## Coordinate Geometry

• Let X’OX and Y’OY be two mutually perpendicular lines through any point O in the plane of the paper. Point O is known as the origin. The line X’OX is called the x-axis or axis of x; the line Y’OY is known as the y-axis or axis of y; and the two lines taken together are called the coordinate axes or the axes of coordinates.

• Any point can be represented on the plane described by the coordinate axes by specifying its x and y coordinates. The x coordinate of the point is also known as the abscissa while the y coordinate is also known as the ordinate.

• Distance Formula : If two points P and Q are such that they are represented by the points (x1, y1) and (x2, y2) on the x-y plane (cartesian plane), then the distance between the points P and Q = (x1x2)2(y1y2)2−−−−−−−−−−−−−−−−−−√

• Section Formula : If any point (x, y) divides the line segment joining the points (x1, y1) and (x2, y2) in the ratio m : n internally, then x = (mx2 + nx1)/(m + n) and y = (my2 + ny1)/(m – n)

Area of a Triangle

• The area of a triangle whose vertices are A (x1, y1), B (x2, y2) and C (x3, y3) is given by

• x1(y2y3)+x2(y3y1)+x3(y1y2)2

• Since the area cannot be negative, we have to take the modulus value given by the above equation

• If one of the vertices of the triangle is at the origin and the other two vertices are A (x1, y1), B (x2, y2) , then the area of triangle is (x1y2x2y1)2

Centre of gravity or centroid of a triangle

• The centroid of a triangle is the point of intersection of its medians (the line joining the vertex to the middle point of the opposite side).

• Centroid divides the medians in the ratio 2 : 1.

• In other words, the CG or the centroid can be viewed as a point at which the whole weight of the triangle is concentrated

• If A (x1, y1), B (x2, y2) and C (x3, y3) are the coordinates of the vertices of a triangle, then the coordinates of the centroid G of that triangle are x = (x1 + x2 + x3)/3 and y = (y1 + y2 + y3)/3

In-centre of a triangle

• The centre of the circle that touches the sides of a triangle is called its Incentre. In other words, if the three sides of the triangle are tangential to the circle then the centre of that circle represents the in-centre of the triangle

• The in-centre is also the point of intersection of the internal bisectors of the angles of the triangle. The distance of the in-centre from the sides of the triangle is the same and this distance is called the in-radius of the triangle.

• IfA (x1, y1), B (x2, y2) and C (x3, y3) are the coordinates of the vertices of a triangle, then the coordinates of its in-centre are

Circumcentre of a triangle

• The point of intersection of the perpendicular bisectors of the sides of a triangle is called its circumcentre

• It is equidistant from the vertices of the triangle.

• It is also known as the centre of the circle which passes through the three vertices of a triangle (or the centre of the circle that circumscribes the triangle

• Let ABC be a triangle. If O is the circumcentre of the triangle ABC, then OA = OB = OC and each of these three represent the circum radius.

Orthocentre of a triangle

• The orthocentre of a triangle is the point of intersection of the perpendiculars drawn from the vertices to the opposite sides of the triangle.

Collinearity of three points

• Three given points A, B and C are said to be collinear, that is, lie on the same straight line, if any of the following conditions occur:

• Area of triangle formed by these three points is zero

• Slope of AB = Slope of AC

• Any one of the three points (say C) lies on the straight line joining the other two points (here A and B).

• General Form : The general form of the equation of a straight line is ax + by + c = 0. (First degree equation in x and y). Where a, b and c are real constants and a, b are not simultaneously equal to zero. In this equation, slope of the line is given by -a/b

• The general form is also given by y = mx + c; where m is the slope and c is the intercept on y-axis. In this equation, slope of the line is given by m.

• Line Parallel to the X-axis : The equation of a straight line parallel to the x-axis and at a distance b from it, is given by y = b. Obviously, the equation of the x-axis is y = 0

• Line Parallel to Y-axis : The equation of a straight line parallel to the y-axis and at a distance a from it, is given by x = a. Obviously, the equation of y-axis is x = 0

• Slope Intercept Form : The equation of a straight line passing through the point A (x1, y1) and having a slope m is given by (y – y1) = m (x – x1)

• Two Points Form : The equation of a straight line passing through two points A (x1, y1), B (x2, y2) is given by (yy1)=(y2y1)(xx1)x2x1. Its slope = y2y1x2x1

• Intercept Form The equation of a straight line making intercepts a and b on the axes of x and y respectively is given by x/a + y/b = 1. If a straight line cuts x-axis at A and the y-axis at B then OA and OB are known as the intercepts of the line on x-axis and y-axis respectively

### Solved Question Papers

Q.
Find the distance between the points (5, 2) and (3, 4).

1.2
2.4
3.2.5
4.2√2

ANS.4

Explanation :
Distance = (5−3)2+(2−4)2−−−−−−−−−−−−−−−√

Q.
Find the point which divides the line segment joining (2, 5) and (1, 2) in the ratio 2 : 1 internally.

1.3,4/3
2.5,4/3
3.2,4/3
4.4/3,3

ANS.1

Explanation :
X = (2.1 + 1.2) /(1 + 2) = 4/3
Y = (2.2 + 1.5)/(2 + 1) = 9/3 = 3

Q.
Find the area of the triangle (0, 4), (3, 6) and (–8, –2).

1.-1
2.1
3.2
4.√2

Ans.2

Explanation :
Area of triangle = |1/2 {0 (6– (–2)) + 3 ((–2) –4) + (–8) (4–6)}|
= |1/2 {(0) + 3 (–6) + (–8) (–2)}|
= |1/2 (–2)| = |–1| = 1 square unit.

Q.
Find the centroid of the triangle whose vertices are (5, 3), (4, 6) and (8, 2)

1.11/3,17/3
2.17/3,11/3
3.7/3,1/3
4.1/3,7/3

ANS.1

Explanation :
X coordinate = (5 + 4 + 8)/3 = 17/3
Y coordinate = (3 + 6 + 2)/3 = 11/3

Q.
What will be the circumcentre of a triangle whose sides are 3x – y + 3 = 0, 3x + 4y + 3 = 0 and x + 3y + 11 = 0?

1.-3, 3
2.3, 3
3.3, –3
4.-3, –3

ANS.3

Explanation :
Let ABC be the triangle whose sides AB, BC and CA have the equations 3x – y + 3 = 0, 3x + 4y + 3 = 0 and x + 3y + 11 = 0 respectively.
Solving the equations, we get the points A, B and C as (–2, –3), (–1, 0) and (7, – 6) respectively.
The equation of a line perpendicular to BC is 4x – 3y + k = 0.
This will pass through (3, –3), the mid-point of BC, if 12 + 9 + k = 0 so k = –21
Putting k1 = –21 in 4x – 3y + k = 0, we get 4x – 3y – 21 = 0
as the equation of the perpendicular bisector of BC.
Again, the equation of a line perpendicular to CA is 3x – y + k1 = 0.
This will pass through (5/2, –9/2), the mid-point of AC if
15/2 + 9/2 + k1 = 0 fi k1 = –12
Putting k1 = –12 in 3x – y + k1 = 0, we get 3x – y – 12 = 0
as the perpendicular bisector of AC.
Solving (i) and (ii), we get x = 3, y = –3.
Hence, the coordinates of the circumcentre of △ABC are (3, –3).

Q.
: Find the orthocentre of the triangle whose sides have the equations y = 15, 3x = 4y, and 5x + 12y = 0.

1.0, 33
2.0, –33
3.3, –33
4.-3, –33

ANS.2

Explanation :
Let ABC be the triangle whose sides BC, CA and AB have the equations y = 15, 3x = 4y, and 5x + 12y = 0 respectively.
Solving these equations pairwise, we get coordinates of A, B and C as (0, 0), (–36, 15) and (20, 15)
respectively.
AD is a line passing through A (0, 0) and perpendicular to y = 15.
So, equation of AD is x = 0.
The equation of any line perpendicular to 3x – 4y = 0 is represented by 4x + 3y + k = 0.
This line will pass through (–36, 15) if –144 + 45 + k = 0 fi k = 99.
So the equation of BE is 4x + 3y + 99 = 0.
Solving the equations of AD and BE we get x = 0, y = – 33.
Hence the coordinates of the orthocentre are (0, –33).

Q.
Select the right option the points (–a, –b), (0, 0) and (a, b) are

1.Collinear
2.Vertices of square
3.Vertices of a rectangle
4.None of these

ANS.1

Explanation :
Let A, B, C are the points whose coordinates are (–a, –b), (0, 0) and (a, b)
Slope of BC = b/a
Slope of AB = b/a
So, the straight line made by points A, B and C is collinear.

Alternative: Draw the points on paper assuming the paper to be a graph paper. This will give you an
indication regarding the nature of points. In the above question, point (a, b) is in first quadrant for a > 0, b
> 0 and point (–a, –b) is directly opposite to the point (a, b) in the third quadrant with the third point (0,
0) in the middle of the straight line joining the points A and B.
You can check this by assuming any value for ‘a’ and ‘b’.
Also, you can use this method for solving any problem involving points and diagrams made by those
points. However you should be fast enough to trace the points on paper. A little practice of tracing points

Q.
Find the equation of a straight line passing through (2, –3) and having a slope of 1 unit.

1.-y – x + 5 = 0
2.y – x + 5 = 0
3.y + x + 5 = 0
4.y – x – 5 = 0

ANS.2

Explanation :
Here slope = 1
And point given is (2, –3).
So, we will use point-slope formula for finding the equation of straight line. This formula is given by:
(y – y1) = m (x – x1)
So, equation of the line will be y – (–3) = 1 (x – 2)
fi y + 3 = x – 2
fi y – x + 5 = 0

Q.
Which of the lines represented by the following equations are parallel to each other?
1. x + 2y = 5
2. 2x – 4y = 6
3. x – 2y = 4
4. 2x + 6y = 8

ANS.2 and 3

Explanation :
Go through the options and check which of the two lines given will satisfy the criteria for two
lines to be parallel. It will be obvious that option c is correct, that is, the line 2x – 4y = 6 is parallel to the
line x – 2y = 4.

Q.
Find the equation of a straight line parallel to the straight line 3x + 4y = 7 and passing through the point (3, –3).

1.-3x + 4y + 3 = 0
2.3x – 4y + 3 = 0
3.3x + 4y – 3 = 0
4.3x + 4y + 3 = 0

ANS.4

Explanation :
Equation of the line parallel to 3x + 4y = 7 will be of the form 3x + 4y = k.
This line passes through (3, –3), so this point will satisfy the equation of straight line 3x + 4y = k.
So, 3*3+ 4*(–3) = k fi k = –3.
Hence, equation of the required straight line will be 3x + 4y + 3 = 0.
Condition for two lines to be perpendicular: Two lines are said to be perpendicular if product of the
slopes of the lines is equal to –1.
For this to happen, the product of the coefficients of x + the product of the coefficients of y should be
equal to zero.

Q.
Which of the following two lines are perpendicular?
1. x + 2y = 5
2. 2x – 4y = 6
3. 2x + 3y = 4
4. 2x – y = 4

1 and 2
2 and 4
2 and 3
1 and 4

Explanation :
Check the equations to get option 4 as the correct answer.

Q.
Find the equation of a straight line perpendicular to the straight line 3x + 4y = 7 and passing through the point (3, –3).

1.-4x – 3y = 21
2.4x + 3y = 21
3.4x – 3y = -21
4.4x – 3y = 21

ANS.4

Explanation :
Equation of the line perpendicular to 3x + 4y = 7 will be of the form 4x – 3y = K.
This line passes through (3, –3), so this point will satisfy the equation of straight line 4x – 3y = K. So, 4.3
– 3.–3 fi K = 21.
Hence, equation of required straight line will be 4x – 3y = 21

Q.
Two sides of a square lie on the lines x + y = 2 and x + y = –2. Find the area of the square formed in this way

1.6
2.7
3.8
4.8√2

ANS.3

Explanation :
Obviously, the difference between the parallel lines will be the side of the square.
To convert it into the form of finding the distance of a point from a line, we will have to find out a point at
which any one of these two lines cut the axes and then we will draw a perpendicular from that point to the
other line, and this distance will be the side of the square.
To find the point at which the equation of the line x + y = 2 cut the axes, we will put once x = 0 and then
again y = 0.
When x = 0, y = 2, so the coordinates of the point where it cuts y-axis is (0, 2).
Now the point is (0, 2), and the equation of line on which perpendicular is to be drawn is x + y = –2.

So, distance = $$\frac{1*0 + 1*2 + 2}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}}$$

Area = 16/2 = 8

Q.
If origin (0, 0) is shifted to (5, 2), what will be the coordinates of the point in the new axis which was represented by (1, 2) in the old axis?

1.–4, -2
2.–4, 2
3.–4, 0
4.4, 0

ANS.3

Explanation :
Let (X, Y) be the coordinates of the point in the new axis.
Then, 1 = X + 5 so X = –4
2 = Y + 2 so Y = 0
So, the new coordinates of the point will be (–4, 0)