For what value of p the point (p, 2) lies on the line 3x + y = 11? -Maths 9th

For what value of p the point (p, 2) lies on the line 3x + y = 11? -Maths 9th
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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.

A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Description : A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Last Answer : (c) 3x - 4y + 15 = 0 Let (m > 0) be the gradient (slope) of the required line. Then, Equation of any line through (-5, 0) having slope = m is y - 0 = m(x - (-5)) or mx - y + 5m = 0 ...(i) Its ... (rac{3}{4}\) (∵ m is +ve)∴ Required equation: y = \(rac{3}{4}\) (x + 5)⇒ 3x - 4y + 15 = 0.

If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3. -Maths 9th

Description : If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3. -Maths 9th

Last Answer : Given, (a−b)3=a3−3ab(a−b)−b3 (3x−2y)3=27x3−8y3−3(3x)(2y)(3x−2y) 113=27x3−8y3−18(12)(11) 27x3−8y3=3707

Find the point which lies on the line y x = -3 having abscissa 3. -Maths 9th

Description : Find the point which lies on the line y x = -3 having abscissa 3. -Maths 9th

Last Answer : Solution :- When x=3 then y= -9,thus the point is (3,-9)

Find the value of the polynomial p(x) = x^3-3x^2-2x+6 at x = underroot 2 -Maths 9th

Description : Find the value of the polynomial p(x) = x^3-3x^2-2x+6 at x = underroot 2 -Maths 9th

Last Answer : In this chapter, we shall proceed with recalling some of the constructions already learnt in the earlier classes and deal with some more. Here in this section, we will construct some of these ... be done? 2. Always explain the construction. Write the sequence of steps that are actually taken.

If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.

Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

Description : Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

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

Draw a graph of the equation x+ y=5 & 3x -2y=0 in the same graph paper find the coordinates of the point whose two two lines intersect. -Maths 9th

Description : Draw a graph of the equation x+ y=5 & 3x -2y=0 in the same graph paper find the coordinates of the point whose two two lines intersect. -Maths 9th

Last Answer : From x + y = 5, If x = 0 0 + y = 5 y = 5 Therefore (0,5) If x = 1 1 + y = 5 y =5 - 1 y = 4 Therefore (1,4) Draw a graph for this And From 3x - 2y = 0 If x = 0 3 (0) - 2y = 0 0 - ... 2y = 0 -2y = -6 y = -6/-2 y = 3 Therefore (2,3) Draw a graph for these points And the point of intersection is (2,3)

Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Description : Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Last Answer : There are two ways to prove it. 1st way: Area of triangle formed by the given points = 0 if they are collinear.∴ \(rac{1}{2}\) [\(x\)(2 - (y + 1) ) + 1((y + 1) - 1) + 0(1 - 2)] = 0⇒ \(rac{1}{2}\) [2\(x\) - \(x\)y - \( ... y - \(x\)y - 1 + \(x\) ⇒ x + y = \(x\)y ⇒ \(rac{1}{x}\) + \(rac{1}{y}\) = 1.

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).

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th

Description : If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th

Last Answer : THINGS WHICH ARE COINCIDE WITH EACH OTHER ARE EQUAL TO ONE ANOTHER PO+OR=PR 2PO=PR PO=OR PO=1/2PR HENCE PROVED

Find the coordinate where the linear equation 3x -4y = 11 meets at x-axis. -Maths 9th

Description : Find the coordinate where the linear equation 3x -4y = 11 meets at x-axis. -Maths 9th

Last Answer : hope it helps

The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

Description : The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

Last Answer : (c) (5, 0) or (1, 4) Let the third vertex of the triangle be P(a, b). Since it lies on the line x + y = 5, a + b = 5 ...(i) Also, given area of triangle formed by the points (2, -1), (3, 2) and (a, b) = 4 ... b) - (-3a + b) = 5 + 15⇒ 4a = 20 ⇒ a = 5 ⇒ b = 0. ∴ The points are (1, 4) and (5, 0).

What are the co-ordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 ? -Maths 9th

Description : What are the co-ordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 ? -Maths 9th

Last Answer : (d) 2x + 9y + 7 = 0PS being the median of ΔPQR, S is the mid-point of QR, i.e., Coordinates of S ≡ \(\bigg(rac{6+7}{2},rac{-1+3}{2}\bigg)\) = \(\bigg(rac{13}{2},1\bigg)\)Slope of line parallel to PS = Slope of PS= \(rac{1 ... y + 1) = \(rac{-2}{9}\)(x - 1), i.e., 9y + 9 = - 2x + 2 ⇒ 2x + 9y + 7 = 0.

If y-coordinate of a point is zero, then this point always lies -Maths 9th

Description : If y-coordinate of a point is zero, then this point always lies -Maths 9th

Last Answer : (c) If y-coordinate of a point is zero, then this point always lies on X-axis. Because perpendicular distance of the point from X-axis measured along Y-axis is zero.

The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Description : The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Last Answer : (C) Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.Hence, the required point is (0, – 5).

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

If y-coordinate of a point is zero, then this point always lies -Maths 9th

Description : If y-coordinate of a point is zero, then this point always lies -Maths 9th

Last Answer : (c) If y-coordinate of a point is zero, then this point always lies on X-axis. Because perpendicular distance of the point from X-axis measured along Y-axis is zero.

The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Description : The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Last Answer : (C) Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.Hence, the required point is (0, – 5).

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

Check whether the point (a ,– a) lies on y=x–a or not. -Maths 9th

Description : Check whether the point (a ,– a) lies on y=x–a or not. -Maths 9th

Last Answer : Solution :-

Find the point whose ordinate is –3 and which lies on y-axis. -Maths 9th

Description : Find the point whose ordinate is –3 and which lies on y-axis. -Maths 9th

Last Answer : Solution :- (0,-3)

p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Description : p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Last Answer : p(x) = x3+3x2+3x+1, g(x) = x+2 g(x) = 0 ⇒ x+2 = 0 ⇒ x = −2 ∴ Zero of g(x) is -2. Now, p(−2) = (−2)3+3(−2)2+3(−2)+1 = −8+12−6+1 = −1 ≠ 0 ∴By factor theorem, g(x) is not a factor of p(x

The point of the form (a, – a) always lies on the line -Maths 9th

Description : The point of the form (a, – a) always lies on the line -Maths 9th

Last Answer : (d) Taking option (d), x + y = a + (-a) = a – a = 0 [since, give point is of the form (a, -a)] Hence, the point (a, – a) always lies on the line x + y = 0.

The point of the form (a, – a) always lies on the line -Maths 9th

Description : The point of the form (a, – a) always lies on the line -Maths 9th

Last Answer : (d) Taking option (d), x + y = a + (-a) = a – a = 0 [since, give point is of the form (a, -a)] Hence, the point (a, – a) always lies on the line x + y = 0.

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 ⇒ 12 = 3a+7 ⇒ 3a = 12 – 7 ⇒ 3a = 5 Hence, the value of a is 5/3.

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 ⇒ 12 = 3a+7 ⇒ 3a = 12 – 7 ⇒ 3a = 5 Hence, the value of a is 5/3.

If the point (2k – 3, k + 2) lies on the graph of the equation 2x + 3y +15 = 0, find the value of k. -Maths 9th

Description : If the point (2k – 3, k + 2) lies on the graph of the equation 2x + 3y +15 = 0, find the value of k. -Maths 9th

Last Answer : Solution :-

Check whether the graph of the equation y = 3x + 5 passes through the origin or not. -Maths 9th

Description : Check whether the graph of the equation y = 3x + 5 passes through the origin or not. -Maths 9th

Last Answer : Solution :-

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

Draw the graph of the following equations (1) Y =3x+2 (b) Y=x -Maths 9th

Description : Draw the graph of the following equations (1) Y =3x+2 (b) Y=x -Maths 9th

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What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Description : What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Last Answer : (c) (8, 6)Let AB be the given line 4x + 3y = 25 Let O′(a, b) be the image of O in the given line AB. Let O O′ cut AB in point P. Also OP ⊥ AB and P is the mid-point of OO′. ∴ Co-ordinates of P are \(\bigg( ... 4 imes6}{3}\) = 8∴ The image of the point O(0, 0) in the line 4x + 3y - 25 = 0 is (8, 6).

The ratio in which the line 3x + 4y = 7 divides the line joining the points (–2, 1) and (1, 2) is -Maths 9th

Description : The ratio in which the line 3x + 4y = 7 divides the line joining the points (–2, 1) and (1, 2) is -Maths 9th

Last Answer : (a) (–24, –2)Co-ordinates of the point of external division are\(\bigg(rac{m_1\,x_2-m_2\,x_1}{m_1-m_2},rac{m_1y_2-m_2y_1}{m_1-m_2}\bigg)\), i.e.,∴ Required point = \(\bigg(rac{3 imes-6-2 imes4}{3-2},rac{3 imes2-2 imes4}{3-2}\bigg)\)= \(\big(rac{-24}{1},rac{-2}{1}\big)\), i.e., (–24, –2).

If x+1 is a factor of the polynomial 3x(square) - kx,then find the value of k. -Maths 9th

Description : If x+1 is a factor of the polynomial 3x(square) - kx,then find the value of k. -Maths 9th

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If x-2 is a factor of x3-3x+5a then find the value of a -Maths 9th

Description : If x-2 is a factor of x3-3x+5a then find the value of a -Maths 9th

Last Answer : -2/5 is value of a

If x-2 is a factor of x3-3x+5a then find the value of a -Maths 9th

Description : If x-2 is a factor of x3-3x+5a then find the value of a -Maths 9th

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If 5^(3x^2 log10 2) = 2^((x+1/2)log10 25), then the value of x is: -Maths 9th

Description : If 5^(3x^2 log10 2) = 2^((x+1/2)log10 25), then the value of x is: -Maths 9th

Last Answer : (d) \(-rac{1}{3}\)\(5^{{3x^2}log_{10}2}\) = 2\(\big(x+rac{1}{2}\big)\)log10 25⇒ \(5^{{3x^2}log_{10}2}\) = 2\(\big(rac{2x+1}{2}\big)\) x log10 5 = 2(2x+1)log10 5⇒ \(5^{{3x^2}log_{10}2}\) = 2(2x+1)log2 5. ... aloga x = x]⇒ 3x2 = 2x + 1 ⇒ 3x2 - 2x - 1 = 0 ⇒ (x - 1) (3x + 1) = 0⇒ x = 1, \(-rac{1}{3}\)

For what value of m will the expression 3x^3 + mx^2 + 4x – 4m be divisible by x + 2 ? -Maths 9th

Description : For what value of m will the expression 3x^3 + mx^2 + 4x – 4m be divisible by x + 2 ? -Maths 9th

Last Answer : f(x) = 3x3 + mx2 + 4x – 4m f(x) is divisible by (x + 2) if f(–2) = 0 Now f(–2) = 3(–2)3 + m(–2)2 + 4(–2) – 4m = – 24 + 4m – ... ; 4m = – 32 ≠ 0 ∴ No such value of m exists for which (x + 2) is a factor of the given expression

For what value of k, will the expression (3x^3 – kx^2 + 4x + 16) be divisible by (x – k/2) ? -Maths 9th

Description : For what value of k, will the expression (3x^3 – kx^2 + 4x + 16) be divisible by (x – k/2) ? -Maths 9th

Last Answer : Given f(x) = 3x³ - kx² + 4x + 16. Since (x - k/2) is a factor of polynomial. This means x = k/2 is the zero of the given polynomial. ⇒ f(k/2) = 3(k/2)³ - k(k/2)² + 4(k/2) + 16 ⇒ 0 ... - 4k + 32) + 4(k² - 4k + 32) ⇒ 0 = (k + 4)(k² - 4k + 32) ⇒ k = -4.

If the polynomials ax^3 + 4x^2 + 3x – 4 and x^3 – 4x + a leave the same remainder when divided by (x – 3), the value of a is : -Maths 9th

Description : If the polynomials ax^3 + 4x^2 + 3x – 4 and x^3 – 4x + a leave the same remainder when divided by (x – 3), the value of a is : -Maths 9th

Last Answer : Given ax^3 + 4x^2 + 3x - 4 and x^3 - 4x + a leave the same remainder when divided by x - 3. Let p(x) = ax^3 + 4x^2 + 3x - 4 and g(x) = x^3 - 4x + a By remainder theorem, if f(x) is divided by (x − a) then ... 4 27a+41 g(3)=27-4(3)+a 15+a f(3)=G(3) 27a+41=15+a 26a=15-41 a=15-41/26 a=-26/26 a=-1

Find the value of 27X3 + 8y3 if (i) 3x + 2y = 14 and xy = 8 (ii) 3x + 2y = 20 and xy = 149 -Maths 9th

Description : Find the value of 27X3 + 8y3 if (i) 3x + 2y = 14 and xy = 8 (ii) 3x + 2y = 20 and xy = 149 -Maths 9th

Last Answer : answer:

If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2 -Maths 9th

Description : If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2 -Maths 9th

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If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y. -Maths 9th

Description : If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y. -Maths 9th

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Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12. -Maths 9th

Description : Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12. -Maths 9th

Last Answer : Consider the equation 3x + 2y = 12 Now, square both sides: (3x + 2y)2 = 122 => 9x2 + 12xy + 4y2 = 144 =>9x2 + 4y2 = 144 – 12xy From the questions, xy = 6 So, 9x2 + 4y2 = 144 – 72 Thus, the value of 9x2 + 4y2 = 72

If cosec |=3x and cot |=3/x. then find the value of (x2-1/x2) -Maths 9th

Description : If cosec |=3x and cot |=3/x. then find the value of (x2-1/x2) -Maths 9th

Last Answer : cosecβ = = 3X and cotβ = 3/X , to find value ( X2 - 1/X2 ) . we know , cosec2β = cot2β +1 putting value , (3X )2 = ( 3/X )2 +1 , OR 9X2 = 9 /X2 + 1 , OR 9X2 – 9/X2 = 1, OR 9( X2 - 1/X2 ) = 1, SO (X2 – 1/X2 ) = 1/9

Find the value of k if the line on 2x + y = k passes through the point (3,5). -Maths 9th

Description : Find the value of k if the line on 2x + y = k passes through the point (3,5). -Maths 9th

Last Answer : Solution :-

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.