Learn Insta try to provide online math tutoring for you. Let k be the ratio in which P( −1, y) divides the line segment joining the points A(−3, 10) and B(6, −8). 0=-4+x2⇒x=4 Hence, the correct answer is option (a). Then, by the section formula: Then Then Area of ∆ACD=12x1y3-y4+x3y4-y1+x4y1-y3                       =1230-19+1419+1+9-1-0                       =12-57+280-9=107 sq. Hence, the correct answer is option (d). Area▱ABCD=12×AC×BD                          =12×42×62=24 sq. Hence, the required relation is 4x − 5y − 3 = 0. In what ratio is the line segment joining the points A(−2, −3) and B(3, 7) divided by the y-axis? So, units Coordinates of midpoint of AB=Px1, y1=2+42,1+32=3, 2Coordinates of midpoint of BC=Qx2, y2=4+22,3+52=3, 4Coordinates of midpoint of AC=Rx3, y3=2+22,1+52=2, 3 That is, (AP)2=(BP)2=(CP)2 Area of ∆ABC=12x1y2-y3+x2y3-y1+x3y1-y2                       =12-3-4+1-2-1+1+4-1+4                       =129-0+12=212 sq. x=3×2+4×-23+4=6-87=-27y=3×-4+4×-23+4=-12-87=-207 AB=5⇒x2-x12+y2-y12=5⇒1-42+0-p2=5⇒-32+-p2=25⇒9+p2=25⇒p2=16⇒p=±16⇒p=±4, View NCERT Solutions for all chapters of Class 10. ⇒x-52+y-32=x-52+y+52  ⇒x2-10x+25+y2-6y+9=x2-10x+25+y2+10y+25  ⇒x2-10x+y2-6y+34=x2-10x+y2+10y+50  ⇒x2-10x+y2-6y-x2+10x-y2-10y = 50-34  ⇒-16y = 16  ⇒y =-1616=-1  And BP2=CP2⇒ x-52+y+52=x-12+y+52⇒ x2-10x+25+y2+10y+25=x2-2x+1+y2+10y+25⇒ x2-10x+y2+10y+50=x2-2x+y2+10y+26⇒ x2-10x+y2+10y-x2+2x-y2-10y = 26-50⇒ -8x = -24 ⇒x =-24-8= 3 On solving equation (iv), using equations (i), (ii) and (iii), we get Let C(1, 2a + 1) be the mid point of AB. Let the point Px, y be equidistant from the points A(7, 1) and B(3, 5). Given: A-4,-1,B-2,-4,C4,0 and D2,3In rectangle the opposite sides are equalUsing d=x2-x12+y2-y12AB=-4+22+-1+42AB=13BC=4+22+0+42BC=52CD=2-42+3-02CD=13AD=2+42+3+12AD=52AB=CD and BC=AD So, the midpoint of AB is (2, 3). Then, by section formula, the coordinates of P are Now Then D1,2 Hence, the correct answer is option (d). Now ⇒7y2-42y-49=0⇒y2-6y-7=0⇒y2-7y+y-7=0⇒yy-7+1y-7=0 x1+x32=8          and          y1+y32=9⇒x1+x3=16     and          y1+y3=18               .....ii Hence, required ratio is 1 : 3 and y=-32. (iii) Join AC and BD, intersecting at O. The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1). Point P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts. 5+y2=6⇒5+y=12⇒y=12-5=7 x=13x1+x2+x3  =13-1+5+8  =4 Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P. Since, the coordinates of A are (1, −4), therefore Thus, the coordinates of B are (0, 3). So using section formula, we get So Find the area of this rhombus. Then Hence, the required coordinates are R4, 0 and P0, 43 or P0, -43. The coordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1 is    [CBSE 2012] Suppose O(2, −5) be the point of intersection of the diagonals AC and BD. The given points are A(3, 0), B(6, 4) and C(− 1, 3). (ii) P(−11, −8) and Q(8, −2). Length of each of its diagonal is Also, diagonal AC = diagonal BD TS Grewal Solutions for Class 12 Commerce, TS Grewal Solutions for Class 11 Commerce, Homework Questions for Class 11 Humanities, Homework Questions for Class 12 Humanities, CBSE Class 10 Board Paper Solutions for Math, CBSE Class 10 Board Paper Solutions for Science, CBSE Class 10 Board Paper Solutions for Social Science, CBSE Class 10 Board Paper Solutions for English, CBSE Class 10 Board Paper Solutions for Hindi, CBSE Class 12 Science Board Paper Solutions for Math, CBSE Class 12 Science Board Paper Solutions for Physics, CBSE Class 12 Science Board Paper Solutions for Chemistry, CBSE Class 12 Science Board Paper Solutions for Biology, CBSE Class 12 Commerce Board Paper Solutions for Economics, CBSE Class 12 Commerce Board Paper Solutions for Accountancy, CBSE Class 12 Commerce Board Paper Solutions for Business Studies, CBSE Class 12 Commerce Board Paper Solutions for Math, CBSE Class 12 Humanities Board Paper Solutions for English. (a) a = b [CBSE 2013]. Let A(1, −1), B(5, 2) and C(9, 5) be the given points. The point on x-axis which is equidistant from points A(−1, 0) and B(5, 0) is                  [CBSE 2013] Therefore, AB=5-12+2+12=42+32=25=5 unitsBC=9-52+5-22=42+32=25=5 unitsAC=9-12+5+12=82+62=100=10 units (c) 2 1+22=-1+x2⇒x-1=3⇒x=1+3=4 Here, x1=2, y1=3, x2=5, y2=k and x3=6, y3=7. Also, the area of the triangle OAB = 34×side2 Therefore, the required ratio is 23 : 1, which is same as 2 : 3. and (AC)2 = 1022=200 Then, the coordinates of its centroid are [CBSE 2015]. Thus, (AB)2+(BC)2 = (AC)2 For what value of k (k > 0) is the area of the triangle with vertices (−2, 5), (k, −4) and The given points are P(a + b, a − b) and Q(a − b, a + b). AB = -5+32+-5-22 = -22+-72 = 4+49 = 53 units.BC = 2+52+-3+52 = 72+22 = 49+4 = 53 units.CD = 4-22+4+32 = 22+72 = 4+49 = 53 units.DA = 4+32+4-22 = 72+22 = 49+4 = 53 units.Therefore, AB = BC = CD = DA = 53 unitsAC = 2+32+-3-22 = 52+-52 =25+25 = 50 =25×2= 52 unitsBD = 4+52+4+52 = 92+92 = 81+81 = 162 =81×2= 92 unitsThus, diagonal AC is not equal to diagonal BD. Prove that the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right triangle. Hence, the point on the y-axis is (0, 9). [CBSE 2015]. [CBSE 2014]. 0=-2+x2⇒x=24=3+y2⇒y=8-3=5 If A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC and AD is median, then the coordinates of D are [CBSE 2014]. It is given that the points A, B and C are collinear. Hence, the given points are collinear. Then x = x1+x22, y =y1+y22⇒x = 3+-52, y =0+42⇒x = -22, y = 42⇒x = -1, y = 2 Therefore, As per the question AB=5-02+0-32    =52+-32    =25+9    =34 units Thus, the x-axis divides the line AB in the ratio 2:3 at the point P. (a) (−6, 7)                                  (b) (6, −7)                                 (c) (4, 2)                                     (d) (5, 3). Hence, the coordinates of P are (73, −2). We have A-2,5 and B2,-3AX=BXAX2=BX2    .....1Using distance formula:d=x2-x12+y2-y12For AXAX=x--22+0-52AX2=x+22+-52AX2=x2+4x+29     .....2 Hence, the required coordinates are A33, 0, B0, 3 and D-33, 0 or A-33, 0, B0, 3 and D33, 0. Hence, the given points are the vertices of an equilateral triangle. So, the area of the triangle ∆DEF is 1 sq. Let G(x, y) be the centroid of ∆ABC. PQ = x2-x12+y2-y12       =a cos α-a sin α2+-a sinα-a cos α2       =a2cos2α +a2 sin2 α - 2a2cos α×sin α+a2 sin2 α+a2cos2α+2a2cos α× sin α       =2a2cos2α +2a2 sin2 α       =2a2cos2α + sin2 α       =2a21                  From the identity cos2α + sin2 α = 1       =2a2       = 2a units, Find the distance of each of the following points from the origin: Also, find the point of division. Thus, PQ = QR = RS = SP and PR≠QS therefore PQRS is a rhombus. Also, find the value of k. Let the point P(−3, k) divide the line AB in the ratio s : 1. Hence, the coordinates of the required point are (1, 3). 7k+2k+1=4  and  8k+3k+1=5                ∵ C4, 5 is given⇒7k+2=4k+4  and  8k+3=5k+5 ⇒3k=2  and  3k=2 By section formula, the coordinate of point P will be So the point P divides the line segment AB in ratio 1: 2. Hence, the correct answer is option (c). ⇒4+a=0 and 6+b=-9 (d) right-angled. Then (x1 = a + b, y1 = a − b) and (x2 = a − b, y2 = a + b) The origin is the midpoint of the base. Learn the basics coordinate geometry questions with the help of our given solved examples that help you to understand the concept in the better way. The midpoint of segment AB is P(0, 4). Find the third vertex of ∆ABC if two of its vertices are B(−3, 1) and C(0, −2) and its centroid is at the origin. The vertices of the rectangle ABCD are A(2, −1), B(5, −1), C(5, 6) and D(2, 6). So, PA = PB. ∵AB=BC=CD=AD=5 and AC≠BD Let F be the midpoint of AB. The given points are collinear if Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus. Hence, the correct answer is (b). A line intersects the y-axis and x-axis  at the points P and Q respectively. Applect Learning Systems Pvt. Hence, x = 2 or −6 and BC=41 units. Then Area∆ABC=12x1y2-y3+x2y3-y1+x3y1-y2⇒6=12k+1-3+k+4-k-1+71+3⇒6=12k2-2k-3-4k-4+28⇒k2-6k+9=0 Then, Also, find the value of y. Area of ∆ADC=12x1y2-y3+x2y3-y1+x3y1-y2                       =1271+1+4-1+3+3-3-1                       =1214+8-12=5 sq. Coordinates of Q=2×6+3×12+3, 2×7+3×22+3                           =12+35, 14+65=3, 4 AB=0+22+1-52=22+-42=20=25 unitsBC=2-02+-3-12=22+-42=20=25 unitsAC=2+22+-3-52=42+-82=80=45 units Area∆ABC=12x1y2-y3+x2y3-y1+x3y1-y2                   =1212-2+32+4-1-4-2                   =120+18+6                   =12 sq. The point  3, a lies on the line 2x-3y=5.If point 3, a lies on the line 2x-3y=5 , then2x-3y=5⇒2×3-3×a=5 Now a2=-6-22⇒a2=-4⇒a=-8 Let k : 1 be the ratio in which the point P34, 512 divides the line segment joining the points A12, 32 and 2,-5. It is given that the points are collinear. Then, by the section formula: Hence, the length of the diagonal is 5 units.. Let the required point be P(x, y). Then the coordinates of its centroid are ∵AB=AC and AB2+AC2=BC2 Let A(a, b) and B(1, 4) be the two end-points of the given diameter AB. x+82=6+92⇒x=7y+22=1+42⇒y=3 AB=0+22+2-02=8=22CB=0-22+2-02=8=22AC=2+22+0-02=4 ⇒6-x2+-1-y2 = 2-x2+3-y2⇒x2-12x+36+y2+2y+1 = x2-4x+4+y2-6y+9⇒x2+y2-12x+2y+37 = x2+y2-4x-6y+13⇒x2+y2-12x+2y-x2-y2+4x+6y =13-37⇒-8x+8y = -24⇒-8x-y = -24⇒x-y =-24-8⇒x-y =3
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