Co ordinate GeometryQ1 If the vertices of a triangle ABC are A = (15, 1) B = (9, 3) & C = (4, 5). Find the product of the slope of line AB?-1/3-2/52/1515/2UndefinedSolution Slope of line of two points (X1, Y1) & (X2, Y2) is given by y2 – Y1 / X2 – X1Slope of line AB = 3-1/9-15 = 2/-6 = -1/3Slope of line BC = 5-3/4-9 = -2/5Therefore the product of slope of line AB & BC = -1/3 X 2/-5 = 2/15A(5,-7)7, 3BX, YQ2 the co-ordinate of one end of a circle is (5,-7). If the centre of the circle has co-ordinate(7,3) the co-ordinate of other end is(13,9)(8,12)(9,13)(-5,7)(-7,-3)Solution We have to find the value of X & Y. The center of the circle is the midpoint of the diameter Therefore X co-ordinate 7 = X + 5 /2 14 =X +5 X = 9 For Y co-ordinate 3 = -7 + Y/2 6 = -7 + Y Y = 13 Thus the co-ordinate of end B is (9, 13)Q 3 the relation between X & Y such that point (X, Y) is equidistant from point (7, 1) & (3, 5) isX –Y = 8X = Y + 2X = -Y + 2Y = X + 2Cannot be determineSolution Let the point P = (X, Y), A = (7, 1) & B = (3, 5) From the given condition AP = PB (AP) 2 = (PB) 2 (X -7)2 + (Y-1)2 = (X - 3)2 + (Y – 5)2 X2 – 14X + 49 + Y2 – 2Y + 1 = X2– 6X + 9 + Y2 – 10Y +25 -14X + 6X – 2Y + 10Y = 34 – 50 -8X + 8Y = -16 X – Y = 2Q4 If points A (6, 1), B (8, 2), C (9, 4) & D (P, 3) are the vertices of a parallelogram, taken in orderA(6, 1) B(8, 2)D (P, 3) C (9, 4)The value of P is7651215/2Solution Diagonal of parallelogram bisect each otherSo co-ordinate of midpoint AC = so co-ordinate of midpoint BD = Equating the X co-ordinate on both side = 15 = 8 + P P = 15 – 8 P = 7Q5 find the area of rhombus if its vertices are (3, 0), (4, 5), (-1, 4), (-2, -1) taken in orderA(3, 10)B(4, 5) C(-1, 4) D(-2,-1) 24363242None of theseSolution Area of rhombus = ½ X diagonal 1 X diagonal2AC = = = = 4BD = = = = √72 = 6√2Hence the area of rhombus = ½ X 4√2 X 6√2 = 24Q6 the equation of a line which cuts of an intercept T on the Y axis & slope 3 isY = 7X + 3Y = 3X + 7X = 3Y + 7X = 7Y + 3Cannot be determineSolution The equation of a straight line which has slope M & cut an intercept ‘C’ on Y axis is given by Y = MX + C M = 3& C = 7 So the equation of the line is Y = 3X + 7Q7 the graphs of which of the following are same?I Y = X/2 + ½II Y+1 = ½(X + 3)III Y – 2 = ½(X - 3)I & IIALL OF THESEI & IIIII & IIISOLUTIONI graph Y = + II graph Y + 1 = Y = – 1 Y = III graph isY - 2 =Y - = + 2 Y = Thus all three graphs I, II & III are the same as they have same slope ½ & same Y intercepts ½.Q8 find the area of triangle whose vertices A, B ,C are respectively (3,4) , (-4,3) ,(8,6)9/29183624Solution If (X1, Y2), (X2, Y2) and (X3, Y3) are the co-ordinate of a triangle then area of triangle is given by½ [ X1 (Y2 – Y3) + X2(Y3 – Y1) + X3(Y1 – Y2)]A(X1, Y1) = (3, 4),B(X2, Y2) = (-4, 3)and C(X3,Y3) = (8,6)So Area of Δ ABC= |½ [ [3-6] + (-4) [6-4] + 8(4-3)] |=|½ [ [3(-3) + (-4) (2) + 8(1)|= |½ [-9 -8 + 8]|= |-9/2|Q9 Two vertices of a triangle are (-1, 4) and (5, 2).If it is centroid is (0,-3) find vertex.A( x1, y1)BX2,y2CX3,y3 G (xo,yo)(-4,-17)(17,4)(4,17)(3,15)Cannot be determinedSolutionLet the third vertex be (X, Y)The centroid of the given triangle is X0 = X1 + X2 + X3 / 3Y0 = Y1 + Y2 +Y3 / 30 = X + 5 + (-1)/3X + 4 = 0X = -4-3 = Y + 4 +2/3-9 = Y + 8 Y = -17So X = -4, Y = -17Q10 The X and Y intercepts for the equation 6X + 3Y = 2 is 6,33, 1⅓ , 2/3½ , 3/21,6SolutionThe equation of a straight line which makes an intercept ‘a’ on x-axis and ‘b’ on y-axis is given by + = 16x + 3y = 2 + = 1 + = 1Thus intercept is 1/3 and y intercept is 2/3Medium Q1 In standard (X, Y) co-ordinate plane, what is the distance in standard units from the point (-4, 3) to the point (2, 3)?14 units12 units6units8 units10 unitsSolution Distance between two points (X1, Y1) & (X2, Y2) is given byLet 0 = (-4, 3) P = (2, 3)OP = = = 6 unitsQ2 if the distance between (9, 3) & (b, a)861214Cannot be determineSolution 10 = = 100 = (b-a) 2 + 36(b-a) 2 =64b-a = 8Q3 what is the slope intercept from equation 5X – 7Y = 30Y = + Y = - Y = + Y = - Y = + SOLUTIONIf a line has an slope m & cuts an intercept ‘c’ on Y axis then Y = mx + CSO 5X – 7Y = 307Y = 5X – 30Y = - Is the slope intercepted from of given equationQ4 if the area of ΔPQR is 12, then the Y co-ordinate R (X, Y) is R (x, y)Q (-6,10)P(0,0)-44-22Cannot be determineSolution Area of ΔPQR = ½ X base X altitudeAltitude is equal to Y co-ordinate of the point RTherefore 12 = ½ X PQ X Y PQ = = 6 12 = ½ X 6X YY = 4 unitsQ5. The area of a circle is (0, 0) is 9π. The circle passes through all of the points except (-3, 0))(3, 0)(0,3)(0,-3)(3, 3)Solution Area of circle = πr2 Where r is radius or circle 9π = πr 2 r = 3Let us take a point (X, Y) on the circle (X-0)2 + (Y-0)2 = (3)2X 2+Y2 =9Hence option a b c d is satisfied by the equation except ‘e’PO (3,3)Q6. If the area of the circle O is kπ. What is the value of k?918612Cannot be determineSolution OP is the radius of the circle ODistances between two points are O(X1, Y1) & P (X2, Y2) OP OP = = =Hence the area of circle =π (OP) 2 = 18π 18π = kπ K = 18-26-2,6Q7. Point (-2,6) is the canter of the circle that is a tangent to the X axis. The co-ordinate of point of tangency is(-2, 0)(0, 4)(-2, -6)(6, 0)(0, -2)Solution Thus from the figure we can see that co-ordinate of tangency is (-2, 0) Q8. The point P (0, -4) is the midpoint of line OPC, where O is at the origin (0, 0). The co-ordinate of C are (2,1)(0,-8)(8,2)(8, 4)(4, 4)Solution P is the midpoint so for X co-ordinate = 0 X = 0For Y co-ordinate = -4 Y = -8Thus the co-ordinate is (0, -8)Q9. If c≠0 & the slope of the line passing through (-c, c) & (3c, a) is perpendicular, which of the following is an expression for a in term of c-3c5c–c/32c3cSolution If two point (X1, Y1) & (X2, Y2) lie on same line slope is given by = 1 = 1 = a-c= 4ca = 5c2112-1-2-1 -2Q10. The following points are joined: (-2, -2) (-1, -1) (1, 1) (2, 2) all of the following correctly describe the result expect The line formed is II to the X axisThe line formed passes through originAny point on line formed is equidistant from X axis & Y axis.A straight line is formed The line formed bisect the right angle formed by the co-ordinate of axisSolution Draw the point in Cartesian co-ordinate systemThus it satisfy all option except (a) Hard Q1. The vertices of a triangle are A (1,1), B (4,5) & C (6,13). Find the value of A(1,1) B(4,5)C(6,13)61/6562/6563/651Cannot be determineSolution = AB = = = = 5AC = = = = 13BC = = = = = = = = 63/ 65Q2.if the distance between (a, 2) & (3, 4) is 8, then find the value of ‘a’3 +√60-3 + √60√603 +√ 303 - √30Solution8 = Squaring both side of the above equation we get(8)2 = (a - 3)2 + (-2)264 = (a-3)2 + 4(a-3)2 = 60(a-3) = √60a = 3 + √60Q3, the vertices of triangle PQR are P = (15,1), Q = (9,3) & R = (4,5). Find the slope of the altitude to side PQ?3-2/31/3-1/34/9Solution If two lines with slope m1 & m2 are perpendicular then M1 = - 1/m2Slope of P (15, 1) & Q (9,3) is M 1= = 2/-6 = -1/3Let the slope of altitude be m2 M1 m2 = -1M 2 = = 3To find the slope of perpendicular using graphing calculator, enter the given slope, then press the reciprocal key & put the ‘-be’ sign.P(0,-4)R(3,-1)S(6,1.5)Q(3,4)Q4 what is the relationship between the area of ΔPQR & ΔSQR in the given figureEqualArea ΔPQR = Area Δ SQRArea ΔPQR >Area Δ QSRArea ΔPQR +1= Area Δ QSRArea ΔPQR =1/3 Area Δ SQRSolution Altitude from P to QR is 3. Using distance formula the value of QR = = √ (-5)2 = 5Area of ΔPQR = ½ X 5X3 = 15/2Altitude from S to QR is 3 & QR = 5Area of Δ QRS = ½ X 5 X3 = 15/2Thus Area ΔPQR = Area Δ SQRQ5 ABCD is a parallelogram such that its co-ordinate are A (1, 1), B (4, 2), C (3, 3) then D is what(-2, -4)(2, -4)(2, 4)(-4, -2)(-4, 2)Solution Diagonal of parallelogram bisect each other. Thus E is midpoint of DB & AC.Let the co-ordinate of D be (X, Y)For diagonal BD E is midpoint For X co-ordinate = 3 4 + X = 6 X = 2For Y co-ordinate = 3 2 +Y = 6 Y = 4Thus the co-ordinate is (2, 4)Q6. If P = (-1,2) & Q = (2,-1) where P & Q are two points in co-ordinate plane then what is the length of segment PQ?3.461234.24Solution PQ = = = = √18 =3√2Using the calculator find the value of √18Q7. Which of the following has the greatest Y intercept?5X + 6Y = 7X + 2Y = 33X + 4Y = 52X + 3Y = 44X + 5Y = 7SOLUTIONThe equation of an straight line which has slope ‘m’ & cut an intercept ‘c’ on y axis is Y = mX + c So for all equation we get Y = + Y = + Y = + Y = + Y = + 4X + 5Y = 7Q8. If the point (b,-2) is on the graph of the equation 2X + Y = 7 then b =4-5-332SolutionPoint (b,-2) lies on 2X + Y = 7 SO We can write 2 X b – 2 = 7 2b = 7+2 b = 9/2 = 4Q9.if points (a,0) (0,b) & (1,1) are collinear then + = 1a + b = 1a + b = 2 + = 2Cannot be determineSolution Area of Δ = If the point is collinear then the area of Δ is zero = 0= a (b-1)-0(0-1) +1(0-b) = 0= ab-a-b = 0= ab = a+b= = 1 + = 1Q10. A parallelogram JKLM has co-ordinate J (-5,2), K(-2,6), L(5,6), M(2,2). What is the area of parallelogram JKLM?21244212Cannot be determineSolution Area of parallelogram = base X altitude Altitude is 3 unitsBase JL = =√ 49 = 7Area of JKLM = 7 X 3 = 21 units