Signs of co-ordinate in the four quadrants:

• x and y co-ordinates are positive in I quadrant.




• x co-ordinate is negative and y co-ordinate is positive in II quadrant.




• x and y co-ordinate are negative in III quadrant.




• x co-ordinate is positive and y co-ordinate is negative in IV quadrant

(x, y), (-x, y), (-x, -y) and (x, -y) will be the co-ordinates of the points in I, II, III and IV quadrants respectively, if x and y are positive numbers.

Let any two points on the plane be A(\(x_{1}, y_{1}\)) and B(\(x_{2}, y_{2}\)).Then the distance between A and B is represented as AB i.e. given by

AB = \(\sqrt{(x_{2} – x_{1})^2 + (y_{2} – y_{1})^2}\)

The distance of the point A(x, y) from the origin (0, 0) is given by

\(\sqrt{x^2 + y^2}\)

Internal division:

The co-ordinates of the point P which divides the line joining AB internally in the ratio m: n are

[\(\frac{mx_{2} + nx_{1}}{m + n}\), \(\frac{my_{2} + ny_{1}}{m + n}\)]

Here, A and B are two points i.e. A(\(x_{1}, y_{1}\)) and B(\(x_{2}, y_{2}\)).

External division:

The co-ordinates of the point P which divides the line joining points A(\(x_{1}, y_{1}\)) and B(\(x_{2}, y_{2}\)) externally in the ratio m: n are

[\(\frac{mx_{2} + nx_{1}}{m + n}\), \(\frac{my_{2} + ny_{1}}{m + n}\)]

Here, the point P is beyond A and B for external division and it can be either close to B or close to A.

Mid point:

If P is the mid point and A(\(x_{1}, y_{1}\)) & B(\(x_{2}, y_{2}\)) are the two points given then m: n = 1: 1 i.e. m = n.

Therefore, the co-ordinates or the point P are given by

(\(\frac{x_{1} + x_{2}}{2}\), \(\frac{y_{1} + y_{2}}{2}\))

Centroid:

Here, from the above figure,

G is the centroid.

A = (\(x_{1}, y_{1}\))

B = (\(x_{2}, y_{2}\))

C = (\(x_{3}, y_{3}\))

If A, B and C are the vertices of a triangle. The three medians of the triangle intersect at its centroid and the centroid divides the median in the ratio 2 : 1.

Co-ordinates of the point G are given by

(\(\frac{x_{1} + x_{2} + x_{3}}{3}\), \(\frac{y_{1} + y_{2} + y_{3}}{3}\))

Straight line:

• A straight line is represented as a linear equation of the first degree in two variables x and y.




• ax + by + c = 0 is the general form of equation of a line.

Eg: 2x + 3y + 7 = 0

Slope of a line:

• How steeply that line climbs or falls as it moves from left to right is a measurement of a line slope.




• If there are two points on a line, once again (\(x_{1}, y_{1}\)) and then the slope of that line can be determined using the formula i.e.
  
  Slope = \(\frac{y_{2} – y_{1}}{x_{2} – x_{1}}\)

Parallel lines: If the slopes of two lines are equal, then the two lines are parallel.

Perpendicular lines: If and only if the product of the two lines slopes is -1, then the two lines are perpendicular.

Equation of lines in standard forms:

1. Line’s general form:

• A linear equation of the first degree in two variables x and y represents a straight line.




• The equation is ax + by + c = 0

2. Line’s slope intercept form: The equation of a straight line having slope ‘m’ and making an intercept ‘c’ with y-axis is

y = mx + c.

3. Line’s point form: The equation of a straight line passing through the points (\(x_{1}, y_{1}\)) and (\(x_{2}, y_{2}\)) is

(\(y – y_{1}\)) = \(\frac{y_{2} – y_{1}}{x_{2} – x_{1}}(x – x_{1}\))

Where, \(x_{1}\) ≠ \(x_{2}\)

4. Line’s point slope form: The equation of a straight line passing through the point (\(x_{1}, y_{1}\)) and having slope m is given by

(\(y – y_{1}\)) = m\((x – x_{1}\))

5. Line’s intercept form: The equation of a line making intercepts a and b on the x and y axis respectively is given by

\(\frac{x}{a} + \frac{y}{b}\) = 1

1. AB = \(\sqrt{(x_{2} – x_{1})^2 + (y_{2} – y_{1})^2}\)

2. The distance of the point A(x, y) from the origin (0, 0) is given by
\(\sqrt{x^2 + y^2}\)

3. The co-ordinates of the point P which divides the line joining AB internally in the ratio m: n are
[\(\frac{mx_{2} + nx_{1}}{m + n}\), \(\frac{my_{2} + ny_{1}}{m + n}\)]

4. The co-ordinates of the point P which divides the line joining points A(\(x_{1}, y_{1}\)) and B(\(x_{2}, y_{2}\)) externally in the ratio m: n are
[\(\frac{mx_{2} + nx_{1}}{m + n}\), \(\frac{my_{2} + ny_{1}}{m + n}\)]

5. The co-ordinates or the point P are given by
(\(\frac{x_{1} + x_{2}}{2}\), \(\frac{y_{1} + y_{2}}{2}\))

6. Co-ordinates of the point G are given by
(\(\frac{x_{1} + x_{2} + x_{3}}{3}\), \(\frac{y_{1} + y_{2} + y_{3}}{3}\))

7. Slope = \(\frac{y_{2} – y_{1}}{x_{2} – x_{1}}\)

8. y = mx + c

9. (\(y – y_{1}\)) = \(\frac{y_{2} – y_{1}}{x_{2} – x_{1}}(x – x_{1}\))

10. (\(y – y_{1}\)) = m\((x – x_{1}\))

11. \(\frac{x}{a} + \frac{y}{b}\) = 1

1. Find the perimeter of the triangle connecting the points with (x, y) co-ordinates (-2, 7), (3, -5) and (-2, -5)?

Solution:

The points A, B and C form a right triangle ACB, by sketching it on a graph.

Given that

A = (-2, 7)

B = (3, -5)

C = (-2, -5)

Now, calculate the lengths of the three sides and add them together to get the perimeter i.e.

Length of AC = 7(-5) = 12

Length of BC = 3-(-2) = 5

Using Pythagorean theorem,

\((AB)^2\) = 12 x 12 + 5 x 5

AB = 13

Therefore, perimeter = 12 + 13 + 5 = 30

2. What is the equation of the line passing through the point (2, 3) and perpendicular to the line 5x + 4y + 6 = 0.

Solution:

Given that,

5x + 4y + 6 = 0

P(\(x_{1}, y_{1}\)) = (2, 3)

Equation to the line passing through the point (\(x_{1}, y_{1}\))) and perpendicular to the line ax + by + c = 0 is

b(\(x – x_{1}\)) + a(\(y – y_{1}\)) = 0

4(x – 2) + 5(y – 3) = 0

4x + 5y – 23 = 0

Therefore, the required line is 4x + 5y – 23 = 0

3. A line is represented by the equation y = 3x + 2.

A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.

Solution:

The given equation is in slope-intercept form, so slope is the co-efficient of the x-term and the y-intercept is the term without a variable.

Here,

The slope is 3.

The y-intercept is 2.

Thus, the slope is the greater quantity.

Hence, correct option is A.

4. The figure of a rhombus is as shown below:

A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.

Solution:

The diagonals of a rhombus are perpendicular to each other.

So, angles AEB and CEB are both right angles.

Thus, AEB is a right triangle.

Hypotenuse of triangle is line segment AB because it is opposite the right angle.

Line segment EB is also part of the same triangle, opposite one of the smaller angles of the triangle.

Therefore, the longest of the three sides of a triangle is hypotenuse always.

So, line segment AB is longer.

Hence, right answer is option B.

5.

A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.

Solution:

Both points are on the y-axis, and are 5 units apart.

The line joining the points on the right hand side forms the hypotenuse of a triangle with sides 4 (parallel to the x-axis), and 3 (parallel to the y-axis).

This means the hypotenuse must be 5 (3-4-5 triangle).

So both columns are equal.

Hence, option C is right answer.