What is distance between the graphs of the equations y = -1 and y = 3? -Maths 9th

What is distance between the graphs of the equations y = -1 and y = 3? -Maths 9th
by

1 Answer

Answer :

Solution   :- 4 units.
by

Related questions

Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Description : Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Last Answer : The given equation is y = x. To draw the graph of this equations, we need atleast two points lying on the given line. For x = 1, y = 1, therefore (1,1) satisfies the linear equation y = x. For x = 4, y = 4, ... of y = - x. We observe that, the line y = x and y = - x intersect at the point 0(0, 0).

Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Description : Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Last Answer : The given equation is y = x. To draw the graph of this equations, we need atleast two points lying on the given line. For x = 1, y = 1, therefore (1,1) satisfies the linear equation y = x. For x = 4, y = 4, ... of y = - x. We observe that, the line y = x and y = - x intersect at the point 0(0, 0).

Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Description : Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Last Answer : Given, x = 2 and y = 1 (i) Given, linear equation is 2x + 5y = 9. On putting x = 2 and y= 1 in LHS, we get LHS = 2x + 5y =2(2) + 5(1) = 4 + 5 = 9 = RHS So, x = 2 and y=1 is a solution of given ... + 3 (1) = 5 + 3 = 8 ≠ 14 ⇒ LHS ≠ RHS So, x = 2 and y = 1 is not a solution of given equation.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Description : In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Last Answer : Following are the rational numbers which represent irrational numbers .

How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Description : How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Last Answer : (c) Let the linear equation be ax + by + c = 0. On putting x = 1 and y = 2, in above equation we get =s a + 2b + c = 0, where a, b and c, are real number Here, different values of a, b and ... a + 2b + c = 0. Hence, infinitely many linear equations in x and yean be satisfied by x = 1 and y = 2.

Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Description : Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Last Answer : Given, x = 2 and y = 1 (i) Given, linear equation is 2x + 5y = 9. On putting x = 2 and y= 1 in LHS, we get LHS = 2x + 5y =2(2) + 5(1) = 4 + 5 = 9 = RHS So, x = 2 and y=1 is a solution of given ... + 3 (1) = 5 + 3 = 8 ≠ 14 ⇒ LHS ≠ RHS So, x = 2 and y = 1 is not a solution of given equation.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Description : In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Last Answer : Following are the rational numbers which represent irrational numbers .

How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Description : How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Last Answer : (c) Let the linear equation be ax + by + c = 0. On putting x = 1 and y = 2, in above equation we get =s a + 2b + c = 0, where a, b and c, are real number Here, different values of a, b and ... a + 2b + c = 0. Hence, infinitely many linear equations in x and yean be satisfied by x = 1 and y = 2.

How many linear equations satisfy x = 2 and y = -3? -Maths 9th

Description : How many linear equations satisfy x = 2 and y = -3? -Maths 9th

Last Answer : Solution :- Infinitely many equations.

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

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

Solve the following equations for x and y. log100 |x+y| = 1/2, -Maths 9th

Description : Solve the following equations for x and y. log100 |x+y| = 1/2, -Maths 9th

Last Answer : (b) \(\bigg(rac{10}{3},rac{20}{3}\bigg)\). (+ 10, 20) log100 |x+y| = \(rac{1}{2}\) ⇒ |x + y| = 100\(^{rac{1}{2}}\)⇒ |x + y| = 10 as (-10 is inadmissible) ...(i) log10y - log10| x | = log1004⇒ log10 ... x < 0, then x = 10.∴ If x = \(rac{10}{3}\), then y = \(rac{20}{3}\) and if x = 10, y = 20.

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).

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

Write three possible linear equations which can pass through point (3, –2). -Maths 9th

Description : Write three possible linear equations which can pass through point (3, –2). -Maths 9th

Last Answer : Solution :-

Give the equations of two lines passing through (4, –2). How many more such lines are there, and why? -Maths 9th

Description : Give the equations of two lines passing through (4, –2). How many more such lines are there, and why? -Maths 9th

Last Answer : Solution :-

Linear Equations in Two Variables Class 9th Formulas -Maths 9th

Description : Linear Equations in Two Variables Class 9th Formulas -Maths 9th

Last Answer : Rational Numbers : Rational Numbers are numbers , which can be expressed in the form p/q, q≠0; where p and q are integers . The collection of all rationals are represented by Q. ∴ Q = {p/q ; p,q ... number of rational numbers x2 , x1, x3,... between any two rational numbers a and b so that a

MCQ Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables with answers -Maths 9th

Last Answer : 1) The linear equation 3x-11y=10 has: a. Unique solution b. Two solutions c. Infinitely many solutions d.No solutions c Explanation: 3x-11y=10 y=(3x-10)/11 Now for infinite values of x, y will ... c=0, always lie in the: a. First quadrant b. Second quadrant c. Third quadrant d. Fourth quadrant a

Examine the nature of the roots of the equations: (i) 2x^2 + 2x + 3 = 0 -Maths 9th

Description : Examine the nature of the roots of the equations: (i) 2x^2 + 2x + 3 = 0 -Maths 9th

Last Answer : answer:

Of the following quadratic equations, which is the one whose roots are 2 and – 15 ? -Maths 9th

Description : Of the following quadratic equations, which is the one whose roots are 2 and – 15 ? -Maths 9th

Last Answer : answer:

Find the values of k for which the equations x^2 – kx – 21 = 0 and x^2 – 3kx + 35 = 0 will have a common root? -Maths 9th

Description : Find the values of k for which the equations x^2 – kx – 21 = 0 and x^2 – 3kx + 35 = 0 will have a common root? -Maths 9th

Last Answer : Let the common root be α ⟹α2−kα−21=0......(1) α2−3kα+35=0........(2) (1)−(2)⟹2kα=56⟹α=k28​Substituting α=k28​ in (1) (k28​)2−28−21=0 ⟹k=±4

If two equations x^2 + a^2 = 1 – 2ax and x^2 + b^2 = 1 – 2bx have only one common root, then -Maths 9th

Description : If two equations x^2 + a^2 = 1 – 2ax and x^2 + b^2 = 1 – 2bx have only one common root, then -Maths 9th

Last Answer : x2+b2=1−2bx ...... (i) x2+b2+2bx=1 (x+b)2=1 x+b= 1 ∴x=1−b,−1−b x2+a2=1−2ax ...... (ii) x2+a2+2ax=1 (x+a)2=1 ∴x=1−a,−1−a It is given that the equations have only one root in ... or −1−a=−b−1 we get a=b but then both roots will be common, which is not possible. Hence, options A,B and C are correct.

Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

Description : Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

Last Answer : answer:

Cbqs (case base study ) of chapter 4 Linear Equations in Two Variables of maths class 9th -Maths 9th

Description : Cbqs (case base study ) of chapter 4 Linear Equations in Two Variables of maths class 9th -Maths 9th

Last Answer : answer:

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).

The perpendicular distance of the point P(3, 4) from the Y-axis is -Maths 9th

Description : The perpendicular distance of the point P(3, 4) from the Y-axis is -Maths 9th

Last Answer : (a) We know that, abscissa or the x-coordinate of a point is its perpendicular distance from the Y-axis. So, perpendicular distance of the point P(3, 4)from Y-axis = Abscissa = 3.

In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th

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

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).

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).

The perpendicular distance of the point P(3, 4) from the Y-axis is -Maths 9th

Description : The perpendicular distance of the point P(3, 4) from the Y-axis is -Maths 9th

Last Answer : (a) We know that, abscissa or the x-coordinate of a point is its perpendicular distance from the Y-axis. So, perpendicular distance of the point P(3, 4)from Y-axis = Abscissa = 3.

In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th

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

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).

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

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

Find the perpendicular distance of the point P(5, 7) from the y-axis. -Maths 9th

Description : Find the perpendicular distance of the point P(5, 7) from the y-axis. -Maths 9th

Last Answer : Solution :- 5

In fig.4.7, LM is a line parallel to the y - axis at a distance of 2 units. -Maths 9th

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

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

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

Last Answer : answer:

Given below graphs an oxygen dissociation curve `:-` Where in the body will haemoglobin be saturation at the percentage shown at points X,Y and Z.

Description : Given below graphs an oxygen dissociation curve `:-` Where in the body will haemoglobin be saturation at the percentage shown at points X,Y and Z.

Last Answer : Given below graphs an oxygen dissociation curve `:-` Where in the body will haemoglobin be ... ventricle , Y-Right ventricle, Z-Systemic artery

A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

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.

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

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.

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Last Answer : (d) Given, AD = 34 cm and AB = 30 cm In figure, draw OL ⊥ AB. Since, the perpendicular from the centre of a circle to a chord bisects the chord.

A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

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.

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

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.