Description : What is the equation of the straight line which passes through (3, 4) and the sum of whose x-intercept and y-intercept is 14 ? -Maths 9th
Last Answer : (a) 4x + 3y = 24 Let the x-intercept = a. Then, y-intercept = 14 - a ∴ Eqn of the straight line is \(rac{x}{a}\) + \(rac{y}{14-a}\) = 1Since it passes through (3, 4), so\(rac{3}{a}\) + \(rac{4}{14-a}\) = 1⇒ 3(14 - ... = 1 ⇒ x + y = 7or \(rac{x}{6}\) + \(rac{y}{8}\) = 1 ⇒ 8x + 6y = 48 ⇒ 4x + 3y = 24.
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 : If x = 0 and y = k is a solution of the equation 5x - 3 y = 0, find the value of k. -Maths 9th
Last Answer : Solution :-
Description : Identify an equation in slope-intercept from for the line parallel to y=4x-9 that passes through (-5,3)?
Last Answer : food
Description : Give the geometric interpretations of 5x + 3 = 3x – 7 as an equation (i) in one variable (ii) in two variables. -Maths 9th
Description : Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th
Last Answer : Solution :- x= -4
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
Description : The equation of a line parallel to the y- axis is of the form ................. -Maths 9th
Last Answer : answer:
Description : The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th
Last Answer : (a) 45º The equations of the given lines are: A\(x\) + By = A + B ⇒ By = -A\(x\) + (A + B) ⇒ y = \(-rac{A}{B}x\) + \(rac{(A+B)}{B}\) ....(i)and A(\(x\) - y) + B(\(x\) ... (ii) = m2 = \(rac{(A+B)}{B-A}\)Let θ be the angle between both the lines, then∴ tan θ = 1 ⇒ θ = tan-1 (1) = 45°.
Description : What is the equation of the line passing through (2, –3) and parallel to the y-axis ? -Maths 9th
Last Answer : (c) x = 2 Slope of y-axis = tan 90º (∵ y-axis ⊥ x-axis) ∴ Equation of line passing through (2, –3) parallel to y-axis is (y + 3) = tan 90º (x – 2) ⇒ (y + 3) = ∞ (x – 2) ⇒ (x – 2) = \(rac{1}{\infty}\) (y + 3) = 0 ⇒ x = 2.
Description : Write a solution of the linear equation 5x + 0y +8 = 0 in two variables. -Maths 9th
Last Answer : Solution : -
Description : For what value of k will the roots of the equation kx^2 – 5x + 6 = 0 be in the ratio 2 : 3 ? -Maths 9th
Last Answer : (b) 1 Let the roots of the equation kx2 - 5x + 6 = 0 be α and β. Then, α + β = \(\frac{5}{k}\) ...(i) αβ = \(\frac{6}{k}\) ...(ii) Given \(\ ... frac{9}{k}\) ⇒ 9k2 - 9k = 0 k(k - 1) = 0 ⇒ k = 0 or 1 But k = 0 does not satisfy the condition, so k = 1.
Description : The equation of the line bisecting the join of (3, – 4) and (5, 2) and having its intercepts on the x-axis and y-axis in the ratio 2 : 1 is -Maths 9th
Last Answer : (d) \(\lambda\) = \(\mu\) Let the equation of the line in the intercept from be\(rac{x}{\lambda}\)+ \(rac{y}{\mu}\) = 1Since it passes through (4, 3) and (2, 5)\(rac{4}{\lambda}\) + \(rac{3}{\mu}\) = 1 ... ) = 1 - \(rac{3}{7}\) = \(rac{4}{7}\) = \(\lambda\) = 7∴ \(\lambda\) = \(\mu\) = 7.
Description : Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th
Last Answer : Given, equation is 5x + 2ky = 3k. On putting x =1 and y =1 in this equation, we get 5(1) + 2k(1) =3k ⇒ 5 + 2k =3k ⇒ 5 = 3k - 2k ⇒ k = 5 Hence, required value of k is 5.
Description : In fig.4.7, LM is a line parallel to the y - axis at a distance of 2 units. -Maths 9th
Last Answer : Given , LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinates of point P=(3,2) [since,its perpendicular distance from X-axis is 2] Coordinate of point ... L=3 , abscissa of point M=3 ∴∴ Difference between the abscissa of the points L and M =3-3=0
Description : What is The equation of a line in slope-intercept form for a line with slope = 2 and y-intercept at (0,3)?
Last Answer : -1
Description : Why equation of the line that passes through point (4,12) and has the y-intercept of -2?
Last Answer : 414
Description : What is the y-intercept of the line given by the equation y= 3x + 4?
Last Answer : The answer is 4
Description : Write the equation in slope-intercept form of the line that has a slope of 2 and y-intercept of -2.?
Last Answer : 4
Description : Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th
Last Answer : (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Here, the highest power of x is 4, Hence, the degree of a polynomial is 4.
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.
Description : Write the equation of a line parallel to x-axis and passing through the point (–3, –4). -Maths 9th
Last Answer : Solution :- y = -4
Description : Write the equation of the line parallel to the x-axis at distance 3 units above x-axis. -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.
Description : Write the equation of line parallel to x- axis. -Maths 9th
Description : Show that the equation of the parallel line midway between the parallel lines -Maths 9th
Last Answer : ∵ Length of perpendicular from point (x1, y1) to line a\(x\) + by + c = 0 = \(rac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\)∴ Length of perpendicular from (0, 0) to \(rac{x}{a}\) + \(rac{y}{b}\) = 1 ⇒ \(rac{\big|rac{1}{a} imes0+ ... {1}{b^2}+rac{1}{b^2}}\) ⇒ \(rac{1}{p^2}\) = \(rac{1}{a^2}\) + \(rac{1}{b^2}\)
Description : Find the equation of normal of the curve `2y= 7x - 5x^(2)` at those points at which the curve intersects the line x = y.
Last Answer : Find the equation of normal of the curve `2y= 7x - 5x^(2)` at those points at which the curve intersects the line x = y.
Description : What is the perpendicular bisector equation of the line y equals 5x plus 10 spanning the parabola y equals x squared plus 4?
Last Answer : If: y = 5x +10 and y = x^2 +4Then: x^2 +4 = 5x +10Transposing terms: x^2 -5x -6 = 0Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x =-1Therefore by substitution endpoints of the line are ... .5 = -1/5(x-2.25) => 5y= -x+114.75Perpendicular bisector equation in its general form: x+5y-114.75= 0
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)
Description : In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th
Last Answer : Given, LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinate of point P = (3, 2) [since, its perpendicular distance from X-axis is 2] Coordinate of ... 3, abscissa of point M = 3 Difference between the abscissa of the points L and M = 3 -3 = 0
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.
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°.
Description : The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th
Last Answer : (a) In natural numbers, there is only one pair i.e., (1, 1) which satisfy the given equation but in positive real numbers, real numbers and rational numbers there are many pairs to satisfy the given linear equation.
Description : Express x in terms of y for the linear equation 23x +4y = -7. -Maths 9th
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 value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th
Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6
Description : One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th
Last Answer : (d) Now, (25x2 -1) + (1 + 5x)2 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] = 50x2 + 10x = 10x (5x+ 1) Hence, one of the factor of given polynomial is 10x.
Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th
Last Answer : Show that p = r.
Description : Find the value of k if (x-2)is a factor of polynomial p(x) = 2x(cube) - 6x(square) + 5x + k. -Maths 9th
Description : What must be added to 2x(square) - 5x + 6 to get x(cube) - 3x(square) + 3x - 5? -Maths 9th