2<\/sup> – 6y
\n\u21d2 61 – 10y = 25 – 6y
\n\u21d2 61 – 25 = – 6y + 10y
\n\u21d2 4y = 36
\n\u21d2 y = 9
\nRequired point P(0, 9)
\nHence choice (a) is correct. (a)<\/p>\nQuestion 7.
\nA point lies in third quadrant. Co-ordinates of this point may be :
\n(a) (5, 5)
\n(b) (5, -5)
\n(c) (-5, -5)
\n(d) (-5, 5).
\nAnswer:
\n(c) (-5, -5)<\/p>\n
Question 8.
\nThe distance between the points (5, – 3) and (8, 1) is:
\n(a) 5 units
\n(b) 6 units
\n(c) 25 units
\n(d) 10 units.
\nAnswer:
\n(a) 5 units<\/p>\n
<\/p>\n
Question 9.
\nThe distance between the points (2, 0) and (-1, 4) is :
\n(a) 5 units
\n(b) 25 units
\n(c)- 25 units
\n(d) 10 units.
\nAnswer:
\n(a) 5 units<\/p>\n
Question 10.
\nThe distance between the points (-6, 5) and (-1, 7) is :
\n(a) 169 units
\n(b) \u221a119 units
\n(c) \u221a13 units
\n(d) None of these.
\nAnswer:
\n(d) None of these.<\/p>\n
Question 11.
\nThe co-ordinates of a point Mare (3, 4). Its distance from the origin is :
\n(a) 7 units
\n(b) 11 units
\n(c) 5 units
\n(d) 12 units.
\nAnswer:
\n(c) 5 units<\/p>\n
Question 12.
\nThe distance between points (7, 3) and (-5, -2) is :
\n(a) 10 units
\n(b) 13 units
\n(c) 16 units
\n(d) 18 units.
\nAnswer:
\n(b) 13 units<\/p>\n
Question 13.
\nThe distance between origin and (15, 8) is :
\n(a) 18 units
\n(b) 15 units
\n(c) 17 units
\n(d) 28 units.
\nAnswer:
\n(c) 17 units<\/p>\n
Question 14.
\nThe distance between (-1, -3) and (3, 0) is :
\n(a) 4 units
\n(b) 5 units
\n(c) 13 units
\n(d) 16 units.
\nAnswer:
\n(b) 5 units<\/p>\n
<\/p>\n
Question 15.
\nThe distance between the points (-3, 4) and (0, 0) will be :
\n(a) 5
\n(b) 4
\n(c) 13
\n(d) 1
\nAnswer:
\n(a) 5<\/p>\n
Question 16.
\nThe third vertex of an equilateral triangle whose other two vertices are (1, 1) and (1, -1) respectively is :
\n(a) (\u221a3, – \u221a3)
\n(b) (- \u221a3, \u221a3)
\n(c) both (a) and (b)
\n(d) none of these.
\nAnswer:
\n(c) both (a) and (b)<\/p>\n
Question 17.
\nThe coordinates of two points are (6, 0) and (0, 8). The coordinates of the mid-point are :
\n(a) (3, 4)
\n(b) (6, 8)
\n(c) (0, 0)
\n(d) None of these.
\nAnswer:
\n(a) (3, 4)<\/p>\n
Question 18.
\nThe co-ordinates of two points are (9, 4) and (3, 8). The co-ordinates of the mid-point are :
\n(a) (6, 0)
\n(b) (0, 6)
\n(c) (6, 6)
\n(d) (-6, -6).
\nAnswer:
\n(c) (6, 6)<\/p>\n
<\/p>\n
Question 19.
\nThe co-ordinates of two points are (-8, 0) and (0, -8). The co-ordinates of the midpoint of the line segment joining them will be :
\n(a) (-8, 4)
\n(b) (4, -8)
\n(c) (-4, -4)
\n(d) (-4, 4)
\nAnswer:
\n(c) (-4, -4)<\/p>\n
Question 20.
\nThe abscissa of the point which divides the line joining points (0, 4) and (0, 8) internally in the ratio of 3 : 1 will be :
\n(a) 0
\n(b) 4
\n(c) 6
\n(d) 5
\nAnswer:
\n(a) 0<\/p>\n
Coordinate Geometry Class 10 Extra Questions Very Short Answer Type<\/h3>\n
Question 1.
\nFind the coordinate of a point A. Where AB is the diameter of a circle whose centre is (2, -3) and coordinate of B is (1, 4).
\nSolution.
\nAs centre is the mid point of AB.
\n
\nLet coordinate of A is (x, y)
\n\u2234 \\(\\frac{x+1}{2}\\) = 2
\n\u21d2 2 x + 1 = 4
\n\u21d2 x = 4 – 1 = 3
\nand \\(\\frac{y+4}{2}\\) = -3 = y + 4 = -6
\ny = – 10
\nHence coordinate of A is (3, -10)<\/p>\n
<\/p>\n
Question 2.
\nWhat are the ordinate of the point whose abscissa is 10 and whose distance from (2, -3) is 10 units?
\nSolution.
\nLet y be the ordinate, then the point is (10, y). The distance between (10, y) and (2, -3)
\n
\n
\n\u21d2 y2<\/sup> + 6y + 73 = 100
\n\u21d2 y2<\/sup> + 6y – 27 = 0
\n\u21d2 (y +9)(y – 3) = 0
\n\u21d2 y = -9, y = 3.<\/p>\nQuestion 3.
\nFind the distance between the points A(a sin \u03b8, a cos \u03b8) and B(a cos \u03b8, – a sin \u03b8).
\nSolution.
\nHere x1 <\/sub>= a sin \u03b8, x2<\/sub> = a cos \u03b8 y1<\/sub> = a cos \u03b8, y2<\/sup> = – a sin AB
\nDistance AB
\n<\/p>\nQuestion 4.
\nProve that the point (4, 1) is the centre of the circle on whose circumference lie the points (-2, 9), (10, -7) and (12, -5).
\nSolution.
\nLet A(-2, 9), B(10, -7), C(12, – 5) and O(4, 1). Then,
\n
\n\u2234 OA = OB = 0C.
\nHence O is the centre and A, B, C are the points on the circumference of the circle.<\/p>\n
<\/p>\n
Question 5.
\nFind that ratio in which the point \\(P\\left(\\frac{3}{4}, \\frac{5}{12}\\right)\\) divides the line segment joining the points \\(A\\left(\\frac{1}{2}, \\frac{3}{2}\\right)\\) and B(2, -5).
\nSolution.
\nLet P divides AB in the ratio \u03bb : 1
\n
\n\u21d2 3\u03bb + 3 = 8\u03bb + 2
\n\u21d2 5\u03bb = 1
\n\u03bb = \\(\\frac {1}{5}\\)
\nHence P divides AB in ratio 1 : 5<\/p>\n
Question 6.
\nFind the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B.
\nSolution.
\nLet M (0, 0) and N (36, 15) be the given points.
\nHere, x1<\/sub> = 0, y1<\/sub> = 0 and x2<\/sub> = 36, y2<\/sub> = 15
\n
\nSince, the position of towns A and B are given (0, 0) and (36, 15), respectively and so, the distance between them is 39 km.<\/p>\nQuestion 7.
\nFind the coordinates of that point which divides the line segment joining two given points (3, 5) and (8, 10) internally in the ratio of 2 : 3.
\nSolution.
\nHere, x1<\/sub> = 3, x2<\/sup> = 8, y1<\/sub> = 5, y2<\/sup> = 10, m = 2, n = 3.
\n
\n
\nHence, the co-ordinates of the required point are (5, 7).<\/p>\n<\/p>\n
Question 8.
\nIf the area of \u2206ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove x + y = 15.
\nSolution.
\nArea of \u2206 = \\(\\frac{1}{2}\\) [x1<\/sub> (y2<\/sub> – y3<\/sub>) + x2<\/sub>(y3<\/sub> – y1<\/sub>)+ x3<\/sub> y1<\/sub> – y2<\/sub>)]
\n\u21d2 6 = \\(\\frac{1}{2}\\)[x(2 – 1) + 1(1 – y) + 2 (y – 2)]
\n\u21d2 12 = x – y + 1 + 2y – 4
\n\u21d2 12 = x + y = 3
\n\u2234 x + y = 15<\/p>\nQuestion 9.
\nFind the coordinates of point which divide line segment AB joining points A(-2, 2) and B( 2, 8) into four equal parts.
\nSolution.
\nClearly point a is the midpoint of AB
\n<\/p>\n
Question 10.
\nThe coordinates of one endpoint of a diameter of a circle are (3, 5). If the coordinates of the centre be (6, 6), find the coordinates of the other endpoint of the diameter.
\nSolution.
\nLet AB be the given diameter and O the centre of the circle. Then, the coordinates of A and O and (3, 5) and (6, 6) respectively. Let (a, b) be the coordinates of B.
\nThen \\(\\frac{3+a}{2}\\) = 6 and \\(\\frac{5+b}{2}\\) = 6
\n\u2234 a = 9 and 6 = 7.
\nHence, the required point is (9, 7).<\/p>\n
<\/p>\n
Question 11.
\nIf the points (6, 9),(0, x)and (-6, -7) are collinear then find the value of x.
\nSolution.
\nx1<\/sub> = 6, y1<\/sub> = 9
\nx2<\/sub> = 0, y2<\/sub> = x
\nx3<\/sub> = -6, y3<\/sub> = -7
\nThe three points are collinear if the area of the triangle formed by them is zero.
\nNow,
\narea = \\(\\frac{1}{2}\\)[x1<\/sub>(y2<\/sub> – y3<\/sub>) + x2<\/sub>(y3<\/sub> – y1<\/sub>)
\n+ x3<\/sub>(y1<\/sub> – y2<\/sub>)] = 0
\n\u21d2 \\(\\frac{1}{2}\\)[6(x + 7) + 0(-7 – 9) + (-6)(9 – x)) = 0
\n\u21d2 6x + 42 – 54 + 6x = 0
\n\u21d2 12x = 12
\n\u21d2 x = 1.<\/p>\nQuestion 12.
\nFind the value of the points (a, 1), (1, -1) and (11, 4) are collinear.
\nSolution.
\nHere, A(a, 1) = (x1<\/sub>, y1<\/sub>) B(1, -1) = (x2<\/sub>, y2<\/sub>,) andC(11, 4) = (x3<\/sub>, y3<\/sub>)
\nCondition for collinear points
\n\u21d2 x1<\/sub>(y2<\/sub> – y3<\/sub>) + x2<\/sub>(y3<\/sub> – y1<\/sub>) + x3<\/sub>(y1<\/sub> – y2<\/sub>)
\n\u21d2 a(-1 -4) + 1(4 – 1) + 11(1 + 1) = 0
\n\u21d2 -5a + 3 + 22 = 0
\n\u21d2 -5a = -25
\n\u21d2 a = 5
\n\u2234 The given three points are collinear.<\/p>\nCoordinate Geometry Class 10 Extra Questions Short Answer Type<\/h3>\n
Question 1.
\nShow that the points (2, 1),(5, 2), (6, 4) and (3, 3) form a parallelogram.
\nSolution.
\nLet A(2, 1), B(5, 2), C(6, 4) and D(3, 3).
\n
\n
\nThus, AB = CD, BC = DA.
\nSince, opposite sides are equal, hence ABCD is a parallelogram.<\/p>\n
<\/p>\n
Question 2.
\nProve that the points A(1, -2), B(3, 0) and C(1, 2) are the vertices of an isosceles right-angled triangle.
\nSolution.
\nLet A = (1, -2), B = (3, 0) and C = (1, 2).
\n
\nCA = 4
\nNow, AB2<\/sup> + BC2<\/sup> = (2\u221a2)2<\/sup> + (2\u221a2)2<\/sup>
\n= 8 + 8 = 16
\nAB2<\/sup> + BC2<\/sup> = (AC)2<\/sup> …(4)
\nHence, from result (1), (2) and (4) \u2206ABC is an isosceles right-angled triangle.<\/p>\nQuestion 3.
\nFind the centre of the circle which passes through the points (-3, -2), (-2, 3) and (3, 2).
\nSolution.
\nLet O(x, y) be the centre of the circle and the points
\nA(-3, – 2), B(-2, 3) and C(3, 2)
\nlie on the circle, Then, we have
\nOA2<\/sup> = (x + 3)2<\/sup> + (y + 2)2<\/sup>
\nOB2<\/sup> = (x + 2)2<\/sup> + (y – 3)2<\/sup>
\nand OC2<\/sup> = (x – 3)2<\/sup> + (y – 2)2<\/sup>
\n\u2235 OA = OB
\n\u21d2 OA2<\/sup> = OB2<\/sup>
\n\u2234(x + 3)2<\/sup> + (y + 2)2<\/sup> = (x + 2)2<\/sup> + (y – 3)2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> + 6x + 4y + 13
\n= x2<\/sup> + y2<\/sup> + 4x – 6y + 13
\n\u21d2 6x + 4y = 4x – 6y
\n\u21d2 2x + 10y = 0
\n\u21d2 x + 5y = 0 ……(i)
\nAgain, OB = OC
\n\u21d2 OB2<\/sup> = OC2<\/sup>
\n\u2234 (x + 2)2<\/sup> + (y – 3)2<\/sup> = (x – 3)2<\/sup> + (y – 2)2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> + 4x – 6y + 13
\n= x2<\/sup> + y2<\/sup> – 6x – 4y + 13
\n\u21d2 4x – 6y = – 6x – 4y
\n\u21d2 10x – 2y = 0
\n\u21d2 5x – y = 0 …..(ii)
\nOn solving (i) and (ii), we get
\nx = 0 and y = 0.
\nHence, the centre of the circle is (0, 0).<\/p>\n<\/p>\n
Question 4.
\nProve that the triangle whose vertices are respectively (a, a), (-a, -a) and (-a\u221a3, a\u221a3) is an equilateral triangle.
\nSolution.
\nLet points P(a, a), Q(-a, -a) and R(-a\u221a3, a\u221a3) are given, then
\nand
\n
\n= a\u221a8 = 2a\u221a2
\nThus, PQ = QR = RP
\n\u2234 \u2206PQR is an equilateral triangle.<\/p>\n
Question 5.
\nFind the point on the X-axis which is equidistant from (2, -5) and (-2, 9).
\nSolution.
\nSince, the point on the X-axis.
\n\u2234 it’s ordinate = 0
\nSo, A(x, 0) is any point on the X-axis.
\nSince, A(x, 0) is equidistant from B(2, -5) and C(-2, 9).
\nAB = AC
\n\u21d2 AB2<\/sup> = AC2<\/sup>
\n\u21d2 (x – 2)2<\/sup> + (0 + 5)2<\/sup> = (x + 2)2<\/sup> + (0 – 9)2<\/sup>
\n
\n\u21d2 x2<\/sup> – 4x + 4 + 25 = x2<\/sup> + 4x + 4 + 81
\n\u21d2 -4x – 4x = 81 – 25
\n\u21d2 -8x = 56
\n\u21d2 x = –\\(\\frac{56}{8}\\) = -7
\nSo, the point equidistant from given points on the x-asis is (-7, 0).<\/p>\n<\/p>\n
Question 6.
\nFind the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
\nSolution.
\nAccording to question,
\nPQ = 10
\n
\nSquaring both sides, we get
\n\u21d2 y2<\/sup> + 6y + 73 = 100
\n\u21d2 y2<\/sup> + 6y – 27 = 0
\n\u21d2 y2<\/sup> + 9y – 3y – 27 = 0
\n\u21d2 y(y + 9) – 3 (y + 9) = 0
\n\u21d2 (y + 9) (y – 3) = 0
\n\u21d2 y = -9 or y = 3
\n\u21d2 y = -9 or 3<\/p>\nQuestion 7.
\nFind a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
\nSolution.
\nLet the point A(x, y) be equidistant from the points B(3, 6) and (-3, 4).
\nAB = AC
\nAB2<\/sup> = AC2<\/sup>
\n\u21d2 (x – 3)2<\/sup> + (y – 6)2<\/sup> = (x + 3)2<\/sup> + (y – 4)2<\/sup>
\n
\n\u21d2 x2<\/sup> – 6x + 9 + y2<\/sup> – 12y + 36
\n\u21d2 x2<\/sup> + 6x + 9 + y2<\/sup> – 8y + 16
\n\u21d2 -6x – 6x – 12y + 8y + 36 – 16 = 0
\n\u21d2 -12x – 4y + 20 = 0
\n\u21d2 -4(3x + y – 5) = 0
\n\u21d2 3x + y – 5 = 0
\n(\u2235 -4 \u2260 0)<\/p>\n<\/p>\n
Question 8.
\nIf the vertices of a \u2206ABC, are (sin2<\/sup> \u03b8, cos2<\/sup> \u03b8), (cos2<\/sup> \u03b8, sin2<\/sup> \u03b8) and (0, 0) then find the coordinates of the centroid of the triangle.
\nSolution.
\nHere A = (sin2<\/sup> \u03b8, cos2<\/sup> \u03b8)
\n= (x1<\/sub>, y1<\/sub>)
\nB = (cos2<\/sup> \u03b8, sin2<\/sup> \u03b8)
\n= (x2<\/sub>, y2<\/sub>)
\nC = (0, 0)
\n= (x3<\/sub>, y3<\/sub>)
\nThe co-ordinates of the centroid be
\n<\/p>\nQuestion 9.
\nFind the ratio in which the point (11, 15) divides the line segment joining the points (15, 5) and (9, 20).
\nSolution.
\nA(x1<\/sub>, y1<\/sub>) = (15, 5)
\nB(x2<\/sub>, y2<\/sub>) = (9, 20)
\nP(x, y) = (11, 15)
\nm1<\/sub> : m2<\/sub> = ?
\n
\n\u21d2 11(m1<\/sub> + m2<\/sub>) = 9m1<\/sub> + 15m2<\/sub>
\n\u21d2 15(m1<\/sub> + m2) = 20m1<\/sub> + 5m2<\/sub>
\n\u21d2 2m1<\/sub> = 4m2<\/sub>
\n\u21d2 5m1<\/sub> = 10m2<\/sub>
\n\u21d2 m1<\/sub> = 2m2<\/sub>
\n\u21d2 m1<\/sub> = 2m2<\/sub>
\n\u21d2 \\(\\frac{m_{1}}{m_{2}}\\) = \\(\\frac{2}{1}\\)
\n= i.e., m1<\/sub> = m2<\/sub> = 2 : 1.<\/p>\nQuestion 10.
\nFind the coordinates of the point which divides the line segment joining the points (a, b) and (b, a) internally in the ratio of (a – b) : (a + b).
\nSolution.
\nFormula,
\n<\/p>\n
Question 11.
\nThe mid-points of the sides of a triangle are (1, 2), (0, -1) and (2, -1). Find the coordinates of the vertices of the triangle.
\nSolution.
\nLet A(x1<\/sub>, y1<\/sub>), B(x2<\/sub>, y2<\/sub>) and C(x3<\/sub>, y3<\/sub>) be the vertices of \u2206ABC. Let (1, 2), (0, -1) and (2, -1) be the mid-points of AB, BC and CA respectively.
\nThen
\n
\nx1<\/sub> + x2<\/sub> = 2 …..(i)
\nx2<\/sup> + x3<\/sub> = 0 …..(ii)
\nx3<\/sub> + x1<\/sub> = 4 …..(iii)
\ny1<\/sub> + y2<\/sub> = 4 ……(iv)
\ny2<\/sub> + y3<\/sub> = -2 …..(v)
\ny3<\/sub> + y1<\/sub> = -2 ……(vi)
\nAdding (i), (ii) and (iii) we get
\nx1<\/sub> + x2<\/sub> + x3<\/sub> = 3.
\n\u2234 x3<\/sub> = 1, x1<\/sub> = 3 and x2<\/sub> = -1.
\nAdding (iv), (v) and (vi) we get
\ny1<\/sub> + y2<\/sub> + y3<\/sub> = 0.
\n\u2234 y3<\/sub>= – 4, y1<\/sub> = 2 and y2<\/sub> = 2.
\nHence the vertices of the triangle are A(3, 2), B(-1, 2) and C(1, -4) respectively.<\/p>\n<\/p>\n
Question 12.
\nThe three vertices of a parallelogram taken in order are (0, -1), (1, 3) and (2, 2) respectively. Find the coordinates of the forth vertex.
\nSolution.
\nLet A(0, -1), B(1, 3), C(2, 2) and D(x, y) be the vertices of a parallelogram. Then the mid-points of the diagonals AC and BD have the same co-ordinates.
\n
\nHence, the forth vertex of the parallelogram is (1, -2).<\/p>\n
Question 13.
\nFind the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
\nSolution.
\nLet the coordinates of a point are (x, y).
\nwe have, x1<\/sub> = -1, y1<\/sub> = 7;
\nx2<\/sub> = 4, y2<\/sub> = -3
\nand m1<\/sub> = 2, m2<\/sub> = 3
\n\u2234 By using section formula,
\n
\nHence, coordinates of the point are (1, 3)<\/p>\nQuestion 14.
\nIf (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
\nSolution.
\nLet A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
\n
\nSince, ABCD is a parallelogram.
\n\u2234 AC and BD will bisect each other.
\nHence, mid-point of AC and mid-point of BD are same point.
\n
\n\u21d2 1 + x = 7 and 8 = 5 + y
\n\u2234 x = 6 and y = 3.<\/p>\n
Question 15.
\nProve that the three points (1, 2), (3, 3) and (-1, 1) are collinear.
\nSolution.
\nThree points are collinear if the area of the triangle formed by them is zero.
\nHere
\nx1<\/sub> = 1, y1<\/sub> = 2
\nx2<\/sub> = 3, Y2<\/sub> = 3
\nx3<\/sub> = -1, y3<\/sub> = 1
\nNow, area
\n
\n\u2234 The given three points are collinear.<\/p>\nQuestion 16.
\nIf the points A(x, y), B(2, 3) and C(3, 4) are collinear, then prove that x + 5y = 17.
\nSolution.
\nHere x1<\/sub> = x, x2<\/sub> = 2, x3<\/sub> = -3, y1<\/sub> = y, y2<\/sub> = 3, y3<\/sub> = 4
\n
\nThe points A, B, C are collinear, therefore \u2206 = 0.
\n\u2234 \\(\\frac{1}{2}\\) (- x – 5y + 17) = 0
\n\u21d2 – x – 5y + 17 = 0
\n\u21d2 x + 5y = 17.<\/p>\n<\/p>\n
Question 17.
\nFor what value of a, the points (1, 4) (a, -2) and (-3, 16) are collinear?
\nSolution.
\nThe given points are collinear if the area of triangle formed by these three points is zero.
\nHere, x1<\/sub> = 1, x2<\/sub> = a, x3<\/sub> = -3
\ny1<\/sub> = 4, y2<\/sub>\u00a0= -2, y3<\/sub> = 16
\n<\/p>\nQuestion 18.
\nFind the area of the triangle whose vertices are
\n(i) (2, 3), (-1, 0), (2, -4)
\n(ii) (-5, -1), (3, -5), (5, 2)
\n
\n<\/p>\n
Question 19.
\nFind the equation of the locus of points equidistant from (-2, 2) and (-4, -2).
\nSolution.
\nLet A(-2, 2) and B(-4, -2) be the given points and let P(x, y) be the variable point, Then.
\nPA = PB PA2<\/sup> = PB2<\/sup>
\n\u21d2 (x + 2)2<\/sup> + (y – 2)2<\/sup> = (x + 4)2<\/sup> + (y + 2)2<\/sup>
\n\u21d2 x2<\/sup> + 2x + 4 + y2<\/sup> – 2y + 4 = x2<\/sup> + 8x + 16 + y2<\/sup> + 4y + 4
\n\u21d2 2x – 2y + 8 = 8x + 4y + 20
\n\u21d2 – 6x – 6y – 12 = 0
\n\u21d2 x + y + 2 = 0
\nwhich is the required equation of the locus.<\/p>\nQuestion 20.
\nFind the equation of the locus of all points equidistant from (3, 5) and the x-axis.
\nSolution.
\nLet P(x, y) be the variable point, then its distance from x-axis is y and its distance from (3, 5) is
\n
\nBy the given condition we have
\n
\nSquaring on both sides we get
\ny2<\/sup> = x