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

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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 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 : 2x + y = 3 passes from origin. Is this statement true or false? -Maths 9th

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Description : The slope of a line perpendicular to the line which passes through the points (–k, h) and (b, – f ) is -Maths 9th

Last Answer : (b) (2, 2)The line 3x + 4y - 24 = 0 cuts the axis at A. To obtain the co-ordinates of A put y = 0, as on x-axis, y = 0. ∴ A ≡ (8, 0) ...(i) Also, on y-axis, x = 0, therefore B ≡ (0, 6 ... 8+10},rac{6 imes0+8 imes6+10 imes0}{6+8+10}\bigg)\)= \(\bigg(rac{48}{24},rac{48}{24}\bigg)\) = (2, 2).

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

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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 : 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 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 : 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 equation of the straight line with a positive gradient which passes through the point (–5, 0) -Maths 9th

Last Answer : (d) Both (a) and (c)Since the line passes through A(a, 0) and B(0, b), it makes intercepts a and b on x-axis and y-axis respectively. Let the equation of this line in the intercept from be \(rac{x}{a}\) + \(rac{y}{a}\) ... \(rac{x}{-12}\) + \(rac{y}{-5}\) = 1⇒ 5x + 12y = 60 and 5x + 12y + 60 = 0.

Description : A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Last Answer : (b) \(rac{\sqrt{17}}{2}\)Equation of the line through the points (5, 0) and (0, 3) y - 0 = \(rac{3-0}{0-5}\) (x - 5)⇒ y = \(rac{-3}{5}\)(x - 5)⇒ 5y + 3x - 15 = 0 ∴ Distance of perpendicular from ... (rac{|20+12-15|}{\sqrt{25+9}{}}\) = \(rac{17}{\sqrt{34}}\) units. = \(rac{\sqrt{17}}{2}\) units.

Description : If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is -Maths 9th

Last Answer : (a) Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation. ⇒ 2 (2) + 3 (0) = k ⇒ k = 4 Hence, the value of k is 4.

Description : If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is -Maths 9th

Last Answer : (a) Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation. ⇒ 2 (2) + 3 (0) = k ⇒ k = 4 Hence, the value of k is 4.

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Description : If x^3 + 5x^2 + 10k leaves remainder – 2x when divided by x^2 + 2, then what is the value of k ? -Maths 9th

Last Answer : x3+5x2+10k =(x2+2)(x+5)+10k−2x−10 ⇒10k−2x−10=−2x ⇒10k−10=0 or k=1.

Description : If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Last Answer : p(x) = (k2 – 14) x2 – 2x – 12 Here a = k2 – 14, b = -2, c = -12 Sum of the zeroes, (α + β) = 1 …[Given] ⇒ − = 1 ⇒ −(−2)2−14 = 1 ⇒ k2 – 14 = 2 ⇒ k2 = 16 ⇒ k = ±4

Description : A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

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

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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 : 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 : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

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Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

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Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y as its solution? -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 : Draw a graph of the equation x - Y = 4 & 2x+ 2y =4 on the same graph paper find the coordinates of the point whose two lines intersect. -Maths 9th

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Description : Find the slope and inclination of the line which passes through the points (1, 2) and (5, 6) ? -Maths 9th

Last Answer : Let A ≡ (x, y), B ≡ (2, 1), C ≡ (3, -2) Area of ΔABC = \(rac{1}{2}\) |{x1 (y2 - y3) + x2(y3 - y1) + x3(y1 - y2)}|= \(rac{1}{2}\) | \(x\)(1 + 2) + 2(-2 - y) + 3(y - 1) | = \(rac{1}{2 ... \(rac{3}{2}\)∴ Co-ordinates of A are \(\bigg(rac{7}{2},rac{13}{2}\bigg)\) or \(\bigg(rac{-3}{2},rac{3}{2}\bigg)\)

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

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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).

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Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

Description : A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of -Maths 9th

Last Answer : (d) \(rac{23}{\sqrt{17}}\)The given lines are:L : \(rac{x}{5}+rac{y}{b}=1\) ....(i)K : \(rac{x}{c}+rac{y}{3}=1.\) ...(ii)Since line L passes through (13, 32),\(rac{13}{ ... between parallel lines ax + by + c1 = 0 and ax + by c2 = 0 is d = \(rac{|c_2-c_1|}{\sqrt{a^2+b^2}}\bigg)\)

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

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.

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Description : Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

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Description : Find (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). -Maths 9th

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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.