Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th
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If the points (x, 1), (1, 2) and (0, y + 1) are collinear show that -Maths 9th

Description : If the points (x, 1), (1, 2) and (0, y + 1) are collinear show that -Maths 9th

Last Answer : Two lines are parallel if their slopes are equal∴ \(rac{0-(-8)}{3-(-5)}\) = \(rac{a-3}{4-6}\) ⇒ \(rac{8}{8}\) = \(rac{a-3}{-2}\) ⇒ a – 3 = –2 ⇒ a = 1.

Plot the following points and check whether they are collinear or not -Maths 9th

Description : Plot the following points and check whether they are collinear or not -Maths 9th

Last Answer : (i) Plotting the points P (1, 3), Q (-1, -1) and R (-2, - 3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear. (ii) Plotting the points ... 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given points are collinear.

Plot the following points and check whether they are collinear or not -Maths 9th

Description : Plot the following points and check whether they are collinear or not -Maths 9th

Last Answer : (i) Plotting the points P (1, 3), Q (-1, -1) and R (-2, - 3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear. (ii) Plotting the points ... 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given points are collinear.

Plot the following points and check whether they are collinear or not: -Maths 9th

Description : Plot the following points and check whether they are collinear or not: -Maths 9th

Last Answer : Solution :-

There are 18 points in a plane such that no three of them are in the same line except five points which are collinear. -Maths 9th

Description : There are 18 points in a plane such that no three of them are in the same line except five points which are collinear. -Maths 9th

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Show that the points (a, b + c), (b, c + a), (c, a + b) are collinear. -Maths 9th

Description : Show that the points (a, b + c), (b, c + a), (c, a + b) are collinear. -Maths 9th

Last Answer : Let A(x1, y1) ≡ (1, 3), B(x2, y2) ≡ (2, 4), C(x3, y3) ≡ (5, 6) be the vertices of ΔABCArea of ΔABC = \(rac{1}{2}\) |{\(x_1\)(y2 – y3) + \(x_2\)(y3 – y1) + \(x​​_3\)(y1 – y2)}|= \(rac{1}{2}\) |{1(4 – 6) + 2(6 – 3) + 5(3 – 4)}| = \(rac{1}{2}\) |{–2 + 6 – 5}| = \(rac{1}{2}\) sq. units.

If the points A(1, 2), B(0, 0) and C(a, b) are collinear, then -Maths 9th

Description : If the points A(1, 2), B(0, 0) and C(a, b) are collinear, then -Maths 9th

Last Answer : (a) - 2For three points to be collinear, area of the triangle formed by the three points should be equal to zero, i.e.\(rac{1}{2}\) [k(3k - 1) + 2k(1 - 2k) + 3(2k - 3k)] = 0⇒ \(rac{1}{2}\) [3k2 - k + ... = 0 or -2 Neglecting k = 0, as then (k, 2k) and (2k, 3k) will be the same point, we take k = -2.

If the three points (k, 2k), (2k, 3k) and (3, 1) are collinear then k is equal to -Maths 9th

Description : If the three points (k, 2k), (2k, 3k) and (3, 1) are collinear then k is equal to -Maths 9th

Last Answer : (d) 3Let (x, y) be the co-ordinates of the third vertex of the triangle. Then\(rac{0+2+x}{3}\) = 1 and \(rac{0+0+y}{3}\) = 1⇒ 2 + \(x\) = 3 and y = 3 ⇒ \(x\) = 1, y = 3. ∴ Co-ordinates of vertices of the triangle ... - y3) + x2 (y2 - y3) + x3(y1 - y2)]= \(rac{1}{2}\) [0+6+0] = \(rac{6}{2}\) = 3.

If the points with the co-ordinates {a, ma}, {b, (m + 1)b}, {c, (m + 2)c} are collinear, then which of the following is correct ? -Maths 9th

Description : If the points with the co-ordinates {a, ma}, {b, (m + 1)b}, {c, (m + 2)c} are collinear, then which of the following is correct ? -Maths 9th

Last Answer : (d) (7, -2)Let the co-ordinates of R be (x, y). As can be easily seen, it is a point of external division Also, PR = 2QR⇒ R divides the join of P and Q externally in the ratio 2:1. ∴ x = \(rac{2 imes2-1 imes-3}{2-1}\), ... }{2-1}\)⇒ x = 4 + 3 = 7 and y = 2 - 4 = -2. ∴ Co-ordinates of R are (7, -2).

If the points A(1, 2), B(2, 4) and C(3, a) are collinear, what is the length of BC ? -Maths 9th

Description : If the points A(1, 2), B(2, 4) and C(3, a) are collinear, what is the length of BC ? -Maths 9th

Last Answer : (c) √5 units Area of Δ ABC = 0 for collinearity of A, B, C.⇒ \(rac{1}{2}\)[1(4 – a) + 2(a – 2) + 3(2 – 4)] = 0 ⇒ 4 – a + 2a – 4 + 6 – 12 = 0 ⇒ a – 6 = 0 ⇒ a = 6. ∴ Point C ≡ (3, 6)⇒ BC = \(\sqrt{(3-2)^2+(6-4)^2}\) = \(\sqrt{1+4}\) = √5 units .

Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

The temperature of a liquid can be measured in Kelvin units as x K or in Fahrenheit units as y°F. The relation between the two system of measurement of temperature is given by the linear equation y = 9/5 ( x - 273 ) + 32 -Maths 9th

Description : The temperature of a liquid can be measured in Kelvin units as x K or in Fahrenheit units as y°F. The relation between the two system of measurement of temperature is given by the linear equation y = 9/5 ( x - 273 ) + 32 -Maths 9th

Last Answer : Solution :-

Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Description : Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Last Answer : R = {a, b : a < b, a ∈ A, b ∈ B}, where A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5}. ∴ R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} Domain of R = {1, 2, 3, 4} Range of R = {3, 5} Codomain of R = {1, 3, 5}.

While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Description : While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Last Answer : (a) Yes , x < y is correct (b) Ð ADB =Ð DBC = y (alternate int. angles) since BC < CD (angle opp. to smaller side is smaller) there for, x < y (c) Truth value

Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th

Description : Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th

Last Answer : Given: A = {All books in a library of a school} R = {(x, y): x and y have the same number of pages} Reflexivity: (x, x) ∈ R ⇒ R is reflexive on A Symmetric: Since books x and y have the same number of pages ... and (y, z) ∈ R ⇒ (x, z) ∈ R ⇒ R is transitive on A. Hence, R is an equivalence relation.

Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Description : Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Last Answer : (a) ReflexiveGiven, {(\(x\), y) : 2\(x\) – y = 10} Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 ∴ Point (10, 10) ∈ N ⇒ R is reflexive.

Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Description : Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Last Answer : (d) R = {(a, b) : b/a is odd} Since (2, 6) ∈R, the relation a and b are odd does not exist. Since (1, 5) and (1, 7) ∈R, the relation a and b are even does not exist. None of the ... \(rac{7}{1}\) = 7, \(rac{6}{2}\) = 3, quotients being all odd numbers, the relation b/a is odd exists.

Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis -Maths 9th

Description : Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis -Maths 9th

Last Answer : answer:

If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Description : If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Last Answer : Solution :- △ ABC ≅ △PQR

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Last Answer : This answer was deleted by our moderators...

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

Plot the points (x, y) given by the following table. -Maths 9th

Description : Plot the points (x, y) given by the following table. -Maths 9th

Last Answer : On plotting the given points on the graph, we get the points P(2,4), Q(4,2) R (-3, 0), S (-2, 5), T (3, – 3)and O (0 0).

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Last Answer : This answer was deleted by our moderators...

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

Plot the points (x, y) given by the following table. -Maths 9th

Description : Plot the points (x, y) given by the following table. -Maths 9th

Last Answer : On plotting the given points on the graph, we get the points P(2,4), Q(4,2) R (-3, 0), S (-2, 5), T (3, – 3)and O (0 0).

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

Draw the graph of the linear equation 3x + 4y = 6. At what points, does the graph cut the x-axis and the y-axis? -Maths 9th

Description : Draw the graph of the linear equation 3x + 4y = 6. At what points, does the graph cut the x-axis and the y-axis? -Maths 9th

Last Answer : hope it helps

Let ABCD be a quadrilateral. Let X and Y be the mid-points of AC and BD respectively and the lines through X and Y respectively -Maths 9th

Description : Let ABCD be a quadrilateral. Let X and Y be the mid-points of AC and BD respectively and the lines through X and Y respectively -Maths 9th

Last Answer : answer:

X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P. CX and BY are joined -Maths 9th

Description : X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P. CX and BY are joined -Maths 9th

Last Answer : answer:

If the points (2, 1) and (1, – 2) are equidistant from the point (x, y), show that x + 3y = 0. -Maths 9th

Description : If the points (2, 1) and (1, – 2) are equidistant from the point (x, y), show that x + 3y = 0. -Maths 9th

Last Answer : (a) The distance d between any two points say P(x1, y1) and Q(x2, y2) is given by:d = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)⇒ d2 = (x2 - x1)2 + (y2 - y1)2 ⇒ d = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)( ... distance of a point P(x1, y1) form the origin= \(\sqrt{(x_2-0)^2+(y_2-0)^2}\) = \(\sqrt{x^2_1+y^2_1}\)

If the sum of the squares of the distances of the point (x, y) from the points (a, 0) and (–a, 0) be 2b^2, then which of the following is correct ? -Maths 9th

Description : If the sum of the squares of the distances of the point (x, y) from the points (a, 0) and (–a, 0) be 2b^2, then which of the following is correct ? -Maths 9th

Last Answer : (d) (0, 0)Let the vertices of Δ ABC be given as: A(0, 0), B(3, 0) and C(0, 4) The orthocentre O is the point of intersection of the altitudes drawn from the vertices of Δ ABC on the opposite sides. Slope of BC = \(rac{ ... ⇒ y = 0 ...(ii) From (i), \(x\) = 0 Hence, the orthocentre is (0, 0).

Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed. -Maths 9th

Description : Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed. -Maths 9th

Last Answer : Let a be the equal distance from origin on both axes. Now, the coordinates of two points on equal distance 'a'on x-axis are Pla, 0) and R(-a, 0). Also, the coordinates of two points on equal distance 'a' on Y-axis are Q(0, a) and S(0, -a). Join all the four points on the graph.

If the roots of the equation x^2 + x + 1 = 0 are in the ratio of m : n, then which one of the following relation holds ? -Maths 9th

Description : If the roots of the equation x^2 + x + 1 = 0 are in the ratio of m : n, then which one of the following relation holds ? -Maths 9th

Last Answer : answer:

Who Four points are always coplaner if:A. They lie on different planesB. They lie on different linesC. They line in the same planeD. They are collinear?

Description : Who Four points are always coplaner if:A. They lie on different planesB. They lie on different linesC. They line in the same planeD. They are collinear?

Last Answer : C. They lie in the same planeD. They are collinear

If the angles of elevation of the top of tower from three collinear points `A`, `B` and `C`, on a line leading to the foot of the tower, are `30^(@)`,

Description : If the angles of elevation of the top of tower from three collinear points `A`, `B` and `C`, on a line leading to the foot of the tower, are `30^(@)`,

Last Answer : If the angles of elevation of the top of tower from three collinear points `A`, `B` and `C`, on a line leading to ... )` C. `1 : sqrt(3)` D. `2 : 3`

Which of the following points lies on Y-axis ? -Maths 9th

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

Which of the following points lies on Y-axis ? -Maths 9th

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

The points A(2, 3), B(3, 5), C(7, 7) and D(5, 6) are such that: -Maths 9th

Description : The points A(2, 3), B(3, 5), C(7, 7) and D(5, 6) are such that: -Maths 9th

Last Answer : (d) 24 unitsAB ⊥ chord PQ ⇒ AB bisects chord PQ ⇒ PQ = 2PB. AB = \(\sqrt{(2-5)^2+(-3-1)^2}\) = \(\sqrt{(-3)^2+(-4)^2}\)= \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5AP = radius of circle = 13 ∴ By Pythagoras' ... = \(\sqrt{AP^2-AB^2}\)= \(\sqrt{169-25}\) = \(\sqrt{144}\) = 12 units∴ PQ = 2 PB = 24 units.

If the area of the quadrilateral whose angular points A, B, C, D taken in order are (1, 2), (–5, 6), (7, – 4) and (–2, k) be zero, -Maths 9th

Description : If the area of the quadrilateral whose angular points A, B, C, D taken in order are (1, 2), (–5, 6), (7, – 4) and (–2, k) be zero, -Maths 9th

Last Answer : (d) 1 : 16Gvien \(rac{AD}{AB}=rac{1}{4}\) ⇒ \(rac{AD}{DB}=rac{1}{3}\),i.e., D divides AB internally in the ratio 1 : 3. ∴ Co-odinates of D are\(\bigg(rac{1+12}{1+3},rac{5+18}{1+3}\bigg)\)i.e.,\(\ ... {Area of}\,\Delta{ADE}}{ ext{Area of}\,\Delta{ABC}}\) = \(rac{rac{15}{32}}{rac{15}{2}}\) = 1 : 16.

Points P (5, -3) is one of the two points of trisection of the line segment joining points A(7, -2) and B(1, -5) near to A. find the coordinates of the other point of trisection. -Maths 9th

Description : Points P (5, -3) is one of the two points of trisection of the line segment joining points A(7, -2) and B(1, -5) near to A. find the coordinates of the other point of trisection. -Maths 9th

Last Answer : answer:

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. -Maths 9th

Description : Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. -Maths 9th

Last Answer : Reflexive: R = {(a, b) : b = a +1} = {(a, a + l) : a, a + 1∈{l, 2, 3, 4, 5, 6}} = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} ⇒ R is not reflexive since (a, a) ∉R for all a. Symmetric: R is not symmetric as (a ... as (a, b) ∈ R and (b, c) ∈ R but (a, c) ∉ R e.g., (1, 2) ∈ R (2, 3) ∈ R but (1, 3) ∉R

Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Description : Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Last Answer : (c) {(2, 3), (5, 3), (5, 6), (11, 3), (11, 6), (11, 10)}

What is the acute angle between the two straight lines y = ( ) 2– 3 x + 5 and y = ( ) 2 3 + x – 7 ? -Maths 9th

Description : What is the acute angle between the two straight lines y = ( ) 2– 3 x + 5 and y = ( ) 2 3 + x – 7 ? -Maths 9th

Last Answer : (a) 60ºThe two lines are: y = \((2-\sqrt3)\) \(x\) + 5 ...(i) y = \((2-\sqrt3)\) \(x\) - 7 ...(ii) Slope of line (i), m1 = \(2-\sqrt3\)Slope of line (ii), m2 = \(2 ... \sqrt3)}{1+(2-\sqrt3)(2+\sqrt3)}\bigg|\)= \(\bigg|rac{-2\sqrt3}{1+1}\bigg|=\sqrt3\)∴ θ = -1 tan ( 3) = 60º.

Find the coordinates of the point which divides the join of the points (8, 9) and (–7, 4) internally in the ratio 2 : 3. -Maths 9th

Description : Find the coordinates of the point which divides the join of the points (8, 9) and (–7, 4) internally in the ratio 2 : 3. -Maths 9th

Last Answer : The circumcentre of a triangle is equidistant from the vertices of a triangle. Let A(3, 0), B(-1, -6), and C(4, -1) be the vertices of ΔABC and P(x, y) be the circumcentre of this triangle. Then, PA = PB = PC ⇒ PA2 = PB2 = ... PC = \(\sqrt{(1-3)^2+(-3-0)^2}\) = \(\sqrt{4+9}\) = \(\sqrt{13}\) units.