Write the linear equation represented by line AB and PQ. Also find the co-ordinate of intersection of line AB and PQ. -Maths 9th

Write the linear equation represented by line AB and PQ. Also find the co-ordinate of intersection of line AB and PQ. -Maths 9th
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Description : Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Last Answer : Let the abscissa of the point be x, According to the question, Ordinate (y) = 3 x Abscissa ⇒ y=3x When x = 1, then y = 3 x 1 = 3 and when x = 2, then y = 3 x 2 = 6. Here, ... the line AB. Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its abscissa.

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Description : Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Last Answer : Let the abscissa of the point be x, According to the question, Ordinate (y) = 3 x Abscissa ⇒ y=3x When x = 1, then y = 3 x 1 = 3 and when x = 2, then y = 3 x 2 = 6. Here, ... the line AB. Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its abscissa.

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Description : If (–5, 4) divides the line segment between the co-ordinate axes in the ratio 1 : 2, then what is its equation ? -Maths 9th

Last Answer : (d) x = yThe equations of the given lines are: 4x + 3y = 12 ...(i) 3x + 4y = 12 ...(ii) Solving the simultaneous equations (i) and (ii), we get\(x\) = \(rac{12}{7}\), y = \(rac{12}{7}\)∴ Point of the ... )isy - 0 = \(\bigg(rac{rac{12}{7}-0}{rac{12}{7}-0}\bigg)\) (x - 0), i.e., y = x.

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Description : Express the given statement in the form of a linear equation in two variables. The sum of the ordinate and abscissa of a point is 6. -Maths 9th

Last Answer : Solution :- x+y = 6

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Description : Determine the point on the graph of the linear equation x + y = 6, whose ordinate is 2 times its abscissa. -Maths 9th

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Description : The ordinate of a point is thrice its abscissa. Express the statement in the form of a linear equation in two variables. -Maths 9th

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P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

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Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

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In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

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

The straight line ax + by + c = 0 and the co-ordinate axes form an isosceles triangle under which of the following conditions ? -Maths 9th

Description : The straight line ax + by + c = 0 and the co-ordinate axes form an isosceles triangle under which of the following conditions ? -Maths 9th

Last Answer : (a) | a | = | b | The equation of line AB, i.e., ax + by + c = 0 in intercept form is ax + by = - c⇒ \(rac{x}{\big(-rac{c}{a}\big)}\) + \(rac{x}{\big(-rac{c}{b}\big)}\) = 1Δ AOB is isosceles Δ if OA = OB, i.e., ... \(rac{-c}{a}\) = \(rac{-c}{a}\) ⇒ \(rac{1}{a}\) = \(rac{1}{a}\) ⇒ | a | = | b |.

Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. -Maths 9th

Description : Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. -Maths 9th

Last Answer : As per question, the sum of the coordinates is 10 units. Let x and y be two coordinates, then we get x + y = 10. For x = 5, y = 5, therefore, (5, 5) lies on the graph of x + y = 10. For x = ... and (3, 7) on the graph paper and joining them by a line, we get graph of the linear equation x + y = 10.

Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. -Maths 9th

Description : Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units. -Maths 9th

Last Answer : As per question, the sum of the coordinates is 10 units. Let x and y be two coordinates, then we get x + y = 10. For x = 5, y = 5, therefore, (5, 5) lies on the graph of x + y = 10. For x = ... and (3, 7) on the graph paper and joining them by a line, we get graph of the linear equation x + y = 10.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Description : If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Last Answer : (b) We know that, if ΔRST is congruent to ΔUVW i.e., ΔRST = ΔUVW, then sides of ΔRST fall on corresponding equal sides of ΔUVW and angles of ΔRST fall on corresponding equal angles of ΔUVW. Here, given AB = ... , or ΔCBA ≅ ΔPRQ, so option (b) is correct, or ΔBCA ≅ ΔRPQ, so option (d) is incorrect.

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Description : If AB = QR, BC = PR and CA = PQ, then -Maths 9th

Last Answer : (b) We know that, if ΔRST is congruent to ΔUVW i.e., ΔRST = ΔUVW, then sides of ΔRST fall on corresponding equal sides of ΔUVW and angles of ΔRST fall on corresponding equal angles of ΔUVW. Here, given AB = ... , or ΔCBA ≅ ΔPRQ, so option (b) is correct, or ΔBCA ≅ ΔRPQ, so option (d) is incorrect.

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

In the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -Maths 9th

Description : In the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -Maths 9th

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Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is -Maths 9th

Description : Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is -Maths 9th

Last Answer : (c) ParallelogramAB = \(\sqrt{(4-7)^2+(5-6)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)BC = \(\sqrt{(7-4)^2+(6-3)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\)CD =\(\sqrt{(4-1)^2+(3-2) ... = \(2\sqrt{13}\)AB = CD, BC = AD and AC ≠ BD ⇒ opposite sides are equal and diagonals are not equal. ⇒ ABCD is a parallelogram.

A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Last Answer : NEED ANSWER

A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

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What is the equation of the line joining the origin with the point of intersection of the lines 4x + 3y = 12 and 3x + 4y = 12 ? -Maths 9th

Description : What is the equation of the line joining the origin with the point of intersection of the lines 4x + 3y = 12 and 3x + 4y = 12 ? -Maths 9th

Last Answer : (b) (5, 6)Let the foot of the perpendicular be M(x1, y1) Slope of line AB, i.e., y = -x + 11 = -1 Slope of line PM = \(rac{y_1-3}{x_1-2}\)Now, PM ⊥ AB⇒ \(\bigg(rac{y_1-3}{x_1-2}\bigg)\) x - ... get 2x1 = 10 ⇒ x1 = 5 Putting x1 in (ii), we get y1 = 6. ∴ Required foot of the perpendicular M is (5, 6).

Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Description : Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Last Answer : (a) 45º 3x + y - 7 = 0 ⇒ y = -3x + 7 ⇒ Slope (m1) = -3 x + 2y + 9 = 0 ⇒ y = \(rac{-x}{2}\) - \(rac{9}{2}\) ⇒ Slope (m2) = \(-rac{1}{2}\)If θ is the angle between the given lines, then tan θ = \(\ ... \bigg|rac{-rac{5}{2}}{1+rac{3}{2}}\bigg|\)= \(\bigg|rac{-rac{5}{2}}{rac{5}{2}}\bigg|\) = 1∴ θ = 45°.

Let cost of a pen and a pencil be “x” and “y” respectively. A girl pays ₹16 for 2 pens and 3 pencils. Write the given data in the form of a linear equation in two variables. Also represent it graphically. -Maths 9th

Description : Let cost of a pen and a pencil be “x” and “y” respectively. A girl pays ₹16 for 2 pens and 3 pencils. Write the given data in the form of a linear equation in two variables. Also represent it graphically. -Maths 9th

Last Answer : Solution :-

What do you mean by Co-ordinate Geometry? -Maths 9th

Description : What do you mean by Co-ordinate Geometry? -Maths 9th

Last Answer : (a) \(rac{2^n}{5^n}\)In any number the last digits can 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. Therefore, the last digit of each number can be chosen in 10 ways. ∴ Exhaustive number of ways = 10n. If the ... , favourable number of ways = 4n ∴ Required probability = \(rac{4^n}{10^n}\) = \(rac{2^n}{5^n}\).

What do you mean by Co-ordinates (Abscissa and Ordinate)? Explain with figure. -Maths 9th

Description : What do you mean by Co-ordinates (Abscissa and Ordinate)? Explain with figure. -Maths 9th

Last Answer : Co-ordinate Geometry is that branch of geometry in which two numbers, called co-ordinates are used to indicate the position of a point in a plane and which make use of algebraic methods in the study of geometric figures.

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Draw the graph of the equation represented by the straight line which is parallel to the x-axis and 3 units above it. -Maths 9th

Description : Draw the graph of the equation represented by the straight line which is parallel to the x-axis and 3 units above it. -Maths 9th

Last Answer : Solution :- Any straight line parallel to x-axis is given by y = a, where a is the distance of the line from the x-axis. Here a = 3. Therefore the equation of the line is y = 3. To draw the graph of this equation plot the points (0, 3) and (3, 3) and join them.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

The cost of a toy horse is same as that of cost of 3 balls. Express this statement as a linear equation in two variables. Also draw its graph. -Maths 9th

Description : The cost of a toy horse is same as that of cost of 3 balls. Express this statement as a linear equation in two variables. Also draw its graph. -Maths 9th

Last Answer : Solution :-

In the linear equation y = 4x + 13, if x is the number of hours a labourer is on work and y are his wages in rupees then draw the graph. Also find the wages when work is done for 6 hours. -Maths 9th

Description : In the linear equation y = 4x + 13, if x is the number of hours a labourer is on work and y are his wages in rupees then draw the graph. Also find the wages when work is done for 6 hours. -Maths 9th

Last Answer : when the work is done for 6 hours x=6 y=4(6)+13 y=24+13 y=37 the labourer gets Rs.37 if he works for 6hrs

The line x – 4y = 6 is the perpendicular bisector of the segment AB and the co-ordinates of B are (1, 3). Find the co-ordinates of A. -Maths 9th

Description : The line x – 4y = 6 is the perpendicular bisector of the segment AB and the co-ordinates of B are (1, 3). Find the co-ordinates of A. -Maths 9th

Last Answer : Co-ordinates of A are \(\bigg(rac{3 imes9+1 imes5}{3+1},rac{3 imes6+1 imes-2}{3+1}\bigg)\) = \(\bigg(rac{32}{4},rac{16}{4}\bigg)\), i.e. (8, 4)Now, \(x\) - 3y + 4 = 0 ⇒ -3y = -\(x\) - 4 ⇒ y = \(rac{x}{3}+rac{4} ... 8) [Using, y - y1 = m (x - x1)]⇒ 3y - 12 = \(x\) - 8 ⇒ 3y - \(x\) = 4.

If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th

Description : If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th

Last Answer : Solution :-

PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Last Answer : hope its clear

Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Description : Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Last Answer : Solution :- Construction: Draw two circles having centres O and O' intersecting at points A and B. Draw a parallel line PQ to OO' ... iii) Again, OO' = MN [As OO' NM is a rectangle] ...(iv) ⇒ 2OO' = PQ Hence proved.

For a ternary mixture, in which equilateral triangular co-ordinate is used in leaching and extraction, a __________ of the equilateral triangular co-ordinates. (A) Binary mixture is represented by the apex (B) Binary mixture is represented by any point inside (C) Ternary mixture is represented by the sides (D) Pure component is represented by the apex

Description : For a ternary mixture, in which equilateral triangular co-ordinate is used in leaching and extraction, a __________ of the equilateral triangular co-ordinates. (A) Binary mixture is ... Ternary mixture is represented by the sides (D) Pure component is represented by the apex

Last Answer : (D) Pure component is represented by the apex

The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Last Answer : (c) Since, the graph of linear equation 2x + 3y = 6 meets the X-axis. So, we put y = 0 in 2x + 3y = 6 ⇒ 2x + 3(0) = 6 = 2x + 0 = 6 ⇒ x = 6/2 ⇒ x = 3 Hence, the coordinate on X-axis is (3, 0).

The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Last Answer : (c) Since, the graph of linear equation 2x + 3y = 6 meets the X-axis. So, we put y = 0 in 2x + 3y = 6 ⇒ 2x + 3(0) = 6 = 2x + 0 = 6 ⇒ x = 6/2 ⇒ x = 3 Hence, the coordinate on X-axis is (3, 0).

At what point does the graph of the linear equation 2x + 3y = 9 meet a line which is parallel to the y-axis, at a distance of 4 units from the origin and on the right of the y-axis? -Maths 9th

Description : At what point does the graph of the linear equation 2x + 3y = 9 meet a line which is parallel to the y-axis, at a distance of 4 units from the origin and on the right of the y-axis? -Maths 9th

Last Answer : hope its clear

The graph of a linear equation in two variables is always a straight line. True/false -Maths 9th

Description : The graph of a linear equation in two variables is always a straight line. True/false -Maths 9th

Last Answer : answer: