`\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} ✍️` 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} ✍️` 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))`

`\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} ✍️` 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`.

`\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."`