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

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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

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Description : Find the coordinate where the linear equation 4x - 23 y = 7 meets at y-axis. -Maths 9th

Last Answer : 4x-2=-7*3y 4x+21y=2 The equation meets y axis when x=0 4.0+21y=2 y=21/2 Hence , the equation meets y-axis at (0,21/2)

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).

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).

Description : Draw the graph of the equation 3x + 4y = 12 and find the co-ordinates of the points of intersection of the equation with the co-ordinate axes. -Maths 9th

Last Answer : Solution :-

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).

Description : Express x in terms of y for the linear equation 23x +4y = -7. -Maths 9th

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Description : A pair of linear equations which has a unique solution x = 2, y = -3 is (a) x + y = -1 ; 2x – 3y = -5 (b) 2x + 5y = -11 ; 4x + 10y = -22(c) 2x – y = 1 ; 3x + 2y = 0 (d) x – 4y – 14 = 0 ;5x – y – 13 = 0

Last Answer : (d) x – 4y – 14 = 0 ;5x – y – 13 = 0

Description : The normal at a point `P` on the ellipse `x^2+4y^2=16` meets the x-axis at `Qdot` If `M` is the midpoint of the line segment `P Q ,` then the locus of

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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).

Description : Find the following products: (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) (ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx) (iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) (iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz) -Maths 9th

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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.

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.

Description : Write any two solutions of the linear equation 3x + 2y =9. -Maths 9th

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Description : Write any four solutions of the linear equation y = 4 x – 11. -Maths 9th

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Description : If y-coordinate of a point is zero, then where will this point lie in the coordinate plane? On the x-axis. -Maths 9th

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Description : The position of a boy on the coordinate plane is given by the point (4,6) . What is the perpendicular distance from the x-axis and the y-axis ? -Maths 9th

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Description : The graph of the linear equation 2x + 3y = 6 cuts the Y-axis at the point. -Maths 9th

Last Answer : (d) Since, the graph of linear equation 2x + 3y = 6 cuts the Y-axis. So, we put x = 0 in the given equation 2x+ 3y = 6, we get 2 x 0+ 3y = 6 ⇒ 3y = 6 y = 2. Hence, at the point (0, 2), the given linear equation cuts the Y-axis.

Description : The graph of the linear equation 2x + 3y = 6 cuts the Y-axis at the point. -Maths 9th

Last Answer : (d) Since, the graph of linear equation 2x + 3y = 6 cuts the Y-axis. So, we put x = 0 in the given equation 2x+ 3y = 6, we get 2 x 0+ 3y = 6 ⇒ 3y = 6 y = 2. Hence, at the point (0, 2), the given linear equation cuts the Y-axis.

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

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Description : Out of the following linear equations in two variables, the two representing a pair of parallel lines graphically are: 3x - 4y = 6 3x + 4y = 6 6x + 8y = 12 6x - 8y + 6 = 0 (a) 3x - 4y = 6 and 6x + 8y = 12 (b) 3x - 4y = 6 ... 6 = 0 (c) 6x + 8y = 12 and 3x + 4y = 6 (d) 6x - 8y + 6 = 0 and 3x + 4y = 6

Last Answer : (b) 3x - 4y = 6 and 6x - 8y + 6 = 0

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Last Answer : (d) any real number except 6

Description : what- shaneese is buying peanuts and cashews. she has $12 to spend. peanuts cost $3 per pound and cashews cost $4 per pound. if she buys x pounds of peanuts and y pounds of cashews, then the equation that represents the amount of each is 3x + 4y =12?

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Description : what- Shaneese is buying peanuts and cashews. She has $12 to spend. Peanuts cost $3 per pound and cashews cost $4 per pound. If she buys x pounds of peanuts and y pounds of cashews, then the equation that represents the amount of each is 3x + 4y = 12.?

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Description : If a + b = 1 ,and the ordered pair (a, b) satisfies the equation 2x + y = , then it also satisfies (a) 2x + y (b) 3x + 4y = 3 (c) x + 2y = (d) 2x + 4y =

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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

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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)

Description : The equation whose roots are the negatives of the roots of the equation x^7 + 3x^5 + x^3 – x^2 + 7x + 2 = 0 is : -Maths 9th

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Description : The equation whose roots are twice the roots of the equation x^2 – 3x + 3 = 0 is -Maths 9th

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Last Answer : Find the equation of normal to the curve `y(x-2)(x-3)-x+7=0` at that point at which the curve meets X-axis.

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.

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Description : Give the geometric interpretations of 5x + 3 = 3x – 7 as an equation (i) in one variable (ii) in two variables. -Maths 9th

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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.

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

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

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Description : In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

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Description : In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

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Description : x = 5 and y = 2 is a solution of the linear equation. -Maths 9th

Last Answer : (c) (a) Take x + 2y, on putting x=5 and y = 2, we get 5 + 2(2) = 5+ 4 = 9≠7. So, (5, 2) is not a solution of x + 2y = 7 (b) Take 5x + 2y, on putting x = 5 and y = 2, we get 5x 5 + 2 x2 =25+ 4 = ... on putting x = 5 and y = 2, we get 5 5 + 2=25 + 2 =27 ≠7 So, (5, 2) is not a solution of 5x + y = 7.

Description : The graph of the linear equation y = x passes through the point. -Maths 9th

Last Answer : (c) The linear equation y = x has same value of x and y-coordinates are same. Therefore, the point (1,1) must lie on the line y = x.

Description : Find the solution of the linear equation x+2y = 8 which represents a point on -Maths 9th

Last Answer : We have, x + 2y = 8 ,..(i) (i) When the point is on the X-axis, then put y = 0 in Eq. (i), we get x+2 (0)=8 ⇒ x = 8 Hence, the required point is (8, 0). (ii) When the point is on the Y-axis, then put x = 0 in Eq. (i), we get 0 + 2y = 8 ⇒ y = 8/2 = 4 Hence, the required point is (0, 4).

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Last Answer : The given linear equation is 2x + cy= 8. …(i) Now, by condition, x and y-coordinate of given linear equation are same, i.e., x = y. Put y = x in Eq. (i), we get

Description : Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Last Answer : Linear equation

Description : x = 5 and y = 2 is a solution of the linear equation. -Maths 9th

Last Answer : (c) (a) Take x + 2y, on putting x=5 and y = 2, we get 5 + 2(2) = 5+ 4 = 9≠7. So, (5, 2) is not a solution of x + 2y = 7 (b) Take 5x + 2y, on putting x = 5 and y = 2, we get 5x 5 + 2 x2 =25+ 4 = ... on putting x = 5 and y = 2, we get 5 5 + 2=25 + 2 =27 ≠7 So, (5, 2) is not a solution of 5x + y = 7.

Description : The graph of the linear equation y = x passes through the point. -Maths 9th

Last Answer : (c) The linear equation y = x has same value of x and y-coordinates are same. Therefore, the point (1,1) must lie on the line y = x.

Description : Find the solution of the linear equation x+2y = 8 which represents a point on -Maths 9th

Last Answer : We have, x + 2y = 8 ,..(i) (i) When the point is on the X-axis, then put y = 0 in Eq. (i), we get x+2 (0)=8 ⇒ x = 8 Hence, the required point is (8, 0). (ii) When the point is on the Y-axis, then put x = 0 in Eq. (i), we get 0 + 2y = 8 ⇒ y = 8/2 = 4 Hence, the required point is (0, 4).

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Last Answer : The given linear equation is 2x + cy= 8. …(i) Now, by condition, x and y-coordinate of given linear equation are same, i.e., x = y. Put y = x in Eq. (i), we get