Coordinate Geometry

Class 10 NCERT

NCERT

1   Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (- a, - b)

Solution :

(i) Distance between the two points is given by$\\$ $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$\\$ therefore distance between (2,3) and(4,1) is given by$\\$ l = $\sqrt{(2 - 4)^2 + (3 - 1)^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{8} = 2\sqrt{2}$$\\$ (ii) Distance between (-5,7) and (-1,3) is given by$\\$ l = $\sqrt{(-5-(-1))^2 + (7 - 3)^2} = \sqrt{(16 + 16)} = \sqrt{32} = 4\sqrt{2}$$\\$ (iii) Distance between(a,b) and (-a,-b) is given by$\\$ l = $\sqrt{(a-(-a))^2 + (b - (-b))^2}$$\\$ = $\sqrt{(2a)^2 + (2b)^2} = \sqrt{(4a^2) + (4b^2)} = 2\sqrt{(a^2) + (b^2)}$$\\$

2   Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2. Answer:

Solution :

Distance between points(0,0) and(36,15)$\\$ $\sqrt{(36-0)^2 + (15-0)^2} = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39$$\\$ Yes, we can find the distance between the given towns A and B.$\\$ Assume town A at origin point (0, 0).$\\$ Therefore, town B will be at point (36, 15) with respect to town A.$\\$ And hence, as calculated above, the distance between town A and B will be 39 km.

3   Determine if the points (1, 5), (2, 3) and (- 2, -11) are collinear.

Solution :

Let the points (1, 5), (2, 3), and (-2, -11) be representing the vertices A, B, and C of the given triangle respectively.$\\$ Let A = (1,5) B =(2,3) C = (-2,-11)$\\$ therefore AB = $\sqrt{(1-2)^2 + (5-3)^2} = \sqrt{5}$$\\$ BC = $\sqrt{(2-(-2) )^2 + (3-(-11))^2} = \sqrt{(4^2 + (14)^2} = \sqrt{16+196} = \sqrt{212} $$\\$ CA = $\sqrt{(1-(-2) )^2 + (5-(-11))^2} = \sqrt{(3^2 + (16)^2} = \sqrt{9+256} = \sqrt{265} $$\\$ since AB +BC$\neq CA,$$\\$ Therefore, the points (1, 5), (2, 3), and (-2, -11) are not collinear.

4   Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.

Solution :

Let the points (5, -2), (6, 4), and (7, -2) are representing the vertices A, B, and C of the given triangle respectively.$\\$ AB= $\sqrt{(5-6) )^2 + (-2-4)^2} = \sqrt{(-1 )^2 + (-6)^2} = \sqrt{(1+ 36)} =\sqrt{37}$$\\$ BC= $\sqrt{(6-7) )^2 + (4-(-2))^2} = \sqrt{(-1 )^2 + (6)^2} = BC = \sqrt{(1+ 36)} =\sqrt{37}$$\\$ CA= $\sqrt{(5-7) )^2 + (-2-(-2))^2} = \sqrt{(-2 )^2 - (-2)^2} =\sqrt{((-2)^2 + 0^2)}$ = 2 $\\$ therefore AB = BC$\\$ As two sides are equal in length, therefore, ABC is an isosceles triangle.

5   In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees.$\\$ Using distance formula, find which of them is correct.

Solution :

It can be observed that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of these 4 friends.$\\$ AB = $\sqrt{(3-6 )^2 + (4-7)^2} = \sqrt{((-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$$\\$ BC = $\sqrt{(6-9 )^2 + (7-4)^2} = \sqrt{((-3)^2 + (3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$$\\$ CB= $\sqrt{(9-6 )^2 + (4-1)^2} = \sqrt{((3)^2 + (3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$$\\$ AD = $\sqrt{(3-6 )^2 + (4-1)^2} = \sqrt{((-3)^2 + (3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$$\\$ diagonal AC = $\sqrt{(3-9)^2 + (4-4)^2} = \sqrt{((-6)^2 + (0)^2} = 6$$\\$ diagonal BD = $\sqrt{(6-6)^2 + (7-1)^2} = \sqrt{((0)^2 + (6)^2} = 6$$\\$

6   Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (-1, - 2), (1, 0), (- 1, 2), (- 3, 0) (ii) (- 3, 5), (3, 1), (0, 3), (- 1, - 4) (iii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution :

(i) Let the points (-1, -2), (1, 0), (-1, 2), and (-3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.$\\$ AB = $\sqrt{(-1-1 )^2 + (-2-0)^2} = \sqrt{((-2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$$\\$ BC= $\sqrt{(-1(-1) )^2 + (0-2)^2} = \sqrt{((2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$$\\$ CD = $\sqrt{(-1-(-3) )^2 + (2-0)^2} = \sqrt{((2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$$\\$ AD = $\sqrt{(-1-(-3))^2 + (-2-0)^2} = \sqrt{((2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$$\\$ DIAGNOL AC = $\sqrt{(-1-(-1))^2 + (-2-2)^2} = \sqrt{(0^2 + (-4)^2} = \sqrt{16} = 4$$\\$ DIAGNOL BD = $\sqrt{(-1-(-3))^2 + (0-0)^2} = \sqrt{(4^2 + 0^2} = \sqrt{16} = 4$$\\$

(iii)Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.$\\$ AB = $\sqrt{(4-7 )^2 + (5-6)^2} = \sqrt{((-3)^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10} $$\\$ BC= $\sqrt{(7-4)^2 + (6-3)^2} = \sqrt{((3)^2 + (3)^2} = \sqrt{9+9} = \sqrt{18}$$\\$ CD = $\sqrt{(4-1 )^2 + (3-2)^2} = \sqrt{(3)^2 + (1)^2} = \sqrt{9+1} = \sqrt{10} $ $\\$ AD = $\sqrt{(4-1)^2 + (5-2)^2} = \sqrt{(3)^2 + (3)^2} = \sqrt{9+9} = \sqrt{18}$$\\$ DIAGNOL AC = $\sqrt{(4-4)^2 + (5-3)^2} = \sqrt{(0^2 + (2)^2} = \sqrt{4} = 2$$\\$ DIAGNOL BD = $\sqrt{(7-1)^2 + (6-2)^2} = \sqrt{(6^2 + 4^2)} = \sqrt{36+16} = 13\sqrt{2}$$\\$ It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.$\\$ (ii)Let the points (-3, 5), (3, 1), (0, 3), and (-1, -4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.$\\$ AB = $\sqrt{(-3-3 )^2 + (5-1)^2} = \sqrt{((-6)^2 + (4)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$$\\$ BC= $\sqrt{(3-0)^2 + (1-3)^2} = \sqrt{((3)^2 + (-2)^2} = \sqrt{9+4} = \sqrt{13}$$\\$ CD = $\sqrt{(0-(-1) )^2 + (3-(-4))^2} = \sqrt{((1)^2 + (7)^2} = \sqrt{1+49} = \sqrt{50} = 5\sqrt{2}$$\\$ AD = $\sqrt{(-3-(-1))^2 + (5-(-4)^2} = \sqrt{((-2)^2 + (9)^2} = \sqrt{4+81} = \sqrt{85}$$\\$ It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.$\\$

7   Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9)

Solution :

We have to find a point on x-axis. Therefore, its y-coordinate will be 0.$\\$ Let the point on x-axis be(x,0)$\\$ Distance between(x,0) and (2,-5) = $\sqrt{(x-2 )^2 + (0-(-5))^2} = \sqrt{((x-2)^2 + (5)^2}$$\\$ Distance between(x,0) and (-2,9) = $\sqrt{(x-(-2) )^2 + (0-(-9))^2} = \sqrt{((x+2)^2 + (9)^2}$$\\$ By the given condition, these distances are equal in measure.$\\$ $\sqrt{(x-2 )^2 + (5)^2} = \sqrt{((x+2)^2 + (9)^2}$$\\$ $(x-2)^2 + 25 = (x+ 2)^2 +81$$\\$ $x^2+4-4x+25=x^2+4+4x+81$$\\$ 8x = 25-81$\\$ 8x=-56$\\$ x = -7$\\$ Therefore, the point is (- 7, 0).

8   Find the values of y for which the distance between the points P (2, - 3) and Q (10, y) is 10 units.

Solution :

It is given that the distance between (2, -3) and (10, y) is 10.$\\$ $\sqrt{(2-10 )^2 + (-3-Y)^2} = 10$$\\$ $\sqrt{(-8 )^2 + (3+Y)^2} = 10$$\\$ $64+(Y+3)^2 = 100$$\\$ $(Y+3)^2 = 36$$\\$ $Y+3=\pm6$$\\$ y+3=6 or y+3=-6$\\$ therefore y= 3 or -9$\\$

9   If Q (0, 1) is equidistant from P (5, -3) and R (x, 6), find the values of x. Also find the distance QR and PR.

Solution :

PQ = QR$\\$ $\sqrt{(5-0)^2 + (-3-1)^2} = \sqrt{(5 )^2 + (-4)^2} = \sqrt{25 + 16} = 41$$\\$ $\sqrt{(0-x )^2 + (1-6)^2} = \sqrt{(-x )^2 + (-5)^2} = \sqrt{(x )^2 + 25}$$\\$ then $x = \pm4$$\\$ Therefore, point R is (4, 6) or (-4, 6). Page$\\$ When point R is (4, 6),$\\$ PR = $\sqrt{(5-4 )^2 + (-3-6)^2} = \sqrt{((1)^2 + (-9)^2} = \sqrt{1+81} = \sqrt{82}$$\\$ QR = $\sqrt{(0-4 )^2 + (1-6)^2} = \sqrt{((-4)^2 + (-5)^2} = \sqrt{16+25} = \sqrt{41}$$\\$ When point R is (-4, 6), QR = $\sqrt{(0-(-4 ))^2 + (1-6)^2} = \sqrt{((4)^2 + (-5)^2} = \sqrt{16+25} = \sqrt{41}$$\\$

10   Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).

Solution :

Point (x, y) is equidistant from (3, 6) and (-3, 4).$\\$ $\sqrt{(x-3 )^2 + (y-6)^2} = \sqrt{(x-3 )^2 + (y-6)^2} = (x-3)^2 + (y-6)^2 = x^2+9-6x+y^2+36-12y = 36-16=20$$\\$ $\sqrt{(x-(-3) )^2 + (y-4)^2} = \sqrt{(x+3 )^2 + (y-4)^2} = (x+3)^2 + (y-4)^2 = x^2+9+6x+y^2+16-8y = 6x+6x+12y-8y=12x+4y=5$$\\$ 20=12x+4y$\\$ 3x+y=5$\\$ 3x+y-5=0$\\$

11   Find the coordinates of the point which divides the join of (- 1, 7) and (4, - 3) in the ratio 2:3.

Solution :

Let P(x, y) be the required point. Using the section formula, we obtain$\\$ x = $\frac{2\times4+3\times(-1)}{2+3} = \frac{8-3}{5} = \frac{5}{5} = 1$$\\$ y = $\frac{2\times(-3)+3\times7}{2+3} = \frac{-6+21}{5} = \frac{15}{5} = 3$$\\$ Therefore, the point is (1, 3).$\\$

12   Find the coordinates of the points of trisection of the line segment joining (4, - 1) and (-2, - 3).

Solution :

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points $\\$i.e., AP = PQ = QB$\\$ Therefore, point P divides AB internally in the ratio 1:2.$\\$ $x_1 = \frac{1\times(-2)+2\times4}{1+2}$$\\$ $y_1 = \frac{1\times(-3)+2\times(-1)}{1+2}$$\\$ $x_1 = \frac{-2+8}{3}=\frac{6}{3} = 2$$\\$ $y_1 = \frac{-3-2}{3} = \frac{-5}{3}$$\\$ therefore $p(x_1,y_1) = (2,-\frac{5}{3})$$\\$ Point Q divides AB internally in the ratio 2:1.$\\$ $\frac{x_2} = \frac{2\times(-2)+1\times}{2+1},$$\\$ $\frac{x_2} = \frac{-4+4}{3}, = 0$$\\$ $\frac{y_2} = \frac{2\times(-3)+1\times(-1)}{2+1},$$\\$ $\frac{y_2} = \frac{-6-1}{3}, = \frac{-7}{3}$$\\$ $Q(x_2,y_2) = (0 - \frac{7}{3})$$\\$

13   To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs$\frac{1}{4}th$ the distance AD on the 2nd line and posts a green flag. Preet runs$\frac{1}{5}th$ the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution :

It can be observed that Niharika posted the green flag at$\frac{1}{4}th$ of the distance AD i.e.,$\\$ $(\frac{1}{4}\times100)m = 25 m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25).$$\\$ Similarly, Preet posted red flag at$\frac{1}{5}$ = 20 of the distance AD i.e.,$(\frac{1}{5}\times100)$$\\$ m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20). Distance between these flags by using distance formula = GR =$\sqrt{(8-2)^2+(25-20)^2} = \sqrt{36+25} = \sqrt{61}$$\\$ The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y).$\\$ $x = \frac{2+8}{2},y = \frac{25+20}{2}$$\\$ x = $\frac{10}{2} = 5,y = \frac{45}{2}=22.5$$\\$ Hence A(x,y) = (5,22.5)$\\$ Therefore, Rashmi should post her blue flag at 22.5m on 5th line

14   Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6).

Solution :

Let the ratio in which the line segment joining (-3, 10) and (6, -8) is divided by point (-1, 6) be k : 1. therefore -1 = $\frac{6k-3}{k+1}$$\\$ -k-1 = 6k-3$\\$ 7k = 2$\\$ k = $\frac{2}{7}$$\\$ therefore the required ratio is 2:7

15   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. Answer:

Solution :

Let the ratio in which the line segment joining A (1, -5) and B (-4, 5) is divided by x-axisbe k:1$\\$ . Therefore, the coordinates of the point of division is$\big(\frac{-4K+1}{k+1},{5K-5}{K+1}\big)$$\\$ . We know that y-coordinate of any point on x-axis is 0. $\\$ therefore$\frac{5k-5}{k+1} = 0$$\\$ k = 1 k = 1 Therefore, x-axis divides it in the ratio 1:1. Division point $\\$ =$\big(\frac{-4(1)+1}{1+1}{5(1)-5}{1+1}\big) = \big(\frac{-4+1}{2}{5-5}{2}\big) = (\frac{-3}{2})$$\\$

16   If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution :

Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals.$\\$ Therefore, O is the mid-point of AC and BD.$\\$ If O is the mid-point of AC, then the coordinates of O are$\\$ $(\frac{1+x}{2})$$\\$ then $(\frac{x+1}{2}4)$$\\$ If O is the mid-point of BD, then the coordinates of O are$\\$ $\big(\frac{4+3}{2},{5+y}{2}\big)$$\\$ Since both the coordinates are of the same point O, therefore $\frac{x+1}{2} = \frac{7}{2} and 4=\frac{5+y}{2}$$\\$ x+1=7 and 5+y = 8$\\$ x= 6 and y = 3

17   Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, -3) and B is (1, 4)

Solution :

Let the coordinates of point A be (x, y).$\\$ Mid-point of AB is (2, -3), which is the center of the circle.$\\$ (2,-3) = $\big(\frac{x+1}{2},{y+4}{2}\big)$$\\$ $\frac{x+1}{2} = 2 and \frac{y+4}{2} = -3$$\\$ x+1 = 4 and y+4=-6$\\$ x=3 and y=-1$\\$ therefore the co ordinates of A are (3,-10)$\\$

18   If A and B are (- 2, - 2) and (2, - 4), respectively, find the coordinates of P such that and P lies on the line segment AB.

Solution :

The coordinates of point A and B are (-2, -2) and (2, -4) respectively.$\\$ Since AP = $\frac{3}{7} AB$$\\$ ,COORDINATESof p = $\big(\frac{3\times2+4(-2)}{3+4},{3\times(-4)+4\times(-2)}{3+4}\big)$$\\$ $\big(\frac{6-8}{7},{-12-8}{7}\big)$$\\$ $\big(\frac{-2}{7}, - {20}{7}\big)$$\\$ Therefore, AP: PB = 3:4 Point P divides the line segment AB in the ratio 3:4.

19   Find the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts.

Solution :

From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.$\\$ Coordinates of p = $\big(\frac{1\times2+3\times(-2)}{1+3},{1\times(8)+3\times(2)}{1+3}\big)$$\\$ =$\big(-1\frac{7}{2}\big)$$\\$ Coordinates of Q = $\big(\frac{2+(-2)}{2},{2+8}{2}\big)$$\\$ =(0.5)$\\$ Coordinates of R = $\big(\frac{3\times2+1\times(-2)}{3+1},{3\times8+1\times2}{3+1}\big)$$\\$ =$(1,\frac{13}{2})$$\\$

20   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 =$\frac{1}{2}$ (product of its diagonals)]

Solution :

Let (3, 0), (4, 5), (-1, 4) and (-2, -1) are the vertices A, B, C, D of a rhombus ABCD.$\\$ length of diagonal AC =$\sqrt{[3+(-1)]^2 +(0-4)^2}$$\\$ $=\sqrt{16+16} =4\sqrt{2}$$\\$ length of diagonal BD =$\sqrt{[4-(-2)]^2 +[5-(-1)]^2}$$\\$ $=\sqrt{36+36} =6\sqrt{2}$$\\$ therefore area of rhoombus ABCD = $\frac{1}{2}\times4\sqrt{2}\times6\sqrt{2}$$\\$ =24 square units.

21   Find the area of the triangle whose vertices are: (i) (2, 3), (- 1, 0), (2, - 4) (ii) (- 5, - 1), (3, - 5), (5, 2)

Solution :

(i) Area of a triangle is given by$\\$ area of triangle = $\frac{1}{2}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}$$\\$ area of the given triangle = $\frac{1}{2}[2{0-(-4)}+(-1){(-4)-(-3)}+(3-0)]$$\\$ =$\frac{1}{2}{8+7+6}$$\\$ $\frac{21}{2}$ square units$\\$ area of given triangle = $\frac{1}{2}[(-5){(-5)-(2)}+3(2-(-1))+5{-1-(-5)}]$$\\$ 3) $\frac{1}{2}{35+9+20}$$\\$ =32 square units$\\$

22   In each of the following find the value of ‘k’, for which the points are collinear.$\\$ (i) (7, - 2), (5, 1), (3, - k) (ii) (8, 1), (k, - 4), (2, - 5)

Solution :

(i) For collinear points, area of triangle formed by them is zero. $\\$ Therefore, for points (7, -2) (5, 1), and (3, k), area = 0$\\$ $\frac{1}{2}[7{1-k}+5{k-(-2)}+3{(-2)-1}]=0$$\\$ 7-7k+5k+10-9=0$\\$ -2k+8=0$\\$ k = 4$\\$ (ii) For collinear points, area of triangle formed by them is zero$\\$ . Therefore, for points (8, 1), (k, -4), and (2, -5), area = 0$\\$ $\frac{1}{2}[8{-4-(-5)}+k{(-5)-(1)}+2{1-(-4}]=0$$\\$ 8-6k+10=0$\\$ 6k=18$\\$ k = 3

23   Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Solution :

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).$\\$ Let D, E, F be the mid-points of the sides of this triangle.$\\$ Coordinates of D, E, and F are given by D = $\big(\frac{0+2}{2},{-1+1}{2}\big) =(1,0)$$\\$ E = $\big(\frac{0+0}{2},{3-1}{2}\big) =(0,1)$$\\$ F = $\big(\frac{2+0}{2},{1+3}{2}\big) =(1,2)$$\\$ Area of triangle = $\frac{1}{2}{x_1(y_2 - y_3) +x_2(y_3 - y_1) +x_3(y_1 - y_2) }$$\\$ area of triangle DEF = $\frac{1}{2}{1(2 - 1) +1(1--0) +0(0- 2) }$$\\$ area of triangle ABC = $\frac{1}{2}[0(1-3) +2{3-(-1)} +0(-1-1) ]$$\\$ = $\frac{1}{2}{8} = 4$ square units$\\$ therefore required atio = 1:4

24   Find the area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3)

Solution :

Let the vertices of the quadrilateral be A (-4, -2), B (-3, -5), C (3, -2), and D (2, 3). Join AC to form two triangles triangle ABC and ACD.$\\$ Area of triangle = $\frac{1}{2}{x_1(y_2 - y_3) +x_2(y_3 - y_1) +x_3(y_1 - y_2) }$$\\$ Area of triangle ABC = $\frac{1}{2}{(-4){(-5)-(-2)} +(-3){(-2)- (-20)} +3{(-2) - (-5)}] }$$\\$ = $\frac{1}{2}(12+0+9) = \frac{21}{2}$ square units$\\$ Area of triangle ACD = $\frac{1}{2}[(-4){(-2)-(-3)} +3{(3)-(-2)}+2{(-2) - (-5)}] $$\\$ = $\frac{1}{2}{20+15+0} = \frac{35}{2} $ square units.$\\$ area of ABCD = area of triangle ABC + AREA of TRIANGLE ACD$\\$ = $\frac{21}{2} + \frac{35}{2} $square units = 28 square units$\\$

25   You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for triangle ABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2)

Solution :

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2). Let D be the mid-point of side BC of triangle ABC. Therefore, AD is the median in triangle ABC. coordinates of point D=$(\frac{3+5}{2},\frac{-2+2}{2}) = (4,0)$$\\$ area of triangle = $\frac{1}{2}{x_1(y_2 - y_3) +x_2(y_3 - y_1) +x_3(y_1 - y_2) }$$\\$ Area of triangle ABCD = $\frac{1}{2}{(4){(-2)-(0)} +(3){(0)- (-6)} +4{(-6) - (-2)}] }$$\\$ =$\frac{1}{2}(-8+18-16) = -3 square units$\\$ However, area cannot be negative. Therefore, area of triangle ADC is 3 square units. Clearly, median AD has divided triangle ABC in two triangles of equal areas.

26   Determine the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A(2, - 2) and B(3, 7)

Solution :

27   Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Solution :

If the given points are collinear, then the area of triangle formed by these points will be 0.$\\$ Area of triangle = $\frac{1}{2}{x_1(y_2 - y_3) +x_2(y_3 - y_1) +x_3(y_1 - y_2) }$$\\$ Area = $\frac{1}{2}{(x){2-0} +(1){0-y} +7{y-2}] }$$\\$ 0 = $\frac{1}{2} [ 2x-y+7y-14]$$\\$ 2x +6y-14=0$\\$ x+3y-7=0$\\$ This is the required relation between x and y.

28   Find the centre of a circle passing through the points (6, - 6), (3, - 7) and (3, 3).

Solution :

Let O (x, y) be the centre of the circle. And let the points (6, -6), (3, -7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.$\\$

29   The two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$. Find the coordinates of the other two vertices.

Solution :

From the Figure,$\\$ Given,$\\$ $\ \ \ \ \ \bullet $ Let $ABCD$ be a square having known vertices$ A (-1, 2) $ and $C (3, 2)$ as vertices $A $and $C$ respectively.$\\$ $\ \ \ \ \ \bullet$ Let $B(x, y)$ be one unknown vertex$\\$ We know that the sides of a square are equal to each other.$\\$ $ \therefore AB = BC$$\\$ By Using Distance formula to find distance between points $AB \& BC,$ $\\$ $\implies \sqrt{\left(x-1\right)^2+\left(y-2\right)^2}\\ \implies \sqrt{\left(x-3\right)^2 + \left(y-2\right)^2} \\ \implies x^2+2x+1+y^2 -4y+4 = x^2 +9-6x+y^2+4-4y $(By Simplifying & Transposing)$\\$ $ \implies 8x =8\\ \implies x=1 $ $\\$ We know that in a square, all interior angles are of $90^o$$\\$ $In \Delta ABC,$$\\$ $AB^2 + BC^2 +AC^2$ [ByPythagoras theorem]$\\$ Distance formula is used to find distance between $AB, BC $and $AC$$\\$ $\implies \left(\sqrt{\left(1+1\right)^2 + \left(y-2 \right)^2}\right)^2 + \left(\sqrt{\left(1-3 \right)^2 + \left(y-2 \right)^2 }\right)^2 $ $\\$$ = \left(\sqrt{\left(3+1\right)^2 +\left(2-2 \right)^2}\right)^2\\ \implies 4+y^2 + 4-4y +4 +y^2 -4y+4 =16\\ \implies 2y^2+16-8y =16\\ \implies y \left(y-4 \right )=0 \\ \implies y = 0 or 4 $ $\\$ Hence the required vertices are$B (1, 0)$ an $D (1, 4)$

30   The class $X$ students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of $1 m$ from each other. There is a triangular grassy lawn in the plot as shown in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.$\\$ (i) Taking A as origin, find the coordinates of the vertices of the triangle.$\\$ (ii) What will be the coordinates of the vertices of $M$ $PQR$ if $C$ is the origin?$\\$ Also calculate the areas of the triangles in these cases. What do you observe?

Solution :

(i)Given,$\\$ $\ \ \ \ \ \bullet$ Taking $A$ as origin, we will take $AD$ as $x-axis$ and $AB$ as $y- axis.$$\\$ $\ \ \ \ \ \bullet$ It can be observed that the coordinates of point $P, Q,$ and $R$ are $(4, 6), (3, 2),$ and $(6, 5)$ respectively.$\\$ $\bullet$ Let $P(x_1, y_1 ) = (4, 6)$$\\$ $\bullet$ Let $Q(x_2, y_2 ) = (3 , 2)$$\\$ $\bullet$ Let $R(x_3, y_3 ) = (6, 5)$$\\$ Area of a triangle =$ \dfrac{1}{2}x_1 \left(y_2-y_3 \right)+x_2 \left(y_3 -y_1 \right)+ x_3 \left(y_1-y_2\right ) \dots $ equation (1) $\\$ By substituting the values of vertices $P, Q, R$ in the Equation (1),$\\$ Area of triangle $ PQR= \dfrac{1}{2}[x_1(y_2-y_3) +x_2(y_3-y_1)+ x_3(y_1-y_2)]$$\\$ $ =\dfrac{1}{2}[4(2-5) +3(5-6)+6(6-2)] \\ =\dfrac{1}{2}\left[-12-3+24\right]\\ =\dfrac{9}{2}$ Square units $\\$ (ii)Given,$\\$ $\ \ \ \ \ \bullet$ Taking $C $as origin, $CB$ as $x-axis$, and $CD$ as $y-axis$$\\$ $\ \ \ \ \ \bullet$ The coordinates of vertices $P, Q,$ and $R$ are $(12, 2), (13, 6),$ and $(10, 3)$ respectively.$\\$ $\ \ \ \ \ \bullet$ Let $P(x_1, y_1 ) = (12, 2)$$\\$ $\ \ \ \ \ \bullet$ Let $Q(x_2, y_2 ) = (13, 6)$$\\$ $\ \ \ \ \ \bullet$ Let $R(x_3, y_3 ) = (10,3)$$\\$ Area of a triangle = $\dfrac{1}{2}x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \dots $Equation (1)$\\$ By substituting the values of vertices $P, Q, R$ in the Equation (1),$\\$ Area of triangle $PQR =\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+ x_3(y_1-y_2)] $$\\$ $= \dfrac{1}{2}[12(6-3)+13(3-2)+10(2-6)]\\ = \dfrac{1}{2}[36-13+40] \\ = \dfrac{9}{2}$ Square units $\\$ It can be observed that the area of the triangle is same in both the cases.

(ii)Given,$\\$ $\ \ \ \ \ \bullet$ Taking $C $as origin, $CB$ as $x-axis$, and $CD$ as $y-axis$$\\$ $\ \ \ \ \ \bullet$ The coordinates of vertices $P, Q,$ and $R$ are $(12, 2), (13, 6),$ and $(10, 3)$ respectively.$\\$ $\ \ \ \ \ \bullet$ Let $P(x_1, y_1 ) = (12, 2)$$\\$ $\ \ \ \ \ \bullet$ Let $Q(x_2, y_2 ) = (13, 6)$$\\$ $\ \ \ \ \ \bullet$ Let $R(x_3, y_3 ) = (10,3)$$\\$ Area of a triangle = $\dfrac{1}{2}x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \dots $Equation (1)$\\$ By substituting the values of vertices $P, Q, R$ in the Equation (1),$\\$ Area of triangle $PQR =\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+ x_3(y_1-y_2)] $$\\$ $= \dfrac{1}{2}[12(6-3)+13(3-2)+10(2-6)]\\ = \dfrac{1}{2}[36-13+40] \\ = \dfrac{9}{2}$ Square units $\\$ It can be observed that the area of the triangle is same in both the cases.

31   The vertices of a $\Delta ABC$ are$ A (4, 6), B (1, 5)$ and $C (7, 2)$. A line is drawn to intersect sides $AB$ and $AC$ at $D$ and $E$ respectively, such that $\dfrac{AD}{AB } =\dfrac{AE}{AC} =\dfrac{1}{4}$. Calculate the area of the $\Delta ADE$ and compare it with the area of $MABC$. (Recall Converse of basic proportionality theorem and Theorem 6.6 related to Ratio of areas of two similar triangles)

Solution :

From the figure,$\\$ Given $ \dfrac{AD}{AB}=\dfrac{AE}{AC} =\dfrac{1}{4}$$\\$ Therefore, $D$ and $E$ are two points on side $AB$ and $AC$ respectively such that they divide side $AB$ and $AC$ in a ratio of $1\colon 3$.$\\$ By Section formula $\\$ $ P(x,y) =\left[\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_2}{m+n}\right] \dots (1)$$\\$ Coordinates of Point D=$\left(\dfrac{1*1+3*4}{1+3},\dfrac{1*5+3*6}{1+3} \right) \\ =\left(\dfrac{13}{4},\dfrac{23}{4}\right) $$\\$ Coordinates of point E = $\left(\dfrac{1*7+3*4}{1+3},\dfrac{1*2+3*6}{1+3}\right) \\ =\left(\dfrac{19}{4},\dfrac{20}{4}\right) $$\\$ Area of triangle =$\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] \dots (2)$$\\$ By substituting the vertices $A, D, E$ in $(2)$,$\\$ Area of $\Delta ADE = \dfrac{1}{2}\left[ 4 \left(\dfrac{23}{4}\right)- \left(\dfrac{20}{4}\right)+ \dfrac{13}{4}\left(\dfrac{20}{4} -6 \right) \dfrac{19}{4}\left(6-\dfrac{23}{4}\right) \right] $$\\$ $ =\dfrac{1}{2}\left[3-\dfrac{13}{4}+\dfrac{19}{16}\right ] = \dfrac{15}{32} $Square units$\\$ By substituting the vertices $A, B, C$ in $(2)$$\\$ Area of $\Delta ABC=\dfrac{1}{2}[4(5-2)+(2-6)+7(6-5)]$$\\$ $ = \dfrac{1}{2}[12-4+7] = \dfrac{15}{2}$Square units $\\$ Clearly, the ratio between the areas of $\Delta ADE$ and $\Delta ABC$ is $1\colon 16$.$\\$ Alternatively, we know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle. These two triangles so formed (here $\Delta ADE$ and $\Delta ABC$) will be similar to each other. Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles.$\\$ Therefore, ratio between the areas of $\Delta ADE $and $\Delta ABC$ =$ \left(\dfrac{1}{4}\right)^2 =\dfrac{1}{16}$

Area of triangle =$\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] \dots (2)$$\\$ By substituting the vertices $A, D, E$ in $(2)$,$\\$ Area of $\Delta ADE = \dfrac{1}{2}\left[ 4 \left(\dfrac{23}{4}\right)- \left(\dfrac{20}{4}\right)+ \dfrac{13}{4}\left(\dfrac{20}{4} -6 \right) \dfrac{19}{4}\left(6-\dfrac{23}{4}\right) \right] $$\\$ $ =\dfrac{1}{2}\left[3-\dfrac{13}{4}+\dfrac{19}{16}\right ] = \dfrac{15}{32} $Square units$\\$ By substituting the vertices $A, B, C$ in $(2)$$\\$ Area of $\Delta ABC=\dfrac{1}{2}[4(5-2)+(2-6)+7(6-5)]$$\\$ $ = \dfrac{1}{2}[12-4+7] = \dfrac{15}{2}$Square units $\\$ Clearly, the ratio between the areas of $\Delta ADE$ and $\Delta ABC$ is $1\colon 16$.$\\$ Alternatively, we know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle. These two triangles so formed (here $\Delta ADE$ and $\Delta ABC$) will be similar to each other. Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles.$\\$ Therefore, ratio between the areas of $\Delta ADE $and $\Delta ABC$ =$ \left(\dfrac{1}{4}\right)^2 =\dfrac{1}{16}$

32   Let $ A (4, 2), B (6, 5)$ and $C (1, 4)$ be the vertices of $\Delta ABC$.$\\$ (i) The median from $A$ meets $BC$ at $D$. Find the coordinates of point $D$.$\\$ (ii) Find the coordinates of the point $P$ on $AD$ such that $AP\colon PD = 2\colon 1$$\\$ (iii) Find the coordinates of point $Q$ and $R$ on medians $BE$ and $CF$ respectively such that $BQ\colon QE = 2\colon 1$ and $CR\colon RF = 2\colon 1$.$\\$ (iv) What do you observe? (v) If $A(x_1 , y_1 ), B(x_2 , y_2 )$, and $C(x_3 , y_3 )$ are the vertices of $\Delta ABC$, find the coordinates of the centroid of the triangle.$\\$

Solution :

(iii) From the figure,$\\$ Median $BE$ of the triangle will divide the side $AC$ in two equal parts. Therefore,$E$ is the mid-point of side $AC$.$\\$ Coordinates of $E = \left(\dfrac{4+1}{2},\dfrac{2+4}{2}\right) =\left(\dfrac{5}{2},3\right)$$\\$ Point $Q$ divides the side $BE$ in a ratio $2\colon 1$.$\\$ Coordinates of $Q=\left(\dfrac{2*\dfrac{5}{2}+1*6}{2+1} ,\dfrac{2*3+1*5}{2+1}\right) =\left(\dfrac{11}{3},\dfrac{11}{3}\right) $$\\$ Point $R$ divides the side $CF$ in a ratio $2\colon 1$$\\$ Coordinates of $R =\left(\dfrac{2*5+1*1}{2+1},\dfrac{2*\dfrac{7}{2}+1*4}{2+1}\right)=\left(\dfrac{11}{3},\dfrac{11}{3}\right) $$\\$ (iv) It can be observed that the coordinates of point $P, Q, R$ are the same. Therefore, all these are representing the same point on the plane i.e., the centroid of the triangle.$\\$ (v) Consider a triangle, $\Delta ABC$, having its vertices as $A(x_1 , y_1 ), B(x_2 , y_2 )$, and $C(x_3 , y_3 )$.$\\$ Median $AD$ of the triangle will divide the side $BC$ in two equal parts. Therefore,$D$ is the mid-point of side $BC$.$\\$ Coordinates of $D = \left(\dfrac{x_2+x_3}{2}, \dfrac{y_2+y_3}{2}\right)$$\\$ Let the centroid of this triangle be $O$.$\\$ Point $O$ divides the side $AD$ in a ratio $2\colon 1$.$\\$ Coordinates of $O=\left(\dfrac{2*\dfrac{x_2+x_3}{2}+1*x_1}{2+1},\dfrac{2*\dfrac{y_2+y_3}{2}+1*y_1}{2+1}\right)$$\\$ $ = \left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$

From the figure,$\\$ Given,$\\$ $\ \ \ \ \ \bullet$ Let $A(x_1, y_1 ) = (4, 2)$$\\$ $\ \ \ \ \ \bullet$ Let $B(x_2, y_2 ) = (6 , 5)$$\\$ $\ \ \ \ \ \bullet$ Let $C(x_3, y_3 ) = (1, 4)$$\\$ (i) $\ \ \ \ \ \bullet$ Let $AD $be the median of the triangle $ABC$$\\$ $\ \ \ \ \ \bullet$ Hence $D$ is the midpoint of $BC$ Coordinates of $D= \left(\dfrac{6+1}{2},\dfrac{5+4}{2}\right) = \left(\dfrac{7}{2},\dfrac{9}{2}\right)$$\\$ (ii) From the Figure,$\\$ Point $P$ divides the side $AD$ in a ratio $m\colon n = 2\colon 1.$$\\$ Coordinates of $P = \left(\dfrac{2*\dfrac{7}{2}+1*4}{2+1} , \dfrac{2*\dfrac{9}{2}+1*2}{2+1}\right) =\left(\dfrac{11}{3},\dfrac{11}{3}\right) $$\\$

(iii) From the figure,$\\$ Median $BE$ of the triangle will divide the side $AC$ in two equal parts. Therefore,$E$ is the mid-point of side $AC$.$\\$ Coordinates of $E = \left(\dfrac{4+1}{2},\dfrac{2+4}{2}\right) =\left(\dfrac{5}{2},3\right)$$\\$ Point $Q$ divides the side $BE$ in a ratio $2\colon 1$.$\\$ Coordinates of $Q=\left(\dfrac{2*\dfrac{5}{2}+1*6}{2+1} ,\dfrac{2*3+1*5}{2+1}\right) =\left(\dfrac{11}{3},\dfrac{11}{3}\right) $$\\$ Point $R$ divides the side $CF$ in a ratio $2\colon 1$$\\$ Coordinates of $R =\left(\dfrac{2*5+1*1}{2+1},\dfrac{2*\dfrac{7}{2}+1*4}{2+1}\right)=\left(\dfrac{11}{3},\dfrac{11}{3}\right) $$\\$ (iv) It can be observed that the coordinates of point $P, Q, R$ are the same. Therefore, all these are representing the same point on the plane i.e., the centroid of the triangle.$\\$ (v) Consider a triangle, $\Delta ABC$, having its vertices as $A(x_1 , y_1 ), B(x_2 , y_2 )$, and $C(x_3 , y_3 )$.$\\$ Median $AD$ of the triangle will divide the side $BC$ in two equal parts. Therefore,$D$ is the mid-point of side $BC$.$\\$ Coordinates of $D = \left(\dfrac{x_2+x_3}{2}, \dfrac{y_2+y_3}{2}\right)$$\\$ Let the centroid of this triangle be $O$.$\\$ Point $O$ divides the side $AD$ in a ratio $2\colon 1$.$\\$ Coordinates of $O=\left(\dfrac{2*\dfrac{x_2+x_3}{2}+1*x_1}{2+1},\dfrac{2*\dfrac{y_2+y_3}{2}+1*y_1}{2+1}\right)$$\\$ $ = \left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$

Coordinates of $P = \left(\dfrac{2*\dfrac{7}{2}+1*4}{2+1} , \dfrac{2*\dfrac{9}{2}+1*2}{2+1}\right) =\left(\dfrac{11}{3},\dfrac{11}{3}\right) $$\\$

33   $ABCD$ is a rectangle formed by the points $A (-1, -1), B (- 1, 4), C (5, 4)$ and $D (5, - 1). P, Q, R$ and $S$ are the mid-points of $AB, BC, CD,$ and $DA$ respectively. Is the quadrilateral $PQRS $ is a square? $A$ rectangle? Or a rhombus? Justify your answer.

Solution :

From the figure below,$\\$ $P$ is the mid-point of side $AB$.$\\$ Therefore, the coordinates of $P$ are $\left(\dfrac{-1-1}{2},\dfrac{-1+4}{2}\right) =\left(-1,\dfrac{3}{2}\right) $$\\$ Similarly, the coordinates of $Q R$ and $S$ are $(2, 4),(5,\dfrac{3}{2}) $ and $(2,-1)$ respectively.$\\$ We know that the distance between the two points is given by the Distance Formula,$\\$ $ \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \dots $Equation (1)$\\$ Distance between two points $P$ and $Q$,$\\$ Length of $PQ =\sqrt{(-1-2)^2+(\dfrac{3}{2}-4)^2} = \sqrt{9+\dfrac{25}{4}} =\sqrt{\dfrac{61}{4}}$$\\$ Distance between two points $Q$ and $R$,$\\$ Length of $QR =\sqrt{(2-5)^2+(4+\dfrac{3}{2})^2} =\sqrt{\dfrac{61}{4}}$$\\$ Distance between two points $R $and $S$,$\\$ Length of $RS = \sqrt{(5-2)^2+(\dfrac{3}{2}+1)^2} =\sqrt{\dfrac{61}{4}} $ $\\$ Distance between two points $S$ and $P$,$\\$ Length of $SP =\sqrt{(2+1)^2+(-1-\dfrac{3}{2})^2} =\sqrt{\dfrac{61}{4}}$$\\$ Distance between two points $P$ and $R$,$\\$ Length of $QS =\sqrt{(2-2)^2+(4+1)^2} =5$$\\$ It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, $PQRS$ is a rhombus.

Distance between two points $R $and $S$,$\\$ Length of $RS = \sqrt{(5-2)^2+(\dfrac{3}{2}+1)^2} =\sqrt{\dfrac{61}{4}} $ $\\$ Distance between two points $S$ and $P$,$\\$ Length of $SP =\sqrt{(2+1)^2+(-1-\dfrac{3}{2})^2} =\sqrt{\dfrac{61}{4}}$$\\$ Distance between two points $P$ and $R$,$\\$ Length of $QS =\sqrt{(2-2)^2+(4+1)^2} =5$$\\$ It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, $PQRS$ is a rhombus.

$ \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \dots $Equation (1)$\\$ Distance between two points $P$ and $Q$,$\\$ Length of $PQ =\sqrt{(-1-2)^2+(\dfrac{3}{2}-4)^2} = \sqrt{9+\dfrac{25}{4}} =\sqrt{\dfrac{61}{4}}$$\\$ Distance between two points $Q$ and $R$,$\\$ Length of $QR =\sqrt{(2-5)^2+(4+\dfrac{3}{2})^2} =\sqrt{\dfrac{61}{4}}$$\\$ Distance between two points $R $and $S$,$\\$ Length of $RS = \sqrt{(5-2)^2+(\dfrac{3}{2}+1)^2} =\sqrt{\dfrac{61}{4}} $ $\\$ Distance between two points $S$ and $P$,$\\$ Length of $SP =\sqrt{(2+1)^2+(-1-\dfrac{3}{2})^2} =\sqrt{\dfrac{61}{4}}$$\\$ Distance between two points $P$ and $R$,$\\$ Length of $QS =\sqrt{(2-2)^2+(4+1)^2} =5$$\\$ It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, $PQRS$ is a rhombus.