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Co-ordinate Geometry - Solved Examples
Answer - A
Explanation
(-2, 3) lies in quadrant II.
Answer - C
Explanation
AB = √(2+4)2 + (-5-7)2 = √62+ (-12)2 = √36+144 = √180 =√36*5 = 6√5 units.
Answer - D
Explanation
OA = √62+ (-6)2 =√36+36 =√72 = √36*2 = 6√2 units.
Q 4 - Show that the points A(0,-2) ,B(3,1) ,C(0,4) and D(-3,1) are the vertices of a square.
Answer - B
Explanation
AB2= (3-0)2+ (1+2)2= (9+9) =18 BC2= (0-3)2+ (4-1)2= (9+9) =18 CD2= (0-3)2+ (1+2)2= (9+9) =18 DA2= (-3-0)2+ (1+2)2= (9+9) =18 ∴ AB= BC=CD=DA = √18 = √9*2 = 3√2 AC2=(0-0)2+(4+2)2= (0+36) =36 BD2= (-3-3)2+ (1-1)2= (36+0) =36 ∴ Diag AC = Diag BD = 6 Thus all sides are equal and the diagonals are equal. ∴ ABCD is a square.
Q 5 - Show that the points P(-4,-1), Q(-2,-4),R(4,0) and S(2,3) are the vertices of a rectangle.
Answer - B
Explanation
PQ2= (-2+4)2+ (-4+1)2= 22+ (-3)2= (4+9) =13 QR2= (4+2)2+ (0+4)2= (62+42) = (36+16) =52 RS2= (2-4)2= (3-0)2= (-2)2+32= (4+9) = 13 SP2= (2+4)2+ (3+1)2= (62+42) = (36+16) = 52 ∴ PQ=RS =√13 AND QR=SP =√52 PR2= (4+4)2+ (0+1)2= (82+12) = (64+1) =65 QS2= (2+2)2+ (3+4)2= (42+72) = (16+49) =65 ∴ Diag PR= Diag QS =√ 65 Thus, opposite sides are equal and diagonals are equal. ∴ ABCD is a rectangle.
Q 6 - Show that the points A (-3, 2), B (-5-5), C (2-3) and D (4, 4) are the vertices of a rhombus.
Answer - B
Explanation
AB2= (-5+3)2+ (-5-2)2= (-2)2+ (-7)2= (4+49) =53 BC2= (2+5)2+ (-3+5)2= (7)2+ (2)2= (49+4) =53 CD2= (4-2)2+ (4+3)2= (22+72) = (4+49) =53 DA2= (4+3)2+ (4-2)2= (72+22) = (49+4) =53 ∴ AB=BC=CD=DA = √53 AC2= (2+3)2+ (-3-2)2= (52) + (-5)2= (25+25) = 50 BD2= (4+5)2+ (4+5)2= (92) + (92) = (81+81) =162 ∴ Diag AC ≠ Diag BD Thus all the sides are equal and diagonals are not equal. ∴ ABCD is a Rhombus.
Q 7 - Discover the region of ABC whose vertices are A (10, - 6), B (2, 5) and C (- 1, 3).
Answer - A
Explanation
Here x1=10, x2=2, X3 = -1 and y1= - 6, y2= 5, y3= 3
∴ ∆= 1/2 {X1(y2-Y3) +x2(Y3-Y1) +X3 (Y1-Y2)}
=1/2 {10(5-3) +2(3+6) - 1(- 6-5) = 1/2 (20+18+11) =49/2 sq.units.
Q 8 - Discover the estimation of h for which the focuses A (- 1, 3), B (2, h) and C (5, - 1) are collinear.
Answer - A
Explanation
Here x1=-1, x2=2, x3=5 and y1=3, y2=h and Y3=-1 Now, ∆=0 ⇒ X1(y2-Y3) +x2(Y3-Y1) +X3(Y1-Y2) = 0 ⇒ -1(h+1) +2(-1-3) +5(3-h) =0 ⇒ -h-1-8+15-5h=0 ⇒ 6h=6 ⇒ h=1
Q 9 - Discover the co-ordinates of the centroid of ∆ABC whose vertices are A (6, - 2) and B (4, - 3) what's more, C (- 1, - 4).
Answer - C
Explanation
The directions of the centroid are
{(6+4-1)/3, (- 2-3-4)/3} i.e. (3, - 3)
Q 10 - Discover the proportion in which the point p (2, - 5) partitions the line portion AB joining A (- 3, 5) what’s more, B (4, - 9).
Answer - B
Explanation
Let the required proportion be x:1. At that point (4x-3/x+1, - 9 x+5/x+1) concurs with p (2, - 5) ∴ 4x-3/ ( x+1) =2 ⇒ 4x-3 = 2x+2 ⇒ 2x=5 ⇒ x=5/2 ∴ required proportion is 5/2:1 i.e. 5:2
Answer - A
Explanation
m= tan 30° = 1/√3
Q 12 -Discover the slant of line whose slop is 1/√3
Answer - A
Explanation
tan x = 1/√3 ⇒ x=30°
Q 13 - Discover the slop of the line which goes through focuses A (- 2, 3) and B (4, - 6).
Answer - B
Explanation
Slop of AB = y2-y1/x2-x1 = - 6-3/4+2 = - 9/6 = - 3/2
Q 14 - Discover the slop of the line whose mathematical statement is 3x+4y-5 = 0.
Answer - B
Explanation
3x+4y-5 = 0 ∴ 4y=-3x+5 ∴ y=-3/4x+5/4 ∴ slop = m =-3/4
Q 15 - Discover the estimation of h for which the line 2x+3y-4 = 0 and hx+6y+5 =0 are parallel.
Answer - C
Explanation
2x+3y - 4 =0 ⇒ 3y= - 2x+4 ⇒y= - 2x/3 +4/3 hx+6y+5 =0 ⇒ 6y =-hx-5 ⇒ y= - hx/6 - 5/6 The line will be parallel if - h/6 -2/3 ⇒ h= (2/3*6) = 4 ∴ h=4
Q 16 - Discover the estimation of h for which the lines 5x+3y +2=0 and 3x-hy+6= 0are perpendicular to each other.
Answer - D
Explanation
5x+3y+2 =0 = -5x-2 ⇒ y= -5x/3-2/3 3x- hy+6 =0 ⇒ hy = 3x+6 ⇒ y =3x/h+6/h The line will be perpendicular to each other if -5/3* 3/h= -1 ⇒ h=5. Hence h= 5.