C1-5+a3, -6+2+b3C-4+a3, -4+b3 BC=4+12+5-42=52+12     =25+1=26 Find the ratio in which the point P34, 512 divides the line segment joining the points (i) A(1, 2), B(−2, 3) and C(−3, −4) (a) 5                                  (b) 4                                 (c) 3                                     (d) 25. Now, we will prove for rectangles because diagnols in a rectangle are also equalAC=4--42+0--12AC=4+42+12AC=65AndBD=2--22+3--42BD=42+72BD=65We can see that diagonals are also equal Therefore, given coordinates are of rectangle. Area of triangle=12x4-5+45-3+33-4⇒4=12x-1+42+3-1⇒4=12-x+8-3⇒4=12-x+5⇒8=-x+5⇒-x+5=8 or -x+5=-8⇒-x=3 or -x=-13⇒x=-3 or x=13 (x1 = 3, y1 = 8), (x2 = −4, y2 = 2) and (x3 = 5, y3 = −1) The coordinates of the point Q are (−4, 0) Copyright © x1y2-y3+x2y3-y1+x3y1-y2=-55-7+57-1+101-5=-5-2+56+10-4=10+30-40=0 Similarly, The given points are A(4, p) and B(1, 0) and AB = 5. But P lies on the x-axis; so, its ordinate is 0. AP=10⇒x-112+0+82=10⇒x-112+82=100                     Squaring both sides⇒x-112=100-64=36 Find the area of this triangle. x = x1+x22, y =y1+y22⇒x = 1+52, y =6+-22⇒x = 62, y = 42⇒x = 3, y = 2 ⇒-10x-2y=2x-10y⇒8y=12x⇒3x=2y The given points are A(3, 0), B(4, 5), C(− 1, 4) and D(− 2, − 1). (d) None of these, (c) 72,92 Hence, AD = BE = 5 units. Find the ratio in which the point P(m, 6) divides the join of A(−4, 3) and B(2, 8). (iii) A(5, 1), B(1, −1) and C(11, 4) If the coordinates of point A are 33, 0, then the coordinates of D are -33, 0. Let (0, y) be the coordinates of B. [CBSE 2012]. (i) Find the coordinates of the point that divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3. Let the point P2411, y divides the line PQ in the ratio k : 1. ∴AB+BC=5+5 units=10 units=AC AC=3+12+0-32=42+-32     =16+9=25=5 Let the coordinates of P be (0, y). Thus, PQ2+PR2=QR2 In Coordinate Geometry of Class 9, we learned what is x and y coordinate of a point. units AB=0-32+2-p2     =-32+2-12                ∵p=1     =9+1     =10 units Then Show that the points A(5, 2), B(2, − 2) and C(− 2, t) are the vertices of a right triangle with ∠B=90∘, then find the value of t. ∵∠B=90∘∴AC2=AB2+BC2⇒5+22+2-t2=5-22+2+22+2+22+-2-t2⇒72+t-22=32+42+42+t+22 Hence, the required ratio is 3 : 2 and y = 6. Calculate its area. (v) P(a + b, a − b) and Q(a −b, a + b) 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. (c) p = ±4 only Therefore, coordinates of points P and Q are (5, -3) and (3, -4) respectively. Hence,  the values of k is –8. x=13x1+x2+x3  =13-1+5+8  =4 AC=6-92+1-42=-32+-32     =9+9=32BD=8-72+2-32=12+-12     =1+1=2 • Then, These solutions for Coordinate Geometry are extremely popular among Class 10 students for Math Coordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. The given points are A(2, 3), B(5, k) and C(6, 7). Therefore, [CBSE 2015]. Find the values of y. Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P. Therefore If the points P(a, −11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS, find the values of a and b. So Therefore, the coordinates of point C are (2, 6). Area of ∆ABD=12x1y2-y3+x2y3-y1+x3y1-y2                       =1273-1+51+3+4-3-3                       =1214+20-24=5 sq. Let the required ratio be k : 1. Hence, the points are: A(11,12), B(1,2) and C(5,6). ∵AB=AC and AB2+AC2=BC2 Let the line​ 2x+y-4=0 divide the line segment in the ratio k : 1 at the point P. (i) A(−4, −1), B(−2, −4) C(4, 0) and D(2, 3)    Also, find the coordinates of the points of division. The given points are A(7, −4) and B(−5, 1). ar(∆ABC)=12(x1)(y2-y3)+(x2)(y3-y1)+(x3)(y1-y2)⇒ar(∆ABC)=12(-7)(-7+8)+(-6)(-8-5)+(-3)(5+7)⇒ar(∆ABC)=12(-7)(1)+(-6)(-13)+(-3)(12)⇒ar(∆ABC)=12(-7)+78+(-36)⇒ar(∆ABC)=1235⇒ar(∆ABC)=352 sq. Solution: True Therefore, (−1, 2) are the coordinates of mid point of AB. In ∆ABC, Hence, the given points are collinear. Let E be the midpoint of AC. The vertices of the rectangle ABCD are A(2, −1), B(5, −1), C(5, 6) and D(2, 6). The given points are A(−6, 7) and B(−1, −5). Therefore, O is the mid-point of AC and BD. So, the condition for three collinear points is Using section formula, coordinates of P are; This show that ∆ABC is right- angled at B. Now Two vertices of ∆ABC are A(−1, 4) and B(5, 2) and its centroid is G(0, −3). ⇒x-11=±6⇒x=11±6⇒x=11-6, 11+6⇒x=5, 17 units. [CBSE 2017]. Then Then, by section formula the coordinates of P are DF=4+42+0-02=8FE=0-42+4-02=42DE=0+42+4-02=42 Let ABCD be the parallelogram with two adjacent vertices A(3, 2) and B(−1, 0). x1+x32=8          and          y1+y32=9⇒x1+x3=16     and          y1+y3=18               .....ii Find a relation between x and y, if the points A(x, y), B(−5, 7) and C(−4, 5) are collinear. Area▱ABCD=12×AC×BD                          =12×32×2=3 sq. [CBSE 2014]. So, the coordinates of D are Hence, the correct answer is option (b). equal and bisect each other. The distance of a point (x, y) from x-axis is y. Let the point P(2, 5) divide AB in the ratio k : 1. x = x1+x22, y =y1+y22⇒x = 6+-22, y =-5+112⇒x = 6-22, y = -5+112⇒x = 42, y = 62⇒x = 2, y = 3 and x = mx2+nx1m+n, y =my2+ny1m+nThe coordinates of P are (-3, k).-3 = -2s-5s+1, k = 3s-4s+1⇒-3s-3 = -2s-5,  ks+1 = 3s-4⇒-3s+2s =-5+3,  ks+1 = 3s-4⇒-s = -2,  ks+1 = 3s-4⇒s =2,  ks+1 = 3s-4Therefore, the point P divides the line AB in the ratio 2 : 1.Now, putting the value of s in the equation ks+1 = 3s-4, we get:k2+1 = 32-4⇒3k = 6-4⇒3k = 2 ⇒k = 23Therefore, the value of k = 23That is, the coordinates of P are (-3, 23). The midpoint of the line segment joining A(2a, 4) and B(−2, 3b) is C(1, 2a + 1). (c) 2a = b [CBSE 2017]. Hence, the given points are the vertices of an equilateral triangle. Then (c) scalene AP=BP⇒0-62+y-52=0+42+y-32⇒62+y-52=42+y-32⇒62+y-52=42+y-32                     Squaring both sides AP=2+22+2-k2      =42+2+12                   =16+9=5 units (iv) Q(a,b)=2×1+1×73,2×-5+1×-23Q(a,b)=93,-123=3,-4 But it is given that the centroid is G0,-3. x1y2-y3+x2y3-y1+x3y1-y2=0⇒2k+3+4-3-3+63-k=0⇒2k+6-24+18-6k=0⇒-4k=0⇒k=0, The distance of the point P(−6, 8) from the origin is                              [CBSE 2013C] Also, find the value of k. Let the point P(−3, k) divide the line AB in the ratio s : 1. Now, ar(quad. Then Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Let A(x1 = 0, y1 = −1), B(x2 = 2, y2 = 1) and C(x3 = 0, y3 = 3) be the given points. AB = 5-22+2-12 = 32+12 = 9+1 =10 unitsBC = 6-52+4-22 = 12+22 = 1+4 =5 unitsCD = 3-62+3-42 =-32+-12 = 9+1 =10 unitsAD = 3-22+3-12 = 12+22 = 1+4 =5 unitsThus, AB = CD = 10 units and BC = AD =5 unitsSo, quadrilateral ABCD is a parallelogramAlso, AC = 6-22+4-12 = 42+32 = 16+9 = 25=5 unitsBD = 3-52+3-22 =-22+12 = 4+1 = 5 units Area∆ABC=12x1y2-y3+x2y3-y1+x3y1-y2                   =12672-3+83-1+71-72                   =1232                   =34 sq. If the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3), find the value of x. x, y=15+221+2,1-8+211+2⇒x, y=5+43,-8+23⇒x, y=93,-63⇒x, y=3,-2 AB = -2-72+5-102 = -92+-52 =81+25 =106BC = 3--22+-4-52 = 52+-92 =25+81 = 106AC = 3-72+-4-102 = -42+-142 = 16+196 =212 ∴AB+BC=5+10 units=15 units=AC Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry with solutions includes all the important topics with detailed explanation that aims to help students to score more marks in Board Exams 2020. x1y2-y3+x2y3-y1+x3y1-y2 = 0⇒ab-1+01-0+10-b=0⇒ab-a-b = oDividing the equation by ab:⇒1-1b-1a = 0⇒1-1a+1b = 0⇒1a+1b=1  [CBSE 2017]. But P lies on the y-axis; so, its abscissa is 0. unit Let the coordinates of A be (x, 0). ⇒8k-4=0 Hence, the points A(2, 4), B(2, 6) and C2+3, 5 are the vertices of an equilateral triangle. (a) 3 units ∵AB=BC=CD=AD=5 and AC≠BD Take a look at the figure below. Let Ax1, y1=A-3, -1, Bx2, y2=B-2, -4, Cx3, y3=C4, -1 and Dx4, y4=D3, 4. Hence, the point on the y-axis is (0, 9). 0=-3+y2⇒y=3 (iii) The given points are P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1). It is given that the points A, B and C​ are collinear. Is this figure a rectangle? Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Using coordinate geometry, it is possible to find the distance between two points, dividing lines in a ratio, finding the mid-point of a line, calculating the area of a triangle in the Cartesian plane, etc. Since, the coordinates of A are (1, −4), therefore Let (x, y) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally.

coordinate geometry class 10 example 6

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coordinate geometry class 10 example 6 2020