Find the ratio in which the x-axes divides the line joining the points (–2, 5) and (1, –9) ? -Maths 9th

1 Answer

Answer :

Let the co-ordinates of the point of internal division A be (x, y). Then,\(x\) = \(rac{2 imes(-7)+3 imes8}{2+3}\) = \(rac{-14+24}{5}\) = \(rac{10}{5}\) = 2y = \(rac{2 imes4+3 imes9}{2+3}\) = \(rac{8+27}{5}\) = \(rac{35}{5}\) = 7∴ Co-ordinates of the point for internal division are (2, 7).

Related questions

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 : The mid-point of the line joining the points (–10, 8) and (–6, 12) divides the line joining the points (4, –2) and (– 2, 4) in the ratio -Maths 9th

Last Answer : (d) 2 : 1 externallyThe mid-point of the line joining the points (-10, 8) and (- 6, 12) is\(\bigg(rac{-10+(-6)}{2},rac{8+12}{2}\bigg)\), i.e., (-8, 10).Let (-8, 10) divide the join of (4 ... 6k = 12 ⇒ k = -2 Since the value of k is negative, it is a case of external division and the ratio is 2 : 1.

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.

Description : Find the coordinates of the point which divides the join of the points (8, 9) and (–7, 4) internally in the ratio 2 : 3. -Maths 9th

Last Answer : The circumcentre of a triangle is equidistant from the vertices of a triangle. Let A(3, 0), B(-1, -6), and C(4, -1) be the vertices of ΔABC and P(x, y) be the circumcentre of this triangle. Then, PA = PB = PC ⇒ PA2 = PB2 = ... PC = \(\sqrt{(1-3)^2+(-3-0)^2}\) = \(\sqrt{4+9}\) = \(\sqrt{13}\) units.

Description : The point (-1,2) divides the line segment joining the points A(2,5) and B(x,y) in the ratio 3:4, then the value of x2 + y2 is : (a) 27 (b) 28 (c) 29 (d) 30

Last Answer : (c) 29

Description : In what ratio is the line joining the points (2, –3) and (5, 6) divided by the x-axis. -Maths 9th

Last Answer : (b) (2, 1) (- 2, 1)Let PQRS be the required square and P(0, -1) and R(0, 3) be its two opposite vertices. Length of diagonal PR = \(\sqrt{(0-0)^2+(3+1)^2}\) = \(\sqrt{16}\) = 4∴ Length of each side = \( ... + 4 = 8 ⇒ a2 = 4 ⇒ a = 2. ∴ The other two vertices of the square are (+2, 1) and (-2, 1).

Description : The ratio in which P ( , B(2,-5) is : (a) 1:5 (b) 5:1 ) divides the line segment joining the points A ( , (c) 1:4 ) and (d) 4:1

Last Answer : (a) 1:5

Description : The point which divides the line segment joining the points (7, –6) and (3, 4) in the ratio 1 : 2 lies in the: (a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant

Last Answer : (d) IV quadrant

Description : In what ratio is the line joining the points A(4, 4) and B(7, 7) divided by P(–1, –1)? -Maths 9th

Last Answer : Let ABCD be the given square and let A ≡ (3, 4) and C ≡ (1, -1). Also let B ≡ (x, y). ABCD being a square,AB = BC ⇒ AB2 = BC2, ∠ABC = 90º⇒ \(\big(\sqrt{(x-3)^2+(y-4)^2}\big)^2\) = \(\big(\sqrt{(x-1)^2+(y+1 ... \(rac{9}{2}\),\(rac{1}{2}\)\(\bigg)\)and \(\bigg(\)\(-rac{1}{2}\), \(rac{5}{2}\)\(\bigg)\)

Description : Find the point of trisection of the line segment joining the points (1, 2) and (11, 9) ? -Maths 9th

Last Answer : Let P divide AB in the ratio k : 1. Then, co-ordinates of P are \(\bigg(rac{7k+4}{k+1},rac{7k+4}{k+1}\bigg)\)But P ≡ (-1, -1)∴ \(rac{7k+4}{k+1}\) = -1 ⇒ 7k + 4 = - k - 1 ⇒ 8k = - ... , it means that the division is external. ∴ AB is divided by P externally in the ratio \(rac{5}{8}\) : 1, i.e. 5 : 8.

Description : The line through the points (4, 3) and (2, 5) cuts off intercepts of lengths λ and μ on the axes. Which one of the following is correct ? -Maths 9th

Last Answer : (c) a, b, c are in H.P. only for all m As the points A(a, ma), B[b, (m + 1)b] and C[c, (m + 2)c] are collinear. Area of Δ ABC should be equal to zero.⇒ \(rac{1}{2}\)[x1(y2 - y3) + x2(y3 - y1) + ... - bc = 0 ⇒ ab + bc = 2ac ⇒ b = \(rac{2ac}{a+c}\)∴ a, b, c are harmonic progression (H.P.) for all m.

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 = \(rac{|c_2-c_1|}{\sqrt{a^2+b^2}}\bigg)\)

Description : Find the coordinates of the point which divides externally the join of the points (3, 4) and (– 6, 2) in the ratio 3 : 2. -Maths 9th

Last Answer : (d) D lies on the boundary of ΔABC∵ Mid-point of BC = \(\bigg(rac{7+3}{2},rac{7+5}{2}\bigg)\), i.e, (5, 6). we can easily show that D lies on the boundary of ΔABC.

Description : The line segment joining P(5, –2) and Q(9, 6) is divided in the ratio 3 : 1 by a point A -Maths 9th

Last Answer : Comparing y = 5\(x\) –7 with y = m\(x\) + c, the slope of given line = m = 5 ∴ Equation of a line parallel to y = 5\(x\) – 7 having y-intercept = –1 is y = 5\(x\) – 1.

Description : Points P (5, -3) is one of the two points of trisection of the line segment joining points A(7, -2) and B(1, -5) near to A. find the coordinates of the other point of trisection. -Maths 9th

Last Answer : answer:

Description : 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Now, In ΔACD, R and S are the mid points of CD and DA respectively. , ... , PQRS is parallelogram. PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

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 : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

Description : Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th

Last Answer : Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

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 : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

Description : Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th

Last Answer : Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

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 segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and equal to half of their difference. -Maths 9th

Last Answer : In a parallelogram ABCD, the bisector of ∠ A also bisects BC at X.Prove that AD = 2AB.

Description : Points P, Q, R and S divide a line segment joining A (2, 6) and B (7, -4) in five equal parts. Find the coordinates of P and R. -Maths 9th

Last Answer : this is the ans hope its clear

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 : On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. -Maths 9th

Last Answer : (b) Here, point 0 (0, 0) is the origin. A(3, 0) lies on positive direction of X-axis, B (3, 4) lies in 1st quadrant and C (0, 4) lines on positive direction of Y-axis. On joining OA AB, BC and CO the figure obtained is a rectangle.

Description : On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. -Maths 9th

Last Answer : (b) Here, point 0 (0, 0) is the origin. A(3, 0) lies on positive direction of X-axis, B (3, 4) lies in 1st quadrant and C (0, 4) lines on positive direction of Y-axis. On joining OA AB, BC and CO the figure obtained is a rectangle.

Description : What is the equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinates axes whose sum is –1 ? -Maths 9th

Last Answer : Diagonals of a rhombus bisect each other at right angles ⇒ Co-ordinates of mid-points of AC and BD are equal∴ 0 = \(\bigg(rac{4+(-2)}{2},rac{-5+(-1)}{2}\bigg)\) = (1, -3)Slope of BD = \(rac{-5+1}{4+2}\) = \(rac{-4}{6}\) ... (rac{3}{2}\) isy + 3 = \(rac{3}{2}\) (x - 1)⇒ 2y + 6 = 3x - 3 ⇒ 2y = 3x - 9.

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

Last Answer : (d) x + 2y = 2Let the required equation make intercept on x-axis = 2a ⇒ intercept made on y-axis = a ∴ Eqn of the given line in the intercept from:\(rac{x}{2a}+rac{y}{a}=1\) ...(i)Since the line ... 1 ⇒ a = 1.∴ Required equation of line : \(rac{x}{2 imes1}+rac{y}{1}=1\) ⇒ x + 2y = 2.

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 |.

Description : Determine the ratio in which 2x +3y – 30 = 0 divides the join of A(3, 4) and B(7, 8) and at what point? -Maths 9th

Last Answer : Let A(1, 2) and B(11, 9) be the given points. Let the points of trisection be P and Q. Then,AP = PQ = QB = k (say)⇒ AQ = AP + PQ = 2k and PB = PQ + QB = 2k ∴ AP : PB = k : 2k = 1 : 2 and AQ ... two points of trisection are \(\big(rac{13}{3},rac{13}{3}\big)\) and \(\big(rac{23}{3},rac{20}{3}\big)\).

Description : Determine the ratio in which the point P(m, 6) divides the join of A(– 4, 3) and B(2, 8). Also find the value of m. -Maths 9th

Last Answer : (b) 1 : 2Any point on the x-axis is (a, 0).Let the point (a, 0) divide the join of A(2, -3) and B(5, 6) in the ratio k : 1. Then the co-ordinates of the point of division are \(\bigg(rac{5k+2}{k+1},rac{6k-3}{k+1}\ ... 6k - 3 = 0 ⇒ k = \(rac{1}{2}\)Required ratio is k : 1 ⇒ \(rac{1}{2}\) : 1 = 1 : 2.

Description : Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Last Answer : Please see Exercise 8.2 - question 1 here in Quadrilaterals.

Description : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

Description : The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if -Maths 9th

Last Answer : According to question the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle,

Description : The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Last Answer : (d) Given, the quadrilateral ABCD is a rhombus. So, sides AB, BC, CD and AD are equal.

Description : The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Last Answer : According to question the mid-points of the sides of a rhombus, taken in order.

Description : The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

Last Answer : According to question mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only.

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

Description : Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Last Answer : Please see Exercise 8.2 - question 1 here in Quadrilaterals.

Description : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

Description : The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if -Maths 9th

Last Answer : According to question the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle,

Description : The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Last Answer : (d) Given, the quadrilateral ABCD is a rhombus. So, sides AB, BC, CD and AD are equal.

Description : The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Last Answer : According to question the mid-points of the sides of a rhombus, taken in order.

Description : The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

Last Answer : According to question mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only.

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.