7.2 Distance between two points and Section Formula:

7.2.2 Dividing the line in the given ratio:  Is it possible to divide a line of length 10cm in 3:4 ratio using only scale? To do this one needs to divide this line into 7 equal parts and then identify points  at  3 parts and 4 parts respectively. For that we need to have exact value of *3. But its value is 4.28571… Hence we cannot  identify a exact point corresponding to this value while using only the scale. However we can do this  based on geometry.

Steps to divide a line segment in the ratio of m:n in general:

 No. Steps Figure 1 Let AB be a line segment (say 11cm). We need to divide it in the ratio of m:n(say 3:2) 2 Draw a line  AX making an acute angle with AB( Ex 300 or 45o) 3 Steps to draw parallel line to AX: i)     With the help of  divider, draw an arc PQ of suitable radius cutting the lines AX and AB ii)    Using the same radius drawn an arc RS  cutting line AB at R iii)   With the length PQ as radius, cut arc RS at T iv)   Join B and T and extend it up to Y Now BY is parallel to AX. 4 Identify points on AX  to make ‘m’ equal parts such that  AA1=A1A2=A2A3=A3A4=…Am-1Am (say m=3) 5 With length as AA1   identify points on BY to make ‘n’ equal parts such that   BB1=B1B2=B2B3=B3B4=…Bn-1Bn(=AA1)( say n=2) 6 Join Am and  Bn ( say A3 and B2). Let this line cut AB at C then   AC:CB=m:n(In the example 3:2). 7 Proof:   In  AAmC and BBnC,  corresponding angles are equal.    AAmCBBnC (AAA postulate)) =(  Corresponding sides are proportional) But = ( By construction)( In the example =) = Note: In the example AB=11Cm;  m=3,  n=2 (Also Am= A3  and Bn= B2)

Steps to construct a triangle whose corresponding sizes are  th of given triangle:

 No. Steps Figure 1 Let ABC be the given triangle. We need to construct another similar triangle whose corresponding sides are  th of the given triangle ABC 2 Below BC draw an another line BX by making an acute angle with BC( Say 300 or 45o) 3 Divide BX such that BB1=B1B2=B2B3=B3B4=…Bx-1Bx(Here x=  maximum of ‘m’ and ‘n)   Now we have two situations Situation 1: <1 ;m1 m>n: then x=m (Ex: construction of   th  of original triangle) i)     Join Bn  and C ii)    From  Bx(=Bm)  draw parallel line to BnC . Let it cut extended BC at Q iii)   From Q draw parallel line to AC. Let it cut extended BA at P   Sides of  BPQ  are   th of  corresponding sides of BAC Note that BPQ is bigger than BAC

Note that when <1 the constructed triangle is smaller than the given triangle and when >1 the constructed triangle is bigger than the given triangle.

7.2.1 Distance between two points:   We have learnt how to plot points on a plane in a graph sheet. There are many instances where we need to find the distance between two points (length of the line segment joining two points).

 We know that any point can be represented in terms of co-ordinates of x and y. Let P (x1,y1) and Q (x2,y2) be the two points. We are required to find the length of the line PQ. Draw PA and QB perpendicular to x axis from P and Q respectively. Note that OA = x1 and OB = x2 Draw PC and QD horizontal to Y axis from P and Q respectively. Note that OC = y1 and OD = y2 Produce CP to meet BQ at R. PR = OB-OA = x2-x1 QR = OD-OC = y2-y1   Since PRQ is a right angled triangle, by Pythagoras theorem PQ2 = PR2+RQ2= (x2-x1)2+ (y2-y1)2  PQ =  {(x2-x1)2+ (y2-y1)2} This  formula is called 'distance formula'

Corollary: What if one point is the origin (0,0) ?

The distance of a point P (x,y) from the origin O(0,0)  is  OP =  (x2+ y2)

7.2 Problem 1: Find the value of k if P(0,2) is equidistant from Q(3,k) and R(k,5)

Solution:

 PQ =  {(3-0)2+ (k-2)2} =  (9 +k2-4k+4) PR =  {(k-0)2+ (5-2)2} =  ( k2+9)   Since PQ=PR it follows that 9 +k2-4k+4 = k2+9 On simplification we get k = 1   The point P(0,2) is equidistant from Q(3,1) and R(1,5).

7.2 Problem 2: State the special property of the triangle formed by three points A(10,-18), B(3,6) and C(-5,2)

Solution:

 AB =  {(3-10)2+ (6-(-18))2} =  (49+ 576) =  (625) =25 AC =  {(-5-10)2+ (2-(-18))2} =  (225+ 400) =  (625) =25 BC =  {(-5-3)2+ (2-6)2} =  (64+ 16) =  (80)     Since AB=AC it is clear that the triangle formed by the given three points is an isosceles triangle

7.2 Problem 3: Using distance formula show that points A(2,5), B(-1,2) and C(4,7) are collinear

 Hint: Show that BA+AC = BC (Verify by plotting points that they are collinear)

7.2 Problem 4:  Find the co-ordinates of the Circumcenter of a triangle ABC whose vertices are A(4,6),B(0,4) and C(6,2)

 Hint: Let O(x,y) be the Circumcenter. Then OA=OB=OC and hence OA2 = OB2 =OC2 Solution: OA2 = (x-4)2+(y-6)2=x2-8x+16+y2-12y+36 OB2 = (x-0)2+(y-4)2=x2+y2-8y+16 OC2 = (x-6)2+(y-2)2=x2-12x+36+y2-4y+4 Equating OA2 = OB2 We get 2x+y =9 Equating OA2 = OC2 We get x-2y = -3 By solving the above two equations we get x=3 and y=3. Hence O(3,3) is the Circumcenter of ABC.

Section Formula:

This section is about finding a point on a line such that the point divides the line in the given ratio.

 Let AB be the line joining the point A (x1, y1) and B(x2, y2). We are required to find a point P(x, y) on the line AB such that it divides AB in the given ratio of m1:m2. From A, P and B draw perpendiculars to the x-axis and let these perpendiculars meet x- axis at C,Q and D respectively. From A and P draw parallel lines to x axis to meet PQ at E and BD at R. If P is a point on AB dividing it in the given ratio of m1:m2, then AP/PB = m1/m2 It is clear from the adjacent figure that,  AEP and PRB are similar (AAA Postulate). Hence AE/PR = PE/BR=AP/PB = m1/m2 --------à(1) AE = OQ-OC = x-x1 : PR = OD-OQ = x2-x PE = QP-QE(=CA) = y-y1 BR = DB-DR = y2-y Thus by substituting values in (1) we get AE/PR = (x-x1)/(x2-x) = PE/PR = (y-y1)/ (y2-y) = m1/m2  --------à(2) By taking first and last expression in (2), we have (x-x1)/(x2-x) = m1/m2  m2(x-x1) = m1(x2-x) (By cross multiplication)  m2x - m2x1 = m1x2- m1x (By expansion)  x(m2+m1) = m1x2+ m2x1(By transposition)  x = (m1x2+ m2x1)/(m2+m1)(By division) Similarly, by taking second and last expression in (2) we get  y = (m1y2+ m2y1)/(m2+m1) Thus the co-ordinates of point P, which divides the line joining points A(x1, y1) and B(x2, y2) in the ratio of m1:m2 are given by: {(m1x2+ m2x1)/(m1+m2), (m1y2+ m2y1)/(m1+m2) } This is known as ‘section formula’.

1. What are the co-ordinates of midpoint of AB (i.e. when m1:m2 = 1:1)?

It is {(x2+x1)/2), (y2+ y1)/2}:  (Mid point formula)

Note: The above formula can be used to show that the quadrilateral formed by joining the midpoints of adjacent sides of a quadrilateral is a parallelogram.

2. What are the co-ordinates of the point which divides the line in the ratio of k:1?

It is {(kx2+x1)/(k+1), (ky2+ y1)/(k+1)}

7.2 Problem 5:  Find the ratio in which the point P(11,15) divides the line segment joining the points A(15,5) and B(9,20)

Solution:

 Let P divide the line AB in the ratio of k:1 x1=15, y1=5, x2=9, y2=20,x=11, y=15 We have seen above that the co-ordinates of the point which divides a line in the ratio of k:1 is {(kx2+x1)/(k+1), (ky2+ y1)/(k+1)}  x = (kx2+x1)/(k+1), y = (ky2+ y1)/(k+1)  x = 9k+15/(k+1) 11 = 9k+15/(k+1)( it is given that x=11)  11k+11 = 9k+15 2k=4 or k=2 Hence the point divides the given line in the ratio of 2:1

7.2 Problem 6:  Find the co-ordinates of points which trisects the line joining A(6,-2) and B(-8,10)

Hint:

 We are required to find the co-ordinates of P and Q such that AP=PQ=QB (1:1:1) This problem needs to be solved in two steps: 1.  Find the co-ordinates of P(x1,y1) such that AP:PB = 1:2. 2.  Find the co-ordinates of Q(x2,y2) such that AQ:QB = 2:1 They are P (4/3,2) and Q (-10/3,6)

7.2 Problem 7:  In triangle ABC, D(-2,5) is mid point of AB, E(2,4) is mid point of BC and F(-1,2) is mid point of AC.

Find the co-ordinates of A,B and C.

 Hint: Let A =(x1,y1), B=(x2,y2) and C=(x3,y3) Since D is mid point of AB, (x1+x2)/2 = -2 and (y1+y2)/2 = 5 Since E is mid point of BC, (x2+x3)/2 = 2 and (y2+y3)/2 = 4 Since F is mid point of AC, (x1+x3)/2 = -1 and (y1+y3)/2 = 2 By solving these three equations we get x1= -5, x2=1, x3= 3 y1= 3, y2=7, y3= 1 Thus the vertices are A(-5,3), B(1,7) and C(3,1)

7.2.3 Area of a triangle given the co-ordinates of a triangle

 As in the adjoining figure let A (x1, y1), B(x2, y2) and C(x3, y3) be the three vertices of a triangle. We are required to find the area of triangle ABC. Let BL, AM and CN be perpendiculars from the vertices B, A and C to x-axis.  OL = x2, OM = x1, ON = x3 and BL = y2, AM = y1, CN = y3   From the figure Area of Triangle ABC = =Area of trapezium BLMA + Area of trapezium AMNC - Area of trapezium BLNC = 1/2(BL+AM)LM + 1/2(AM+NC)MN - 1/2(BL+NC)LN = 1/2(y2+ y1) (x1- x2) + 1/2(y1+ y3) (x3- x1) - 1/2(y2+ y3) (x3- x2) = 1/2[x1(y2- y3) + x2(y3- y1) +x3(y1- y2)] --------- (By rearranging the terms)     Note that if A B and C are collinear then area is zero.

7.2 Problem 8: If D(3,-1), E(2,6), F(-5,7) are the mid points of the sides of ABC, find the area of ABC.

 Let us calculate the area of DEF first Area of DEF = 1/2[3(6-7)+2(7-(-1))+(-5)(-1-6)] =1/2[-3+16+35] =1/2(48) = 24 sq units   Since the area of ABC is four times the area of DEF (By BPT theorem – Refer 6.13) Area of ABC = 4*24 = 96 sq units

7.2 Problem 9: If the consecutive verticies of   triangle are (-4,-2), (-3,-5),(3,-2)  and (2,3) then what is its area?

 Draw a rough diagram of quadrilateral ABCD. Join A and C, then we get two triangles          Area of triangle =  [x1(y2- y3) + x2(y3- y1) +x3(y1- y2)] ABC area = [ -4*(-5-(-2))+ -3*(-2-(-2))+ 3*(-2-(-5))] = [12+0+9]= *21 ACD area =[ -4*(3-(-2))+ 2*(-2-(-2))+ 3*(-2-3)] =[-20+0-15]= *(-35)=*(+35) (Area cannot be –ve) Area of quadrilateral  = 2 Area of 2 triangles = (21+35) = 28 Sq units

7.2 Summary of learning

 No Points to remember 1 The distance between points  P (x1,y1) and Q (x2,y2)  is =  {(x2-x1)2+ (y2-y1)2} 2 The  co-ordinates of the point which divides A(x1,y1) and B (x2,y2) in the given ratio of m1:m2  are

Co-ordinates of centroid:

 We have learnt that the centroid divides the median in the ratio of 2:1 (Refer to Section 6.4) Given the three vertices of a triangle let us calculate the co-ordinates of the centroid. In the adjoining figure A (x1, y1), B(x2, y2) and C(x3, y3) AD is one median and G is the centroid which divides AD in the ratio 2:1. We are required to find the co-ordinates of G. Since AD is median BD = DC The co-ordinates of D are {(x2+x3)/2), (y2+ y3)/2} Since G(x,y) divides AD in 2:1 (m1=2 and m2=1) We have x = {2(x2+x3)/2)+1. x1}/(2+1) = (x2+x3+x1)/3 y = {2(y2+y3)/2)+1. y1}/(2+1) = (y2+y3+y1)/3   Thus the co-ordinates of centroid are

7.2 Problem 8: Find the third vertex of a triangle if two of its vertices are at A(-3,1) and B(0,-2) and the centroid is at the origin.

Solution:

Let C(x3, y3) be the third vertex

We know that the co-ordinates of centroid are

x = (x1+x2+x3)/3 and

y = (y1+y2+y3)/3.

Since the co-ordinates of origin are (0,0)

It follows that

x1+x2+x3 = 0  and y1+y2+y3 = 0

By substituting the values for co–ordinates we have

x1+x2+x3= -3+0+ x3=0

x3 = 3

y1+y2+y3= 1-2+ y3=0

y3 = 1

Thus (3,1) is the third vertex.