NCERT10th MathematicsExercise 7.2Topic: 7 Coordinate geometry
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EXERCISE 7.2
Q 1. Find the coordinates of the point which divides the join of
(–1, 7) and (4, –3) in the ratio 2 : 3.
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) The point P(x, y) divides the segment joining the points A(– 1, 7) and B(4, – 3) inthe ratio 2 : 3, so,
1) The point P(x, y) divides the segment joining the points A(– 1, 7) and B(4, – 3) in
the ratio 2 : 3, so,
a) first we will find x-coordinate
x = (mx2 + nx1)/(m + n)
x = ((2) (4) + (3) (– 1))/(2 + 3)x = (8 – 3)/(5)x = (5)/(5)x = 1
b) now we will find y-coordinate
y = (my2 + ny1)/(m + n)
y = ((2) (– 3) + (3) (7))/(2 + 3)
y = (– 6 + 21)/(5)
y = (15)/(5)
y = 3
2) So, the point P(1, 3) divides the segment joining the points
A(– 1, 7) and B(4, – 3) in the ratio 2 : 3.
Q 2. Find the coordinates of the points of trisection of the line segment
joining (4, –1) and (–2, –3).
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) The point P(x1, y1) divides the segment joining the points
1) The point P(x1, y1) divides the segment joining the points
A(4, – 1) and B(– 2, – 3) in the ratio 1 : 2, so,
a) first we will find x-coordinate of point P(x1, y1)
x1 = (mx2 + nx3)/(m + n)
x1 = ((1) (– 2) + (2) (4))/(1 + 2)
x1 = (– 2 + 8)/(3)
x1 = (6)/(3)
x1 = 2
b) now we will find y-coordinate
y1 = (my2 + ny3)/(m + n)
y1 = ((1) (– 3) + (2) (– 1))/(1 + 2)
y1 = (– 3 – 2)/(3)
y1 = (– 5)/(3)
y1 = – 5/3
2) So, the coordinates of the point P(x1, y1) is P(2, – 5/3)
3) The point Q(x2, y2) divides the segment joining the points
A(4, – 1) and B(– 2, – 3) in the ratio 2 : 1, so,
a) first we will find x-coordinate of point Q(x2, y2)
x2 = (mx1 + nx3)/(m + n)
x2 = ((2) (– 2) + (1) (4))/(2 + 1)
x2 = (– 4 + 4)/(3)
x2 = (0)/(3)
x2 = 0
b) now we will find y-coordinate
y2 = (my1 + ny3)/(m + n)
y2 = ((2) (– 3) + (1) (– 1))/(2 + 1)
y2 = (– 6 – 1)/(3)
y2 = (– 7)/(3)
y2 = – 7/3
4) So, the coordinates of the point Q(x2, y2) is P(0, – 7/3).
Q 3. To conduct Sports Day activities, in your rectangular-shaped school
ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in the following fig., Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the distance between both flags? If Rashmi has to post a blue flag halfway between the line segment joining the two flags, where should she post her flag?
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) According to the problem, the total distance of AD is 100 m.2) Niharika covers 1/4th of the distance of AD. i.e. (1/4)(100) = 25 on 2nd line. So
Niharika posts a green flag at the coordinates P(2, 25).
3) In the same way, Preet covers 1/5th of the distance of AD. i.e. (1/5)(100) = 20
on 8th line. So Preet posts a red flag at the coordinates Q(8, 20).
4) so using the distance formula, we can find the distance between two flags as
5) The coordinates of P and Q are P(2, 25) and Q(8, 20), so herefollows:
6) We know that:a) x1 = 2
b) y1 = 25c) x2 = 8
d) y2 = 20
(PQ) = √[(x1 – x2)2 + (y1 – y2)2](PQ) = √[(2 – 8)2 + (25 – 20)2]
(PQ) = √[(– 6)2 + (5)2]
(PQ) = √[36 + 25]
(PQ) = √61
7) The distance between the two flags is √61 m.
8) As Rashmi puts the blue flag in the middle of the green and red flag, i.e.,
R(x, y) = ((x1 + x2)/2, (y1 + y2)/2)R(x, y) = ((2 + 8)/2, (25 + 20)/2)R(x, y) = ((10)/2, (45)/2)
R(x, y) = (5, 22.5)
9) Therefore, Rashmi should post her blue flag at 22.5m on the 5th line.
Q 4. Find the ratio in which the line segment joining the points (– 3, 10)
and (6, – 8) is divided by (– 1, 6).
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) Let the point P(– 1, 6) divides segment A(– 3, 10) B(6, – 8) in the ratio k : 1, so
using section formula, we have,
x = (mx2 + nx1)/(m + n)
(6k – 3)/(k + 1) = – 1
(6k – 3) = – 1(k + 1)
(6k – 3) = – k – 1
(6k + k) = – 1 + 3
7k = 2
k = 2/7
2) The ratio is 2:7.
Q 5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) Let the line segment joining the points A(1, – 5) and B(– 4, 5) get divided by the
x-axis in the ratio k : 1.
2) We know that the y-coordinate of every point on the x-axis is 0, so first we will
find the y-coordinate.
3) Using the section formula, we have,
y = (my2 + ny1)/(m + n)
y = (5k – 5)/(k + 1)
0 = (5k – 5)/(k + 1)
0 (k + 1) = (5k – 5)
(5k – 5) = 0
5k = 5
k = 5/5
k = 1
4) The x-axis divides the line segment joining the points A(1, – 5) and B(– 4, 5) in
1:1 ratio.
5) Now we will find the x-coordinate point of division with the ratio 1:1
Using the section formula, we have,
x = (mx2 + nx1)/(m + n)
x = (x2 + x1)/(1 + 1)
x = (– 4 + 1)/(2)
x = (– 3)/(2)
x = – 3/2
6) So the coordinates of the point of division are (– 3/2, 0).
Q 6. If (1, 2), (4, y), (x, 6), and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) As ABCD is the parallelogram, the diagonals bisect each other.2) Point O is the midpoint of diagonal AC and BD.3) Now we will find the mid-point of AC
Mid-point of AC = ((1 + x)/2, (2 + 6)/2)
Mid-point of AC = ((1 + x)/2, (8)/2)
Mid-point of AC = ((1 + x)/2, 4) --------- equation 1
4) Now we will find the mid-point of BD
Mid-point of BD = ((3 + 4)/2, (5 + y)/2)
Mid-point of BD = ((7)/2, (y + 5)/2)
Mid-point of BD = (7/2, (y + 5)/2) --------- equation 2
5) From equations 1 and 2, we will get the x-coordinate,
(1 + x)/2 = 7/2(1 + x) = 7
x = 7 – 1
x = 6 --------- equation 3
6) From equations 1 and 2, we will get the y-coordinate,
(y + 5)/2 = 4
(y + 5) = 4 (2)
(y + 5) = 8y = 8 – 5
y = 3 --------- equation 4
7) From equations 3 and 4, x = 6 and y = 3.
Q 7. Find the coordinates of point A, where AB is the diameter of a circle whose center is (2, – 3) and B is (1, 4).
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) Let the coordinates of point A be (x, y).2) AB is the diameter of the circle with center O(2, – 3).3) So, point O(2, – 3) is the mid-point of the segment joining the points
A(x, y) and B(1, 4).
4) So, using the mid-point form, we will find the x-coordinate of point A,
(x + 1)/2 = 2
(x + 1) = 2 (2)
(x + 1) = 4
x = 4 – 1
x = 3 --------- equation 1
5) So, using the mid-point form, we will find the y-coordinate of point A,
(y + 4)/2 = – 3
(y + 4) = – 3 (2)
(y + 4) = – 6
y = – 6 – 4
y = – 10 --------- equation 2
6) From equations 1 and 2, x = 3 and y = – 10. So, the coordinates of
A are A(3, – 10).
Q 8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = (3/7) AB and P lies on the line segment AB.
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) Let the coordinates of point P be (x, y).2) Point P divides AB such that, AP = (3/7) AB, so,
(3/7) AB = AP
(AB)/(AP) = 7/3
(AB - AP)/(AP) = (7 - 3)/3
(PB)/(AP) = (4)/3
(AP)/(PB) = 3/4
3) Here, point P(x, y) divides the segment joining point A(-2, -2) and B(2, -4), So,
we now, will find x coordinate
x = (3 (2) + 4 (- 2))/(3 + 4)x = (6 - 8)/(7)x = (- 2)/(7)x = - 2/7 --------- equation 1
we now, will find y coordinate
y = (3 (- 4) + 4 (- 2))/(3 + 4)
y = (- 12 - 8)/(7)
y = (- 20)/(7)
y = - 20/7 --------- equation 2
4) From equations 1 and 2, x = - 2/7 and y = - 20/7. So, the coordinates of
P are P(- 2/7, - 20/7).
Q 9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) The point Q(x2, y2) the mid-point of the segment joining the points
A(- 2, 2), B(2, 8), so the coordinates of point Q will be:
Q(x2, y2) = ((- 2 + 2)/2, (2 + 8)/2)
Q(x2, y2) = ((0)/2, (10)/2)
Q(x2, y2) = (0, 5) -------- equation 1
2) The point P(x1, y1) the mid-point of the segment joining the points
A(- 2, 2), Q(0, 5), so the coordinates of point Q will be:
P(x1, y1) = ((- 2 + 0)/2, (2 + 5)/2)
P(x1, y1) = ((- 2)/2, (7)/2)
P(x1, y1) = (- 1, 7/2) -------- equation 2
3) The point R(x3, y3) the mid-point of the segment joining the points
Q(0, 5), B(2, 8), so the coordinates of point Q will be:
R(x3, y3) = ((0 + 2)/2, (5 + 8)/2)R(x3, y3) = ((2)/2, (13)/2)R(x3, y3) = (1, 13/2) -------- equation 3
4) From equations 1, 2, and 3, the coordinates of points P, Q, and R are as follows.
P(x1, y1) = P(- 1, 7/2)
Q(x2, y2) = Q(0, 5)
R(x3, y3) = R(1, 13/2).
Q 10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4),
and (– 2, – 1) taken in order.
[Hint: Area of a rhombus = (1/2) (product of its diagonals)].
Solution:
1) We know that the area of the rhombus = (1/2) (product of diagonals).2) So first we will find AC and BD using the distance formula.3) First we will find AC with A(3, 0), C(- 1, 4)a) x1 = 3
b) y1 = 0
c) x2 = - 1
d) y2 = 4
4) We know that:(AC) = √[(x1 – x2)2 + (y1 – y2)2]
(AC) = √[(3 – (– 1))2 + (0 – 4)2]
(AC) = √[(4)2 + (– 4)2]
(AC) = √[16 + 16]
(AC) = 4√2 ------------- equation 1
5) First we will find BD with B(4, 5), C(- 2, - 1)a) x1 = 4
b) y1 = 5
c) x2 = - 2
d) y2 = - 1
6) We know that:(BD) = √[(x1 – x2)2 + (y1 – y2)2]
(BD) = √[(4 – (– 2))2 + (5 – (– 1))2]
(BD) = √[(6)2 + (6)2]
(BD) = √[36 + 36]
(BD) = 6√2 ------------- equation 2
7) From equations 1 and 2, we can find the area of the rhombus as follows.Area of the rhombus = (1/2) (product of diagonals)
1) We know that the area of the rhombus = (1/2) (product of diagonals).
2) So first we will find AC and BD using the distance formula.
3) First we will find AC with A(3, 0), C(- 1, 4)
a) x1 = 3
b) y1 = 0
c) x2 = - 1
d) y2 = 4
4) We know that:
(AC) = √[(x1 – x2)2 + (y1 – y2)2]
(AC) = √[(3 – (– 1))2 + (0 – 4)2]
(AC) = √[(4)2 + (– 4)2]
(AC) = √[16 + 16]
(AC) = 4√2 ------------- equation 1
5) First we will find BD with B(4, 5), C(- 2, - 1)
a) x1 = 4
b) y1 = 5
c) x2 = - 2
d) y2 = - 1
6) We know that:
(BD) = √[(x1 – x2)2 + (y1 – y2)2]
(BD) = √[(4 – (– 2))2 + (5 – (– 1))2]
(BD) = √[(6)2 + (6)2]
(BD) = √[36 + 36]
(BD) = 6√2 ------------- equation 2
7) From equations 1 and 2, we can find the area of the rhombus as follows.
Area of the rhombus = (1/2) (product of diagonals)
Area of the rhombus = (1/2) (4√2) (6√2)
Area of the rhombus = (2√2) (6√2)
Area of the rhombus = (2) (6) (√2) (√2)
Area of the rhombus = (12) (2)
Area of the rhombus = 24 square units.
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