CHAPTER – 7
COORDINATE GEOMETRY
DISTANCE FORMULA
The distance between any two points A(x1, y1) and B(x2, y2) is given by
AB ( x2 x1 )2 ( y2 y1 )2
or AB (difference of abscissae)2 (difference of ordinates) 2
Distance of a point from origin
The distance of a point P(x, y) from origin O is given by OP = x2 y 2
Problems based on geometrical figure
To show that a given figure is a
Parallelogram – prove that the opposite sides are equal
Rectangle – prove that the opposite sides are equal and the diagonals are equal.
Parallelogram but not rectangle – prove that the opposite sides are equal and the diagonals are
not equal.
Rhombus – prove that the four sides are equal
Square – prove that the four sides are equal and the diagonals are equal.
Rhombus but not square – prove that the four sides are equal and the diagonals are not equal.
Isosceles triangle – prove any two sides are equal.
Equilateral triangle – prove that all three sides are equal.
Right triangle – prove that sides of triangle satisfies Pythagoras theorem.
IMPORTANT QUESTIONS
Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.
Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points.
AB (1 4) 2 (7 2)2 9 25 34
BC (4 1)2 (2 1) 2 25 9 34
CD (1 4)2 (1 4) 2 9 25 34
DA (1 4)2 (7 4)2 25 9 34
AC (1 1) 2 (7 1) 2 4 64 68
BD (4 4)2 (2 4) 2 64 4 68
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal
and its diagonals AC and BD are also equal. Therefore, ABCD is a square.
Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).
Solution : We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be
equidistant from A and B. Then AP2 = BP2
(6 – 0)2 + (5 – y)2 = (– 4 – 0)2 + (3 – y)2
36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y 4y = 36 y = 9
So, the required point is (0, 9).
, Questions for practice
1. Show that the points A(1, 2), B(5, 4), C(3, 8) and D(–1, 6) are vertices of a square.
2. Show that the points A(5, 6), B(1, 5), C(2, 1) and D(6, 2) are vertices of a square.
3. Show that the points A(1, –3), B(13, 9), C(10, 12) and D(–2, 0) are vertices of a rectangle.
4. Show that the points A(1, 0), B(5, 3), C(2, 7) and D(–2, 4) are vertices of a rhombus.
5. Prove that the points A(–2, –1), B(1, 0), C(4, 3) and D(1, 2) are vertices of a parallelogram.
6. Find the point on x-axis which is equidistant from (7, 6) and (–3, 4).
7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
8. Find a point on the y-axis which is equidistant from the points A(5, 2) and B(– 4, 3).
9. Find a point on the y-axis which is equidistant from the points A(5, – 2) and B(– 3, 2).
10. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
11. Find the value of a , if the distance between the points A (–3, –14) and B (a, –5) is 9 units.
12. If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find
distance PQ.
Section formula
The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and
B(x2, y2), internally, in the ratio m1 : m2 are
m1 x2 m2 x1 m1 y2 m2 y1
,
m1 m2 m1 m2
Mid-point formula
The coordinates of the point P(x, y) which is the midpoint of the line segment joining the points
x x y y2
A(x1, y1) and B(x2, y2), are 1 2 , 1
2 2
IMPORTANT QUESTIONS
Find the coordinates of the point which divides the line segment joining the points (4, – 3) and
(8, 5) in the ratio 3 : 1 internally.
Solution : Let P(x, y) be the required point.
m x m1 x2 m y m1 y2
Using the section formula, x 2 1 ,y 2 1 we get
m1 m2 m1 m2
3(8) 1(4) 3(5) 1(3)
x 7, y 3
3 1 3 1
Therefore, (7, 3) is the required point.
In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and
B(3, – 8)?
Solution : Let (– 4, 6) divide AB internally in the ratio k : 1.
m x m1 x2 m y m1 y2
Using the section formula, x 2 1 ,y 2 1 we get
m1 m2 m1 m2
k (8) 1(10)
y 6
k 1
8k 10 6k 6 8k 6k 6 10
4 2
14k 4 k
14 7
Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in
the ratio 2 : 7.
COORDINATE GEOMETRY
DISTANCE FORMULA
The distance between any two points A(x1, y1) and B(x2, y2) is given by
AB ( x2 x1 )2 ( y2 y1 )2
or AB (difference of abscissae)2 (difference of ordinates) 2
Distance of a point from origin
The distance of a point P(x, y) from origin O is given by OP = x2 y 2
Problems based on geometrical figure
To show that a given figure is a
Parallelogram – prove that the opposite sides are equal
Rectangle – prove that the opposite sides are equal and the diagonals are equal.
Parallelogram but not rectangle – prove that the opposite sides are equal and the diagonals are
not equal.
Rhombus – prove that the four sides are equal
Square – prove that the four sides are equal and the diagonals are equal.
Rhombus but not square – prove that the four sides are equal and the diagonals are not equal.
Isosceles triangle – prove any two sides are equal.
Equilateral triangle – prove that all three sides are equal.
Right triangle – prove that sides of triangle satisfies Pythagoras theorem.
IMPORTANT QUESTIONS
Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.
Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points.
AB (1 4) 2 (7 2)2 9 25 34
BC (4 1)2 (2 1) 2 25 9 34
CD (1 4)2 (1 4) 2 9 25 34
DA (1 4)2 (7 4)2 25 9 34
AC (1 1) 2 (7 1) 2 4 64 68
BD (4 4)2 (2 4) 2 64 4 68
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal
and its diagonals AC and BD are also equal. Therefore, ABCD is a square.
Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).
Solution : We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be
equidistant from A and B. Then AP2 = BP2
(6 – 0)2 + (5 – y)2 = (– 4 – 0)2 + (3 – y)2
36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y 4y = 36 y = 9
So, the required point is (0, 9).
, Questions for practice
1. Show that the points A(1, 2), B(5, 4), C(3, 8) and D(–1, 6) are vertices of a square.
2. Show that the points A(5, 6), B(1, 5), C(2, 1) and D(6, 2) are vertices of a square.
3. Show that the points A(1, –3), B(13, 9), C(10, 12) and D(–2, 0) are vertices of a rectangle.
4. Show that the points A(1, 0), B(5, 3), C(2, 7) and D(–2, 4) are vertices of a rhombus.
5. Prove that the points A(–2, –1), B(1, 0), C(4, 3) and D(1, 2) are vertices of a parallelogram.
6. Find the point on x-axis which is equidistant from (7, 6) and (–3, 4).
7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
8. Find a point on the y-axis which is equidistant from the points A(5, 2) and B(– 4, 3).
9. Find a point on the y-axis which is equidistant from the points A(5, – 2) and B(– 3, 2).
10. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
11. Find the value of a , if the distance between the points A (–3, –14) and B (a, –5) is 9 units.
12. If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find
distance PQ.
Section formula
The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and
B(x2, y2), internally, in the ratio m1 : m2 are
m1 x2 m2 x1 m1 y2 m2 y1
,
m1 m2 m1 m2
Mid-point formula
The coordinates of the point P(x, y) which is the midpoint of the line segment joining the points
x x y y2
A(x1, y1) and B(x2, y2), are 1 2 , 1
2 2
IMPORTANT QUESTIONS
Find the coordinates of the point which divides the line segment joining the points (4, – 3) and
(8, 5) in the ratio 3 : 1 internally.
Solution : Let P(x, y) be the required point.
m x m1 x2 m y m1 y2
Using the section formula, x 2 1 ,y 2 1 we get
m1 m2 m1 m2
3(8) 1(4) 3(5) 1(3)
x 7, y 3
3 1 3 1
Therefore, (7, 3) is the required point.
In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and
B(3, – 8)?
Solution : Let (– 4, 6) divide AB internally in the ratio k : 1.
m x m1 x2 m y m1 y2
Using the section formula, x 2 1 ,y 2 1 we get
m1 m2 m1 m2
k (8) 1(10)
y 6
k 1
8k 10 6k 6 8k 6k 6 10
4 2
14k 4 k
14 7
Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in
the ratio 2 : 7.
