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**Introduction**

- A coordinate graph is a rectangular grid with two number lines called
**axes**. Theis the horizontal number line and the*x*-axisis the vertical number line.*y*-axis - The axes intersect at the
**origin**which is the point (0,0).

**Coordinate Geometry Formulas**

- Now, Let us have a look at some formulas for coordinate geometry. We will use the below picture as a reference for the formulas.
- Slope of PQ = m =
- Equation of PQ is as below:

or y = mx + c - The product of the slopes of two perpendicular lines is –1.
- The slopes of two parallel lines are always equal. If m1 and m2 are slopes of two parallel lines, then m1=m2.
- The distance between the points (x
_{1}, y_{1}) and (x_{2}, y_{2}) is - If point P(x, y) divides the segment AB, where A (x
_{1}, y_{1}) and B (x_{2}, y_{2}), internally in the ratio m: n, then,

x= (mx_{2}+ nx_{1})/(m+n) and y= (my_{2}+ ny_{1})/(m+n) - If P is the midpoint then,
- If G (x, y) is the
**centroid**of triangle ABC, A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3}), then,

x = (x_{1}+ x_{2 }+ x_{3})/3 and y = (y_{1}+ y_{2}+ y_{3})/3 - If I (x, y) is the
**in-center**of triangle ABC, A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3}), then, - where a, b and c are the lengths of the BC, AC and AB respectively.
- The equation of a straight line is y = mx + c, where m is the slope and c is the y-intercept (tan = m, where is the angle that the line makes with the positive X-axis).
- If two intersecting lines have slopes m1 and m2 then the angle between two lines θ will be tan θ = (m1−m2) / (1+m1m2)
- The length of perpendicular from a point (x1 ,y1 ) on the line AX+BY+C = 0 is

P = (Ax1+By1+C) / (A^{2}+B^{2})

**Equations of a Line**

- General equation of a line Ax + By = C
- Slope intercept form y = mx + c (c is y intercept)
- Point-slope form y – y1 = m (x – x1) (m is the slope of the line)
- Intercept form x / a + y / b = 1 (where a and b are x and y intercepts respectively)
- Two point form: (y−y1) / (y2−y1) = (x−x1) / (x2−x1)

**Knowledge of Quadrants**

- Quadrant I ⇒ X is Positive Y is Positive
- Quadrant II ⇒ X is Negative Y is Positive
- Quadrant III ⇒ X is Negative Y is Negative
- Quadrant IV ⇒ X is Positive Y is Negative

Try yourself:Angles between 180 ° and 270 ° lies in:

View Solution

**Solved Examples**

**Example 1: ****What is the distance between the points A (3,8) and B(-2,-7) ?****a)** 5√2**b)** 5**c)** 5√10**d)** 10√2

- The distance between 2 points (x1, y1) and (x2, y2) is given as
- Sqrt ((x2-x1)
^{2}+ (y2-y1)^{2})- Hence, required distance = sqrt((-2-3)
^{2}+ (-7-8)^{2}) = 5√10

**Example 2: ****The points of intersection of three lines 2X + 3Y – 5=0 and 5X – 7Y + 2=0 and 9X – 5Y – 4 = 0****a)** Form a triangle**b)** Are on lines perpendicular to each other**c)** Are on lines parallel to each other**d)** Are coincident

To solve the question above, we should remember the properties of the lines for being parallel, perpendicular or intersecting:

- Two lines are parallel to each other if their slopes are equal
- Two lines are perpendicular if the product of slopes is -1
- Lines are coincident if they at least have one point which satisfies all the equation.
- The three lines can be expressed in the y=mx + c format as:

Y = (5/3) – (2X/3), Y = (5X/7) + (2/7) , Y = (9X/5) – (4/5)- Therefore, the slopes of the three lines are -2/3, 5/7, 9/5 and their Y intercepts are 5/3, 2/7 and 4/5 respectively.
- We see above that the product of slopes of none of the lines is -1. Thus, lines are not perpendicular to each other.
- Also, slopes of the no two lines is same. Thus, lines are not parallel to each other.
- Solving the first two equations we get X=1 and Y = 1. If we substitute (1,1) in the third equation Y=(9X/5 – 4/5), we find that it also satisfies the equation. This suggests that the three lines intersect at a common point and hence coincident.

**Example 3: ****The area of the triangle whose vertices are (a + 1, a + 1), (a, a) and (a+2, a) is****a)** a^{3}**b)** 1**c)** 2a**d)** 2^{1/2}

- Let a = 0, Thus the three vertices of the triangle becomes (1, 1) (0, 0) and (2, 0)
- If we look at the below figure, Area = ½ * base * height = ½ * 2 * 1 = 1
- Imp: The main point to note here is that area will be independent of a.

**Example 4: ****Consider a triangle drawn on the X – Y plane with its three vertices of (41,0) , (0,41) and (0,0), each vertex being represented by its (X,Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is:****a)** 12**b)** 8**c)** 6**d)** 4

- Equation of the line will be of the form => x + y = 41.
- Now, we know that if the x,y coordinates of a point are integer, the sum will also be an integer

X + y = k (k, a variable)- As per the question we need to exclude all the values lying on the boundary of triangle, k can take all values from 1 to 40 only. K = 0 is rejected as at k =0 will give the point at A which is also not allowed.
- With K = 40, x + y = 40; this will be satisfied by points (1, 39), (2, 38), (3,37) …… (38, 2), (39, 1). That is a total of 39 points
- Similarly x + y = 38, will be satisfied by 37 points.

X + Y = 37, will be satisfied by 36 points

X + Y = 3 will be satisfied by 2 points

X + Y = 2 will be satisfied by 1 point

X + Y = 1 will not be satisfied by any point- So, the total number of all such points is: 39 + 38 + 37 + 36 + ……………………. + 3 + 2 + 1 = n(n+1)/2 points = (39*40) / 2 = 780 points

**Example 5: ****Two lines P and Q intersect at point (3, 2) in the x-y plane. The slope of line P is 45 degrees and line Q is parallel to the X axis. What is the area (in sq. units) of the triangle formed by P, Q and a line perpendicular to P passing through point (5, 4) ?****a) **12**b)** 8**c)** 6**d)** 4

- Let us look at the image below:
- As slope of line P is 45 degree. Therefore, ∠ABC = 45 degree
- In triangle ABC, length of AB =
SQRT[(5-3)^{2 }+ (4-2)^{2}] = 2√2 units- Therefore, length of line AC = 2
√2 units (Since ABC is an isosceles triangle. Thus AB = AC)- Thus, required area = ½ * 2
√2 * 2√2 = 4 sq. units

**Example 6: ****The line ****√3 Y = x is the radius of the circle. It meets the circle o=centered at origin O at point M (√****3, 1). If PQ is the tangent to the circle at M as shown, find the length of the PQ.**

**a)** (5/2)**√**3 units**b)** 3 **√**3 units**c)** 2**√**3 units**d)** 8/**√**3 units

- PQ is perpendicular to line Y = X /
√3 (Since, radius of a circle is perpendicular to the tangent of the circle)- Therefore, slope of PQ = -1 / (1/
√3) = –√3 (Since, product of slopes of line perpendicular to each other is -1)- Therefore, Let equation of the line PQ be y = –
√3x + c- Now at the point M, when x =
√3, y = 1- Putting the above values of x and y in the above equation, we get c = 4 The equation of the line becomes, Y = –
√3x + 4- Thus, by using the above equation, we get:
- Coordinates of point P = (0, 4) and coordinates of point Q = (4/
√3, 0) (Putting x = 0 in above equation, we find value of P and putting Y = 0 in above equation, we find value of Q)- Hence PQ = sqrt [(4/
√3)^{2 }+ 4^{2}] = 8/√3 units.

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