## Show that points P(1,-2), Q(5,2), R(3,-1), S(-1,-5) are the vertices of a parallelogram.

ProofGiven points are P(1,-2), Q(5,2), R(3,-1), S(-1,-5).By distance formula PQ = √[(x2-x1)2+(y2-y1)2]PQ = √[(5-1)2+(2-(-2))2]= √[(4)2+(4)2]= √[(16+16)]= √32PQ = √32 ………………(i)By distance formula QR = √[(x2-x1)2+(y2-y1)2]QR = √[(3-5)2+(-1-2)2]= √[(-2)2+(-3)2]= √[(4+9)]= √13QR = √13 ………………(ii)By distance formula RS = √[(x2-x1)2+(y2-y1)2]RS = √[(-1-3)2+(-5-(-1))2]= √[(-4)2+(-4)2]= √[(16+16)]= √32RS = √32 ………………(iii)By distance formula PS = √[(x2-x1)2+(y2-y1)2]PS = √[(-1-1)2+(-5-(-2))2]= √[(-2)2+(-3)2]= √[(4+9)]= […]

## Show that the line joining the points A(4, 8) and B(5, 5) is parallel to the line joining the points C(2,4) and D(1,7).

ProofGiven co-ordinates of A = (4,8)Co-ordinates of B = (5,5)Slope of line AB = (y2-y1)/ (x2-x1)= (5-8)/(5-4)= -3/1 = -3Given co-ordinates of C = (2,4)Co-ordinates of D = (1,7)Slope of line CD = (y2-y1)/ (x2-x1)= (7-4)/(1-2)= 3/-1 = -3Slope of line AB = Slope of line CDLine AB is parallel to line CD.Hence proved.

## Find k if the line passing through points P(-12,-3) and Q(4, k)has slope 1/2 .

Given points are P(-12,-3) and Q(4, k).x1 = -12 , y1 = -3 , x2 = 4 , y2 = kSlope of line PQ = (y2-y1)/ (x2-x1)= (k-(-3))/(4-(-12)= (k+3)/16Given slope of line passing through P and Q is 1/2.½ = (k+3)/162(k+3) = 16k+3 = 8k = 8-3 = 5Hence the value of k is 5.

## In the following examples, can the segment joining the given points form a triangle ? If triangle is formed, state the type of the triangle considering sides of the triangle.

(1) Given points are L(6,4) , M(-5,-3) , N(-6,8).By distance formula LM = √[(x2-x1)2+(y2-y1)2]= √[(-5-6)2+(-3-4)2]= √[(-11)2+(-7)2]= √[121+49]= √170LM = √170 …………(i)By distance formula MN = √[(x2-x1)2+(y2-y1)2]= √[(-6-(-5))2+(8-(-3))2]= √[(-1)2+(11)2]= √(1+121)= √(122)MN = √122 ……………(ii)By distance formula LN = √[(x2-x1)2+(y2-y1)2]= √[(-6-6)2+(8-4)2]= √[(-122+(4)2]= √(144+16)= √(160)LN = √160 …………..(iii)(MN +LN ) LMThese points are not collinear.We can construct a […]

## Find the coordinates of the circumcentre of a triangle whose vertices are (-3,1), (0,-2) and (1,3).

Let A(-3,1), B(0,-2) and C(1,3) be the vertices of the triangle.Let O(h,k) be the circumcentre of the triangle. OA , OB and OC are the radii of the circle.OA = OB………………….(i)By distance formula, OA = √[(x2-x1)2+(y2-y1)2]√[(h–3)2+(k-1)2] = √[(h+3)2+(k-1)2] ……..(ii)OB = √[(h-0)2+(k-(-2))2]= √[(h2+(k+2)2] …………………………..(iii)Equating (ii) and (iii)√[(h+3)2+(k-1)2] = √[(h2+(k+2)2]Squaring both sides[(h+3)2+(k-1)2] = [(h2+(k+2)2]h2+6h+9+k2-2k+1 = h2+k2+4k+46h-2k+10 = […]

## Find the distances between the following points.

(i) Given points are A(a,0) and B(0,a)x1 = a , y1= 0, x2 = 0, y2 = aBy Distance formula,d(A,B) = √[(x2-x1)2+(y2-y1)2]d(A,B) = √[(0-a)2+(a-0)2]= √(a2+a2)= √(2a2)= a√2 units.Hence the distance between the points A and B is a√2 units. (ii) Given points are P(-6,-3) and Q(-1,9)x1 = -6 , y1= -3, x2 = -1, y2 = 9By Distance formula,d(P,Q) = √[(x2-x1)2+(y2-y1)2]d(P,Q) […]

## Find the point on X-axis which is equidistant from P(2,-5) and Q(-2,9).

Let M be the point on X-axis which is equidistant from P(2,-5) and Q(-2,9).Since the point M is on X-axis, its y co-ordinate is zero.M = (x,0)Since M is equidistant from P and Q,PM = QM ………..(i)by Distance formula, PM = √[(x2-x1)2+(y2-y1)2]= √[(x-2)2+(0-(-5))2]= √[(x-2)2+(5)2]= √(x2-4x+4+25)= √(x2-4x+29)by Distance formula, QM = √[(x2-x1)2+(y2-y1)2]= √[(x-(-2))2+(0-9)2]= √[(x+2)2+(9)2]= √(x2+4x+85)From (i)√(x2-4x+29) […]

## Find the ratio in which the line segment joining the points A(3,8) and B(-9, 3) is divided by the Y- axis.

Suppose, P be the point on Y axis divides segment AB in the ratio m:n.Since P lies on Y axis, its x co-oordinate is zero.Let P = (0,y)By Section formula, x = (mx2+nx1)/(m+n)Given A(x1,y1) = (3,8)B(x2,y2) = (-9,3)0 = m×-9+n×3/(m+n)0 = (-9m+3n)/(m+n)-9m+3n = 0-9m = -3nm/n = 3/9 = 1/3The required ratio m:n = 1:3.

## Find the coordinates of the midpoint of the line segment joining P(0,6) and Q(12,20).

Given P(x1,y1) = (0,6)Q(x2,y2) = (12,20)Let co-ordinate of midpoint be M(x,y)By Midpoint formula x = (x1+x2)/2 and y = (y1+y2)/2x = (0+12)/2 = 6y = (6+20)/2 = 26/2 = 13Hence co-ordinates of midpoint of PQ are (6,13).

## Determine whether the given points are collinear.

(1) A(0,2) , B(1,-0.5), C(2,-3) are the given points.Slope of line AB = (y2-y1)/ (x2-x1)= (-0.5-2)/(1-0))= -2.5/1 = -2.5Slope of line BC = (y2-y1)/ (x2-x1)= (-3-(-0.5))/(2-1)= -2.5/1 = -2.5Slope of line AB and BC are equal.Point B lies on both lines.Point A,B,C are collinear. (2) P(1, 2) , Q(2, 8/5 ) , R(3, 6/5 ) […]