41 + Interview Questions in CO-ORDINATE-GEOMETRY IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Student of a school are standing in rows and columns in their playground for a drill practice. A, B, C, D are the positions of four students as shown in the figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from eachl of the four students A, B, C and D? If so, what should be his position ?
Answer» Co-ordinates of A, B, c and D fron graph are A(3, 5), B(7, 9), C(11, 5) and D(7, 1). To find the shape of `squareABCD` : `" "AB^(2)=(7-3)^(2)+(9-5)^(2)=4^(2)+4^(2)=32` `rArr" "AB=4sqrt(2)` units `" "BC^(2)=(11-7)^(2)+(5-9)^(2)=(4)^(2)+(-4)^(2)=32` `rArr" "BC=4sqrt(2)` units `" "CD^(2)=(7-11)^(2)+(1-5)^(2)=(-4)^(2)+(-4)^(2)=32` `rArr" "CD=4sqrt(2)` units `" "DA^(2)=(7-3)^(2)+(1-5)^(2)=4^(2)+(-4)^(2)=32` `rArr" "DA=sqrt(4^(2)(1+1))=4sqrt(2)` units `therefore" "AB=BC=CD=DA=4sqrt(2)` units So, ABCD will be either square or rhombus. Now, `" ""Diagonal "AC=sqrt((11-3)^(2)+(5-5)^(2))` `rArr" "AC=sqrt((8)^(2)+(0)^(2))" "rArrAC=8` units and `" ""diagonal "BD=sqrt((7-7)^(2)+(1-9)^(2))=sqrt((0)^(2)+(-8)^(2))=sqrt(8^(2))` `rArr" "BD=8` units `therefore " ""Diagonal "AC="Diagonal "BD` So, given quadrilateral ABCD is a squre. The point which is equidistant from point A, B, C, D of a square ABCD will be at the intersecting point of diagonals and diagonals bisects each other. Hence, the required point O equidistant from A, B, C, D is mid-point of any diagonal `=((7+7)/(2), (9+1)/(2))=((14)/(2), (10)/(2))=(7, 5)`. Hence, the required point is (7, 5).

Discussion

2.	Wrtie the slope of X-axis and Y-axis .
Answer» Correct Answer - D

Discussion

3.	Seg AB is parallel to Y-axis and Co-ordinates of point A are `(1,3)` then co-ordinates of point B can be ………….. A)`(3,1)` B)`(5,3)` C)`(3,0)` D)`(1,-3)`A. `(3,1)`B. `(5,3)`C. `(3,0)`D. `(1,-3)`
Answer» Correct Answer - D

Discussion

4.	Out of the following, Point _________ lies to the right of the origin on X- axis.A. `(-2,0)`B. `(0,2)`C. `(2,3)`D. `(2,0)`
Answer» Correct Answer - D

Discussion

5.	The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p.
Answer» In `DeltaABC` `" "angleB= 90^(@)` `therefore" "AB^(2)+BC^(2)=AC^(2)` `rArr(p-4)^(2)+(3-7)^(2)+(p-7)^(2)+(3-3)^(2)=(7-4)^(2)+(3-7)^(2)` `rArr" "p^(2)-8p+16+p^(2)-14p+49+0=9` `rArr" "2p^(2)-22p+56=0` `rArr" "p^(2)-11p+28=0` `rArr" "p^(2)-7p-4p+28=0` `rArr" "p(p-7)-4(p-7)=0` `rArr" "(p-7)(p-4)=0` `rArr" "p-7=0 or p-4=0` `rArr" "p=7 or p=4` when p=7, then the points B and C coincide and so no triangle is formed. `therefore" "p ne7` Hence, `p` =4

Discussion

6.	Prove that the points (5, -2), (-4, 3) and (10, 7) are the vertices of an isosceles right-angled triangle.
Answer» Let the points are `A(5,-2), B(-4, 3) and C(10, 7)`. `therefore" "AB^(2)=(-4-5)^(2)+(3+2)^(2)=(-9)^(2)+(5)^(2)= 81+ 25=106` `" "BC^(2)=(10+4)^(2)+(7-3)^(2)=(14)^(2)+(4)^(2)=196+16=212` `" "AC^(2)=(10-5)^(2)+(7+2)^(2) =(5)^(2)+(9)^(2)=25+81=106 ` Therefore, `" "AB=AC=sqrt(106)` `and " "AB^(2)+AC^(2)=BC^(2)` `thereforeDeltaABC` is an isosceles right-angled triangle.

Discussion

7.	Find the area of the triangle, whose vertices are (a,c+a), (a,c) and (-a,c-a).
Answer» Area of triangle `=1/2[x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))]` `=1/2[a(c-c+a)+a(c-a-a)-a(c+a-c)]` `=1/2[a^(2)-2a^(2)-a^(2)]=-a^(2)` But area of triangle connot be negative `therefore` Area of triangle `=a^(2)` square units

Discussion

8.	Find the area of triangle, whose vertices are (2,3), (7,5) and (-7,-5).
Answer» Area of triangle `=1/2[x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(2)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))` `=1/2[2(5+5)+7(-5-3)-7(3-5)]=1/2(20-56+14)=-11` But the area of triangle connot be negative `therefore` Area of triangle = 11 square units

Discussion

9.	Find the co-ordinates of a point which divides the line joining the points `A(5, -2) and B(4, 6)` in the ratio 1 : 2 externally.
Answer» Let ` A(x_(1), y_(1))=(5, -2)and B(x_(2), y_(2))=(4, 6 )` `" "m:n =1:2` Let `P(x, y)` be the point which divides the line segment AB in the ratio `m : n ` externally. `therefore" "x=(1(4)-2(5))/(1-2)=6` `" "y=(1(6)-2(-2))/(1-2)=-10` `therefore` Co-ordinates of required point = (6, - 10)

Discussion

10.	If (x,y) be any point on the line segment joinijng the points (a,0)and (0,b) then prove that `x/b+y/b=1.`
Answer» Given three points are collinear `implies" ""Area of" Delta=0` `implies" "1/2[x(0-b)+a(b-y)+0(y-0)]=0` `implies" "-bx+ab-ay=0` Divide both sides by ab `" "-x/a+1-y/b=0implies" "x/a+y/b=1`

Discussion

11.	If a `nebne0,` prove that the points `(a,a^(2)),(b,b^(2)),(0,00)` will not be colliear.
Answer» Let the 3 points A `-=(a,a^(2)),B-=(b,b^(2))and C-=(0,0)` from a triangle ABC. `therefore` Area of `DeltaABC=1/2\|a(b^(2)-0)+b(0-a^(2))+0(a^(2)-b^(2))\|` `=1/2[ab^(2)-a^(2)b]=1/2ab(b-a)ne0" "(becauseanebne0)` So, `DeltaABC` will be formed. Therefore 3 points A, B and C will not be collinear.

Discussion

12.	If the points A (x,y), B (1,4) and C (-2,5) are collinear, then shown that x + 3y = 13.
Answer» Given points are collinear `therefore" ""Area of triangle"=0` `implies" "1/2[x(4-5)+1(5-y)-2(y-4)]=0` `implies" "-x+5-y-2y+8=0` `implies" "x+3y=13`

Discussion

13.	Prove that the points (6,4) (4,5) and (2,6) are collinear.
Answer» Area of triangle `=1/2[x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))]` `=1/2[6(5-6)+4(6-4)+2(4-5)]=1/2[-6+8-3]=0` Therefore, the given points are collinear.

Discussion

14.	If the points `A(-1,-4),B(b,c)and C(5,-1)` are collinear and 3b + c = 4, find the values of b and c.
Answer» Points `A(-1,-4),B(b,c)andC(5,-1)` are collinear `therefore" ""area of"DeltaABC=0` `implies" "-1(c+1)+b(-1+4)+5(-4-c)=0` `implies" "-c-1+3b-20-5c=0` `implies" "3b-6c=21` `implies" "b-2c=7" "...(1)` `"Given that,"2b+c=4" "...(2)` from eqs. (1) and (2), we get `b=3andc=2`

Discussion

15.	Find the points on the X-axis which are at distance of `2sqrt(5)` from the point (7,-4) . How many such points are there ?
Answer» Let P (x,0) be the required point on X-axis. Let A be the point (7,-4) Given that, `" "PA=2sqrt5` `implies" "PA^(2)=20` `implies" "(x-7)^(2)+(0+4)=20` `implies" "(x-7)^(2)=20-16=4` `implies" "x-7=+-2` `implies" "x=9or x=5` `therefore` There are two such points. and points are `(9,0) and(5,0).`

Discussion

16.	Find the value of m , if the points (5,1), (-2,-3) and (8,2m) are collinear .
Answer» Points `A(5, 1), B(-2, -3), C(8, 2m)` will be collinear. if `" "5(-3-2m)-2(2m-1)+8(1+3)=0` `rArr" "-15-10m-4m+2+32=0` `rArr" "14m=19rArr" "m=(19)/(14)`

Discussion

17.	If ` A(2,2), B(-2, -2) , C(-2sqrt(3), 2sqrt(3)) and D(-4-2sqrt(3), 4+2sqrt(3))` are the co-ordinates of 4 points. What can be said about these four points ?
Answer» `" "AB=sqrt((-2-2)^(2)+(-2-2)^(2))=4sqrt(2)` units `" "BC=sqrt((-2+2sqrt(3))^(2)+(-2-2sqrt(3))^(2))` `" "=sqrt (4+12-8sqrt(3)+4+12+8sqrt(3))=4sqrt(2) ` units `" "CD=sqrt((-2sqrt3+4+2sqrt(3))^(2)+(2sqrt(3)-4-2sqrt(3))^(2))` ` " " =sqrt(16+16)=4sqrt(2)` units `" "AC=sqrt((2+2sqrt(3))^(2)+(2- 2sqrt(3))^(2))` `" "=sqrt(4+12+8sqrt(3)+4+12-8sqrt(3))=4sqrt(2)` units ` " "AD=sqrt((2+4+2sqrt(3))^(2)+(2-4-2sqrt(3))^(2))` `" "=sqrt(36+12+24sqrt(3)+12+4+8sqrt(3))=sqrt(64+32sqrt(3))` units `" "BD=sqrt((-2+4+2sqrt(3))^(2)+(-2-4-2sqrt3)^(2))` `" "=sqrt(4+12+8sqrt(3)+36+12 +24sqrt(3))=sqrt(64+32sqrt( 3))` units Here, `AB =BC=CD=AC` and also, `AD=BD` So, in first view it seems to be the vertices of a square. `" "BUT" "` Here, ` AB, BC, CD and DA` are not equal. (order of `A, B, C and D` must be cyclic in case of square). Also AD and BD are equal but they cannot be the diagonals. So, they do not form a square. Actually, A, B and D lie on a circle with C as the centre (as CA=CB=CD i.e., C is equidistant from A, B and D).

Discussion

18.	What point on the X-axis is equidistant from (7, 6) and (-3, 4) ?
Answer» We know that `y`-co-ordinate of a point on X-axis is always 0. So, let a point on X-axis be `P(x, 0)` and let two given points be `A(7, 6) and B(-3, 4).` According to the condition, `" "PA=PB` `rArr" "sqrt((x-7)^(2)+(0-6)^(2))=sqrt((x+3)^(2)+(0-4)^(2))` Squaring both sides, we have `" "x^(2)-14x+49 +36=x^(2)+6x+9 +16` `rArr" "20x=60" "rArr" "x=3` `therefore` Required point is (3, 0)

Discussion

19.	Find the coordinates of points on the line joiningthe pointthat is twice as far fromas from
Answer» Let `P(x, 3x+3)` be a point on `3x-y+3=0` (as this satisfies the given equation) which is twice as far from `A(3, 2)` as from `B(1, 1)` So, `" "PA=2PB` `rArr" "sqrt((x-3)^(2)+(3x+1)^(2))=2sqrt((x-1)^(2) +(3x+2)^(2))` Squaring both sides, we have ` " "x^(2)- 6x+ 9+9x^(2)+6x+1=4(x^(2)-2x+1+9x^(2)+12x+4)` `rArr" "30x^(2) +40x+10=0" "rArr" "3x^(2)+4x+1 =0` `rArr" "(3x+1)(x+1)=0` `therefore" "x=-1 " "or" "x=-(1)/(3)` When `x=- 1 rArr y=3 (-1) +3 =0` `" "\|"when " x=-(1)/(3)rArry=3(-(1)/(3))+3=2` So, required points are `(-1, 0) and (-(1)/(3), 2)`.

Discussion

20.	Find apoint on y-axis which is equidistant from the points `(5, -2)`and `(-3, 2)`.
Answer» Let the required point on X-axis be P(x,0) and the given points be A (5,-2) and B(-3, 2). Now, given that `" "PA=PB" "rArr" "PA^(2)=PB^(2)` `rArr" "(x-5)^(2)+(0+2)^(2)=(x+3)^(2)+(0-2)^(2)` `rArr" "x^(2)-10x+25+4=x^(2)+6x+9+4` `rArr" "-16x=-16" "rArr" "x=1` `therefore` Required point is (1, 0)

Discussion

21.	Find a point which is equidistant from the points `A(-5,4)andB(-1,6)` How many such points are there ?
Answer» Let P (x,y) be the required point. Given that, `" " PA-PB` `implies" "PA^(2)=PB^(2)` `implies" "(x+5)^(2)+(y-4)^(2)=(x+1)^(2)+(y-6)^(2)` `implies" "x^(2)+10x+25+y^(2)-8y+16=x^(2)+2x+1+y^(2)-12y+36` `implies" "8x+4y+4=0` `implies" "2x+y+1=0` It shows that infinite points are equidistant from AB because all points on perpendicular bisector of AB will be equidistant from AB. One such point is the mid-point of AB. which is `((-5-1)/(2),(4+6)/(2))=(-3,5)`

Discussion

22.	Find the equation of the set of all points which are twice as far from (3, 2) as from (1, 1).
Answer» Let `P(x, y)` be a point and let `A( 3,2) and B(1,1)` be two other points on the plane, such that `" " PA=2PB` `rArr" "sqrt((x-3)^(2)+(y-2)^(2))=2sqrt((x-1)^(2)+(y -1)^(2))` Squaring both sides, we have `" "x^(2)-6x+9+y^(2)-4y+4=4(x^(2)-2x+1+y^(2)-2y+1)` `rArr" "3x^(2)+3y^(2)-2x-4y-5=0` Which is the required equation.

Discussion

23.	Find the equation of the set of points such that the sum of its distances from (0, 3) and (0, -3) is 8.
Answer» Let `P(x, y)` be any general point and let ` A-=(0, 3) and B-=(0,-3)` such that `" "PA+PB=8`. `rArr" "sqrt((x-0)^(2)+(y-3)^(2))+sqrt((x-0)^(2)+(y+3)^(2))=8` `rArr" "sqrt(x^(2)+y ^(2)-6y+9)=8-sqrt(x^(2)+y^(2)+6y+9)` Squaring both sides, we have `rArr" "x^(2)+y^(2)-6y+9=64+x^(2)+y^(2)+6y+9- 16sqrt(x^(2)+y^(2)+6y+9)` `rArr" "12y+64 =16sqrt(x^(2)+y^(2) +6y+9)` ` rArr" "3y+16=4sqrt(x^(2)+y^(2)+6y+9)` Squaring again, we have `therefore" "9y^(2)+256+96y=16(x^(2)+y^(2)+6y+9)` `rArr" "16x^(2)+ 7y^(2)=112" "or" " (x^(2))/(7)+( y^(2))/(9)=1` which is the required equation.

Discussion

24.	Find the distance between the following points : (i) (3, 4) and (5, 2) (ii) (0, 2) and (4, -1) (iii) (a, 2a) and (-a, -2a) (iv) (4, -3) and (- 6, 5)
Answer» (i) Distance between the points (3, 4) and (5, 2) ` " "=sqrt((x_(2)-x_(1))^(2)+( y_(2)-y_(1))^(2))=sqrt((5-3)^(2)+(2 -4)^(2))` `" "=sqrt(4+4)=sqrt(8)=2sqrt(2)` units (ii) Distance between the points (0, 2) and ( 4, -1) `" "=sqrt((x_(2)-x_(1))^(2 )+(y_(2) -y_(1))^(2))=sqrt((4-0)^(2)+(-1-2)^(2))` `" "=sqrt((4)^(2)+(-3)^(2))=sqrt(16+9)=sqrt(25)= 5` units (iii) Distance between the points (a, 2a) and ( - a, -2a) `" "=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))=sqrt((-a -a)^(2)+(- 2a -2a)^(2))` `" " =sqrt((-2a)^(2)+(-4a)^(2))=sqrt(4a^(2)+16a^(2))=sqrt(20a^(2))=2sqrt(5)a` units (iv) Distance between the points (4, -3) and (-6, 5) ` " "=sqrt((x_(2)-x_(1))^(2)+( y_(2)-y_(1))^(2))=sqrt((-6-4)^(2)+(5+3)^(2))` ` " "=sqrt((-10)^(2 )+(8)^(2))=sqrt(100+64)=sqrt(164)=2sqrt(41)` units

Discussion

25.	Find the distance between the points (5, 8) and (-3, 2).
Answer» Distance between the points (5, 8) and (-3, 2) `" "=sqrt((x_(2)-x_(1))^(2)+(y_(2 )-y_(1)) ^(2)) =sqrt((-3-5)^(2)+(2-8)^(2))` `" "=sqrt((-8)^(2) +(-6)^(2))=sqrt(64+36)=sqrt (100)=10 ` units

Discussion

26.	If the distances of `P(x, y)` from `A(5,1) and B(-1, 5)` are equal, then prove that `3x=2y.
Answer» Since, `P(x,y)` is equidistant from `A(5,1) and B( -1, 5)`. therefore`" "PA=PB` `rArr" "sqrt ((x- 5)^(2)+(y-1)^(2))=sqrt((x+1)^(2)+(y-5)^(2))" "` (by using distance formula) Squaring both sides, we get ` " "(x-5)^(2)+(y -1)^(2)=(x+1)^(2)+(y-5)^(2)` `rArr " "x^(2)-10x+25+y^(2)-2y+1=x+2x+1+y^(2)-10y+25` `" "-1 0x-2x=-10y+2y rArr" "12x=8y` `rArr" "3x=2y`

Discussion

27.	If the distance between the points (-2, -5) and (-6, y) is 5 units, find the value of y.
Answer» Distance between the points (-2, -5) and (-6, y) `" "=sqrt((-6 +2)^(2)+(y+5)^(2))=sqrt(16+y^(2)+10y + 25)` `" "=sqrt(y^(2)+10y+41) ` Given that, ` " "sqrt(y^(2)+10y+41)=5rArr" "y^(2)+10y+41=25` ` rArr" "y ^(2)+10y+ 16=0 rArr" "y^(2)+ 2y+8y+16=0` `rArr" "y(y+2)+8(y+2)=0rArr" "(y+2)(y+8)=0` `rArr" "y+2=0 or y+8=0` `" "y=-2 or y=-8`

Discussion

28.	Find the distance between the points (2, 6) and (0, 9).
Answer» Distance between the points (2, 6) and (0,9) `" "=sqrt((x_(2)-x_(1))^(2)+(y_(2 )-y_(1)) ^(2))=sqrt((0-2)^(2)+(9-6)^(2))` ` " "=sqrt( 4+ 9)=sqrt(13)` units

Discussion

29.	Distance of point (-3, 4) from the origin is ________A. 7B. 1C. 5D. `-5`
Answer» Correct Answer - C

Discussion

30.	Plot the points `A(2,0), B(8,0),C(8,0),D(8,4)`. Complete the rectangle ABCD and find the co-ordinates of point D.
Answer» Correct Answer - (2,4)

Discussion

31.	Find the distance of the point (3, 4) from the origin.
Answer» Distance of the point (3, 4) to the origin `" "=sqrt((3-0)^(2)+(4-0)^(2)) =sqrt(9+16)=sqrt(25)=5` units

Discussion

32.	Find the distance between the points `( acostheta, asintheta) ` from the origin.
Answer» Distance between the points `( acostheta, asintheta) ` and origin (0, 0) ` " "=sqrt((acostheta-0 )^(2)+(asintheta-0)^(2))` `" "=sqrt (a^(2)cos^(2)theta+a^(2)sin^(2)theta)=sqrt(a^(2) (cos^(2)theta+sin^(2)theta)) ` `" "=sqrt(a^(2))` `" "=a" units"" "(because cos^(2)theta+sin^(2)theta=1)`

Discussion

33.	Find the area of the triangle, whose vertices are (2,1), (4,5) and (6,3).
Answer» Area of triangle `=1/2[x_(x)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))]` `=1/2[2(5-3)+4(3-1)+6(1-5)]=1/2(4+8-24)= -6` But the area of triangle connot be negative `therefore" "` Area of triangle = 6 square units

Discussion

34.	Find the co-ordinates of a point which divides the line joining the points `A(3, 4) and B(-2, -1)` in the ratio 3 : 2 externally.
Answer» Let `A(x_(1), y_(1))=(3, 4) and B(x_(2), y_(2))= (-2, -1)` `" "m:n=3:2` Let ` P(x, y)` be the point which divides the line segment AB in the ratio `m : n` externally. `therefore " "x=(3(-2)-2(3))/(3-2)=-12` ` and" "y=(3(-1)-2(4))/(3-2)=-11` `therefore` Co-ordinates of required point = (-12, -11)

Discussion

35.	If a point `P` lies on the line segment joining points `A(-3, 4) and B(-2, -6)` such that `" "2AP=3BP` then, find the co-ordinates of point P.
Answer» Given that, `" "2AP=3BP` ` rArr" " (AP)/(BP)=(3)/(2)` `rArr" "m:n=3:2` `" "(x_(1), y_(2))=(-3, 4) and (x_(2),y_(2))=(-2, -6)` Now, let the co-ordinates of point P are `(x, y)` ` therefore" "x=(3(-2)+2(-3 ))/(3+2)=(-12)/(5)` `and " " y=(3(-6)+2(4))/( 3+2)=-2` `therefore` Co-ordinates of point ` P =((-12)/(5),-2)`

Discussion

36.	Find the co-ordinates of the point P which bisects seg having co- ordinates `(3, 2) and (5, -2)` A)`(-3,5) ` B)`(0,4)` C)`(4,0)` D)`(5,-3)`A. `(-3,5) `B. `(0,4)`C. `(4,0)`D. `(5,-3)`
Answer» Correct Answer - C

Discussion

37.	Find the co-ordinates of the points of trisection of the line segment joining the points `A(-5, 6) and B (4, -3)`.
Answer» Let P and Q be the points of trisection of AB, then P divides AB in the ratio 1 : 2. `therefore` Co-ordinates of point `P-=((-5xx2+4xx1)/(1+2), (6xx2+1xx-3)/(1+2))-=(-2,3)` and point Q divides AB in the ratio 2 : 1. `therefore` Co-ordinates of point `Q-=((-5xx1+4xx2)/(2+1),(6xx1+2xx-3)/(1+2))-=(1,0)` `therefore` Co-ordinates of the point of trisection are (-2, 3) and (1, 0).

Discussion

38.	In what ratio does the point `(1, 6)` divide the line segment joinigng the points `(3, 6) and (-5, 6)` ?A. `1 : 3`B. `2 : 3`C. `3 : 1`D. `3 : 2`
Answer» Correct Answer - A

Discussion

39.	On which axis do the given points lie ?
Answer» (i) In (0,5) , we have abscissa =0 ltbr) `therefore` The point (0,5) lies on the positive y-axis. (ii) In (-5,0), we have ordinate =0 `therefore` The point (-5,0) lies on the negative x-axis.

Discussion

40.	Prove that the points (4, 8), (7, 5) , (1, -1) and (-2, 2) are the vertices of a parallelogram.
Answer» Let the points are `A( 4, 8), B(7, 5), C(1, - 1) and D(-2, 2)`. ` therefore" "AB^(2)=(7-4)^(2)+(5-8 )^(2)=(3)^(2)+(-3)^(2)=9+9=18` `rArr" "AB=3sqrt(2)` `" "BC^(2)=(1-7)^(2)+(-1 -5)^(2)=(-6)^(2)+(-6)^(2)= 36+36=72 ` `rArr" "BC=6sqrt(2)` `" "CD^(2)=(-2-1)^2+(2+1)^(2)=(-3)^(2)+(3)^(2)=9+9=18` `rArr" " CD=3sqrt(2)` `" "DA^(2)=(4+2)^(2)+(8-2)^(2)=6^(2)+6^(2)=36+36 =72` `rArr" "DA=6sqrt(2)` `because" "AB=CD and BC=DA`

Discussion

41.	`A(-2,4),C(4,10)` and `D(-2,10)` are the vertices of a square ABCD. Use the graphical method to find the co-ordinates of fourth vertex B. Also find. (i) the co-ordinates of mid-point of BC. (ii) the co-ordinates of point of intersection of the diagonals of the square ABCD.
Answer» Correct Answer - B(4,4) (i) (4,7) (ii) (1,7)

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Wrtie the slope of X-axis and Y-axis .

Seg AB is parallel to Y-axis and Co-ordinates of point A are `(1,3)` then co-ordinates of point B can be ………….. A)`(3,1)` B)`(5,3)` C)`(3,0)` D)`(1,-3)`A. `(3,1)`B. `(5,3)`C. `(3,0)`D. `(1,-3)`

Out of the following, Point _________ lies to the right of the origin on X- axis.A. `(-2,0)`B. `(0,2)`C. `(2,3)`D. `(2,0)`

The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p.

Prove that the points (5, -2), (-4, 3) and (10, 7) are the vertices of an isosceles right-angled triangle.

Find the area of the triangle, whose vertices are (a,c+a), (a,c) and (-a,c-a).

Find the area of triangle, whose vertices are (2,3), (7,5) and (-7,-5).

Find the co-ordinates of a point which divides the line joining the points `A(5, -2) and B(4, 6)` in the ratio 1 : 2 externally.

If (x,y) be any point on the line segment joinijng the points (a,0)and (0,b) then prove that `x/b+y/b=1.`

If a `nebne0,` prove that the points `(a,a^(2)),(b,b^(2)),(0,00)` will not be colliear.

If the points A (x,y), B (1,4) and C (-2,5) are collinear, then shown that x + 3y = 13.

Prove that the points (6,4) (4,5) and (2,6) are collinear.

If the points `A(-1,-4),B(b,c)and C(5,-1)` are collinear and 3b + c = 4, find the values of b and c.

Find the points on the X-axis which are at distance of `2sqrt(5)` from the point (7,-4) . How many such points are there ?

Find the value of m , if the points (5,1), (-2,-3) and (8,2m) are collinear .

If ` A(2,2), B(-2, -2) , C(-2sqrt(3), 2sqrt(3)) and D(-4-2sqrt(3), 4+2sqrt(3))` are the co-ordinates of 4 points. What can be said about these four points ?

What point on the X-axis is equidistant from (7, 6) and (-3, 4) ?

Find the coordinates of points on the line joiningthe pointthat is twice as far fromas from

Find apoint on y-axis which is equidistant from the points `(5, -2)`and `(-3, 2)`.

Find a point which is equidistant from the points `A(-5,4)andB(-1,6)` How many such points are there ?

Find the equation of the set of all points which are twice as far from (3, 2) as from (1, 1).

Find the equation of the set of points such that the sum of its distances from (0, 3) and (0, -3) is 8.

Find the distance between the following points : (i) (3, 4) and (5, 2) (ii) (0, 2) and (4, -1) (iii) (a, 2a) and (-a, -2a) (iv) (4, -3) and (- 6, 5)

Find the distance between the points (5, 8) and (-3, 2).

If the distances of `P(x, y)` from `A(5,1) and B(-1, 5)` are equal, then prove that `3x=2y.

If the distance between the points (-2, -5) and (-6, y) is 5 units, find the value of y.

Find the distance between the points (2, 6) and (0, 9).

Distance of point (-3, 4) from the origin is ________A. 7B. 1C. 5D. `-5`

Plot the points `A(2,0), B(8,0),C(8,0),D(8,4)`. Complete the rectangle ABCD and find the co-ordinates of point D.

Find the distance of the point (3, 4) from the origin.

Find the distance between the points `( acostheta, asintheta) ` from the origin.

Find the area of the triangle, whose vertices are (2,1), (4,5) and (6,3).

Find the co-ordinates of a point which divides the line joining the points `A(3, 4) and B(-2, -1)` in the ratio 3 : 2 externally.

If a point `P` lies on the line segment joining points `A(-3, 4) and B(-2, -6)` such that `" "2AP=3BP` then, find the co-ordinates of point P.

Find the co-ordinates of the point P which bisects seg having co- ordinates `(3, 2) and (5, -2)` A)`(-3,5) ` B)`(0,4)` C)`(4,0)` D)`(5,-3)`A. `(-3,5) `B. `(0,4)`C. `(4,0)`D. `(5,-3)`

Find the co-ordinates of the points of trisection of the line segment joining the points `A(-5, 6) and B (4, -3)`.

In what ratio does the point `(1, 6)` divide the line segment joinigng the points `(3, 6) and (-5, 6)` ?A. `1 : 3`B. `2 : 3`C. `3 : 1`D. `3 : 2`

On which axis do the given points lie ?

Prove that the points (4, 8), (7, 5) , (1, -1) and (-2, 2) are the vertices of a parallelogram.

`A(-2,4),C(4,10)` and `D(-2,10)` are the vertices of a square ABCD. Use the graphical method to find the co-ordinates of fourth vertex B. Also find. (i) the co-ordinates of mid-point of BC. (ii) the co-ordinates of point of intersection of the diagonals of the square ABCD.