Real life situation of concept of slope.
(i) While building the roads, need to consider the slope.
(ii) Wheel - chair ramp in Hospitals.
(iii) While constructing Bridges.

A(0, 5), B (5, 0) and C (- 4,-7)
Centroid of a triangle \(=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right) \)
\(=\left(\frac{0+5-4}{3}, \frac{5+0-7}{3}\right) \)
\(=\left(\frac{1}{3}, \frac{-2}{3}\right) \)

The vertices Q(3, - 2) and R(- 5, 4)
slope of the line QR \(=\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \)
\(=\frac{-2-4}{3+5}=\frac{-6}{8}=\frac{-3}{4} \)
Slope of the line parallel to QR is \(-\frac{3}{4}\)
Equation of the line passing through
P(- 5, 2) and having slope \(-\frac{3}{4}\) is
y - y₁ = m(x - x₁)
y - 2 = \(-\frac{3}{4}(x+5)\)
4y - 8 = -3x - 15
3x + 4y + 7 = 0

Let A be the location of the
cat and B be the point where bottle of milk is kept.
Given A (-6, - 4), B (5, 11)
Equation of line passing through (x₁, y₁) and (x₂, y₂) is
\(\frac{y-y_{1}}{y_{2}-y_{1}} =\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y+4}{11+4} =\frac{x+6}{5+6} \)
11(y + a) = 15(x + 6)
11y + 44 = 15x + 90
15x - 11y + 46 = 0

Given points \(\left( 5,\sqrt { 5 } \right) \) and (0, 0)
Slope of a line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}(\text { or }) \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
\(=\frac{\sqrt{5}-0}{5-0}=\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)

Given angle of inclination θ = 90⁰
Slope of a line = tan θ
= tan90⁰ = ∝ (undefined)

Given points are (2, 3), (4, a) and (6, - 3)
Since the points are colinear, Area of triangle is zero
\(\text { i.e., } \frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=0\)
2(a + 3) + 4(- 3 -3) + 6(3 - a) = 0
2a + 6 - 24 + 18 - 6a = 0
-4a + 0 = 0
-4a = 0
a = 0

Given points are \((-\frac12 ,3)\), (- 5, 6) and (-8, 8)
Let us use area of triangle formula
Area of triangle = \(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+\right. \left.\quad x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \)
\(=\frac{1}{2}\left[-\frac{1}{2}(6-8)-5(8-3)-8(3-6)\right] \)
\(=\frac{1}{2}\left[-\frac{1}{2}(-2)-5(5)-8(-3)\right] \)
\(=\frac{1}{2}[1-25+24]=\frac{1}{2}(0)=0 \)
Since, the area of triangle is zero, the given points are collinear.

Let P be (3, 8) and Q (a, b) be the image of P
Slope of PQ = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \)
= \(\frac{8-b}{3-a}=m_{1} \)
Slope of the line x+ 3y = 7 is \(\frac{-1}{3}=\mathrm{m}_{2}\)
Since PQ is perpendicular to the given line, then m₁ x m₂ = -1
\(\frac{8-b}{3-a} \times\left(\frac{-1}{3}\right)=-1\)
8 - b = 9 - 3a
3a - b = 1
Mid point of PQ \(=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \)
\(=\left(\frac{3+a}{2}, \frac{8+b}{2}\right) \)
Since the line x + 3y = 7 is the perpendicular bisector of PQ, the midpoint of PQ lies on the line
\(\left(\frac{3+a}{2}\right)+3\left(\frac{8+b}{2}\right)=7\)
3 + a + 24 + 3b = 14
a + 3b = -13       ...(2)
Solving (1) and (2)

Substituting in (1)
b = -4
Image of (3, 8) is (- 1, - 4).

Given sides of a triangle are
3x + y - 2 = 0 = 3x + y = 2       ....(1)
5x + 2y - 3 = 0 = 5x + 2y = 3        ......(2)
2x - y - 3 = 0 = 2x - y = 3       ..........(3)
Solving (1) and (2)

Substituting in (1)
y = 2 - 3 = -1
Point of intersection of (1) and (2) is (1, - 1)
Now, solving (2) and (3)

x = 1
Substituting in (3)  y = -1
Point of intersection of (2) and (3) is ( 1, - 1). Since the point of intersection of (t), (2) and (2), (3) is same, No such triangle is possible.
Hence, area of triangle is zero.

Given vertices are A(- 3, 0), B(10, - 2) and, C(12, 3).
Slope of BC = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{-2-3}{10-12}=\frac{-5}{-2}=\frac{5}{2}\)
Altitude AD is perpendicular to BC and passing through A(- 3, 0)
Slope of AD = \(-\frac{2}{5}\)
Equation of AD y - y₁ = m(x - x₁)
\(y-0=-\frac{2}{5}(x+3)\)
5y = -2x - 6
2x + 5y + 6 = 0
Slope of AC \(=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{0-3}{-3-12}=\frac{-3}{-15}=\frac{1}{5}\)
Altitude BE is perpendicular to AC and passing through B(10, - 2). Slope of BE = 5
Equation of BE y - y₁ = m(x - x₁)
y + 2 = - 5(x - 10)
y + 2 = -5x + 50
5x + y - 48 = 0

Given, vertices of a quadrilateral arc
A(- 4, - 2), B(5, - 1), C(6, 5) and D(- 7 ,6).
Let B Q, R and S be the mid points of the sides
AB, BC, CD and AD respectively
Mid point of 
\(\mathrm{AB}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) =\mathrm{P}\left(\frac{-4+5}{2}, \frac{-2-1}{2}\right) \)
\(=\mathrm{P}\left(\frac{1}{2},-\frac{3}{2}\right) \)
Mid point of BC
\(=Q\left(\frac{5+6}{2}, \frac{-1+5}{2}\right)=Q\left(\frac{11}{2}, 2\right)\)
Mid point of CD
\(=R\left(\frac{6-7}{2}, \frac{5+6}{2}\right)=R\left(-\frac{1}{2}, \frac{11}{2}\right)\)
Mid point of AD
\(=S\left(\frac{-4-7}{2}, \frac{-2+6}{2}\right)=S\left(-\frac{11}{2}, 2\right)\)
Slope of PQ \(=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{-\frac{3}{2}-2}{\frac{1}{2}-\frac{11}{2}}=\frac{-\frac{7}{2}}{-\frac{10}{2}}=\frac{7}{10}\)
Slope of QR  \(= \frac{2-\frac{11}{2}}{\frac{11}{2}+\frac{1}{2}}=\frac{-\frac{7}{2}}{\frac{12}{2}}=-\frac{7}{12} \)
Slope of RS  \(= \frac{\frac{11}{2}-2}{-\frac{1}{2}+\frac{11}{2}}=\frac{\frac{7}{2}}{\frac{10}{2}}=\frac{7}{10} \)
Slope of PS  \(= \frac{-\frac{3}{2}-2}{\frac{1}{2}+\frac{11}{2}}=\frac{-\frac{7}{2}}{\frac{12}{2}}=-\frac{7}{12} \)
Slope of PQ = Slope of RS = PQ || RS
Slope of QR = Slope of PS = QR || PS
Hence, the mid points form a parallelogram.

From the table, we found that as x increases, y also increases. Thus, the variation is a direct variation.
Let y = kx, where k is a constant of proportionality.
From the given values, we have
\(k=\frac{3.1}{1}=\frac{6.2}{2}=\frac{9.3}{3}=\frac{12.4}{4}=\ldots=3.1\)
When you plot the points (1, 3.1) (2, 6.2) (3, 9.3), (4, 12.4), (5, 15.5), you find the relation y = (3.1)x forms a straight-line graph.
Clearly, from the graph, when diameter is 6 cm, its circumference is 18.6 cm.

Given, radius = 4 cm
Construction
Step 1 : With O as the centre, draw a circle of radius 4 cm.
Step 2 : Take a point L on the circle. Through L draw any chord LM.
Step 3 : Take a point M distinct from L and N on the circle, so that L, M and N are in anti clockwise direction. Join LN and NM.
Step 4 : Through L draw a tangent TT' such that \(\angle\)TLM =\(\angle\)MNL
Step 5 : TT' is the required tangent.