan α_2 - tan α_1 )/(1+ tan α_1 tan α_2) = (m_2 - m_1)/(1+ m_1 m_2))` (as `1 + m_1 m_2 ≠ 0`)

and `color(blue)(φ = 180° – θ)` so that

`color(red)(tan φ = tan (180^o - θ ) = - tan θ = -(m_2 - m_1 )/( 1+ m_1 m_2)) ` , as `1 + m_1 m_2 ≠ 0`

Now, there arise two cases:

`color(red)(text(Case I)) :` If ` color(green)((m_2 - m_1 )/( 1+ m_1 m_2))` is `color(blue)("positive"),` then `tan θ` will be positive and `tan φ ` will be negative, which means `θ` will be acute and `φ` will be obtuse.

`color(red)(text(Case II)) :` If ` color(green)((m_2 - m_1 )/(1+ m_1 m_2))` is `color(blue)("negative")`, then `tan θ` will be negative and `tan φ` will be positive, which means tha t` θ` will be obtuse and `φ` will be acute.

Thus, the acute angle (say `θ`) between lines `L_1` and `L_2 `with slopes `m_1` and `m_2`, respectively, is given by

`color(red)(tan theta = | (m_2- m_1 )/(1+ m_1 m_2) |)` , as `1 + m_1 m_2 ≠ 0` .... (1)

The obtuse angle (say `φ`) can be found by using `φ =180^o – θ`.

Q 3116245170

If the angle between two lines is ` pi/4` and slope of one of the lines is `1/2` , find the slope of the other line.

We know that the acute angle θ between two lines with slopes `m_1` and `m_2`

is given by ` tan theta = | (m_2-m_1)/( 1+ m_1 m_2) |` .............(1)

Let `m_1 = 1/2 , m_2 = m ` and `theta = pi/4`

Now, putting these values in (1), we get

`tan pi/4 = | ( m-1/2 )/(1+1/2 m) | ` or ` 1 = | (m-1/2)/( 1+1/2 m ) |` ,

which gives ` (m-1/2)/(1+1/2 m ) =1` or ` - ( m -1/2)/(1+1/2 m) = -1` .

Therefore ` m =3` or ` m = -1/3`

Hence, slope of the other line is

`3 ` or `-1/3` Fig 10.7 explains the

reason of two answers.

Q 3166245175

Line through the points (–2, 6) and (4, 8) is perpendicular to the line

through the points (8, 12) and (x, 24). Find the value of x.

through the points (8, 12) and (x, 24). Find the value of x.

Slope of the line through the points (– 2, 6) and (4, 8) is

` m_1 = (8-6)/(4- (-2) ) = 2/6 = 1/3`

Slope of the line through the points (8, 12) and (x, 24) is

`m_2= (24-12)/(x-8) = 12/(x-8)`

Since two lines are perpendicular,

`m_1 m_2 = –1`, which gives

`1/3 xx 12/(x-8) = -1` or `x =4`

`\color{green} ✍️` We know that slopes of two parallel lines are equal. If two lines having the same slope pass through a common point, then two lines will coincide.

`\color{green} ✍️` Hence, if `A, B` and `C` are three points in the `XY`-plane, then they will lie on a line, i.e.,

`color(blue)("three points are collinear")` (Fig 10.8)

if and only if `color(red)"slope of AB = slope of BC."`

`\color{green} ✍️` Hence, if `A, B` and `C` are three points in the `XY`-plane, then they will lie on a line, i.e.,

`color(blue)("three points are collinear")` (Fig 10.8)

if and only if `color(red)"slope of AB = slope of BC."`

Q 3186245177

Three points `P (h, k), Q (x_1, y_1)` and `R (x_2, y_2)` lie on a line. Show that

`(h – x_1) (y_2 – y_1) = (k – y_1) (x_2 – x_1)` .

`(h – x_1) (y_2 – y_1) = (k – y_1) (x_2 – x_1)` .

Since points P, Q and R are collinear, we have

Slope of PQ = Slope of QR, i.e., ` (y_1 k)/(x_1 - h) = (y_2 - y_1 )/(x_2 - x_1)`

or ` (k-y_1)/(h-x_1) = (y_2 - y_1)/( x_2 - x_1)` ,

or ` (h-x_1) (y_2 - y_1) = (k- y_1) (x_2 -x_1)`

Q 3176345276

In Fig 10.9, time and distance graph of a linear motion is given.

Two positions of time and distance are recorded as, when T = 0, D = 2 and when

T = 3, D = 8. Using the concept of slope, find law of motion, i.e., how distance

depends upon time.

Two positions of time and distance are recorded as, when T = 0, D = 2 and when

T = 3, D = 8. Using the concept of slope, find law of motion, i.e., how distance

depends upon time.

Let (T, D) be any point on the line, where D denotes the distance at time

T. Therefore, points (0, 2), (3, 8) and (T, D) are collinear so that

` (8-2)/(3-0) = (D-8)/(T-3)` or ` 6 (T-3) = 3 (D-8)`

or `D = 2 (T+1)`

which is the required relation.