Point on Y-axis is equidistant from 5,4 and - 2,3 is -Maths 9th

Point on Y-axis is equidistant from 5,4 and - 2,3 is -Maths 9th
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Point on Y-axis is equidistant from 5,4 and - 2,3 is -Maths 9th

Description : Point on Y-axis is equidistant from 5,4 and - 2,3 is -Maths 9th

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

If the points (2, 1) and (1, – 2) are equidistant from the point (x, y), show that x + 3y = 0. -Maths 9th

Description : If the points (2, 1) and (1, – 2) are equidistant from the point (x, y), show that x + 3y = 0. -Maths 9th

Last Answer : (a) The distance d between any two points say P(x1, y1) and Q(x2, y2) is given by:d = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)⇒ d2 = (x2 - x1)2 + (y2 - y1)2 ⇒ d = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)( ... distance of a point P(x1, y1) form the origin= \(\sqrt{(x_2-0)^2+(y_2-0)^2}\) = \(\sqrt{x^2_1+y^2_1}\)

Find a point on the y-axis equidistant from (-5, 2) and (9, -2). -Maths 10th

Description : Find a point on the y-axis equidistant from (-5, 2) and (9, -2). -Maths 10th

Last Answer : answer:

If M(x, y) is equidistant from A(a + b, b – a) and B(a – b, a + b), then -Maths 9th

Description : If M(x, y) is equidistant from A(a + b, b – a) and B(a – b, a + b), then -Maths 9th

Last Answer : (b)10 + \(5\sqrt2\)Perimeter of ΔABC = AB + BC + CA= \(\sqrt{(0+4)^2+(-1-2)^2}\) + \(\sqrt{(3-0)^2+(3+1)^2}\) + \(\sqrt{(3-4)^2+(3-2)^2}\)= \(\sqrt{16+9}\) + \(\sqrt{9+16}\) +\(\sqrt{49+1}\)= \(\sqrt{25}\) + \(\sqrt{25}\) + \(\sqrt{50}\) = 5 + 5 + \(5\sqrt2\) = 10 + \(5\sqrt2\)

The point whose abscissa is equal to its ordinate and which is equidistant from A(–1, 0) and B(0, 5) is -Maths 9th

Description : The point whose abscissa is equal to its ordinate and which is equidistant from A(–1, 0) and B(0, 5) is -Maths 9th

Last Answer : Putting \(x\) = 0 in equation of one of the lines say 9\(x\) + 40y -20 = 0, we get y = \(rac{1}{2}\)∴ A point on 9\(x\) + 40y - 20 = 0 is \(\big(0,rac{1}{2}\big)\)∴ Distance of \(\big(0,rac{1}{2}\big) ... imesrac{1}{2}+21\big|}{\sqrt{9^2+40^2}}\) = \(rac{|41|}{\sqrt{1681}}\) = \(rac{41}{41}\) = 1.

The point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1). Then PB is equal to : -Maths 9th

Description : The point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1). Then PB is equal to : -Maths 9th

Last Answer : (b) (2, 2)Let the point be P whose abscissa = ordinate = a. ∴ P ≡ (a, a) Given, PA = PB ⇒ (a + 1)2 + a2 = a2 + (a – 5)2 ⇒ 2a2 + 2a + 1 = 2a2 – 10a + 25 ⇒ 12a = 24 ⇒ a = 2. ∴ The point is (2, 2).

If point A (0,2) is equidistant from the point B (3, p)and C (p, 5), find p. -Maths 9th

Description : If point A (0,2) is equidistant from the point B (3, p)and C (p, 5), find p. -Maths 9th

Last Answer : Given, AB=AC (AB)2=(AC)2 Distance between two points=(x2​−x1​)2+(y2​−y1​)2​(AB)2=(AC)2⟹(0−3)2+(2−p)2=(0−p)2+(2−5)2 9+4+p2−4p=p2+9 p=1 Distance=(0−3)2+(2−1)2​Distance=10​

The locus of a point in rhombus ABCD which is equidistant from A and C is -Maths 9th

Description : The locus of a point in rhombus ABCD which is equidistant from A and C is -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 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).

Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -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

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

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.

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

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.

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.

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

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.

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

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

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

Last Answer : Solution :- On the x-axis.

Find the point whose ordinate is –3 and which lies on y-axis. -Maths 9th

Description : Find the point whose ordinate is –3 and which lies on y-axis. -Maths 9th

Last Answer : Solution :- (0,-3)

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:

The neutral axis of a beam cross-section must (A) Pass through the centroid of the section (B) Be equidistant from the top of bottom films (C) Be an axis of symmetry of the section (D) None of these

Description : The neutral axis of a beam cross-section must (A) Pass through the centroid of the section (B) Be equidistant from the top of bottom films (C) Be an axis of symmetry of the section (D) None of these

Last Answer : (A) Pass through the centroid of the section

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

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.

Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis -Maths 9th

Description : Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis -Maths 9th

Last Answer : answer:

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

Which of the following points lies on Y-axis ? -Maths 9th

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

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

Which of the following points lies on Y-axis ? -Maths 9th

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

Write the equation representing y-axis. -Maths 9th

Description : Write the equation representing y-axis. -Maths 9th

Last Answer : Solution :- x =0

Find the coordinate where the linear equation 4x - 23 y = 7 meets at y-axis. -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)

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

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

Last Answer : hope it helps

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

Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the longer side lies on the y-axis and one of the vertices lies in the second quadrant. -Maths 9th

Description : Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the longer side lies on the y-axis and one of the vertices lies in the second quadrant. -Maths 9th

Last Answer : Solution :-

The equation of y-axis is .......... -Maths 9th

Description : The equation of y-axis is .......... -Maths 9th

Last Answer : answer:

The equation of a line parallel to the y- axis is of the form ................. -Maths 9th

Description : The equation of a line parallel to the y- axis is of the form ................. -Maths 9th

Last Answer : answer:

What is the equation of the line passing through (2, –3) and parallel to the y-axis ? -Maths 9th

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.

Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed. -Maths 9th

Description : Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed. -Maths 9th

Last Answer : Let a be the equal distance from origin on both axes. Now, the coordinates of two points on equal distance 'a'on x-axis are Pla, 0) and R(-a, 0). Also, the coordinates of two points on equal distance 'a' on Y-axis are Q(0, a) and S(0, -a). Join all the four points on the graph.

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.

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.

The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Description : The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Last Answer : (b) Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

Any point on the X-axis is of the form -Maths 9th

Description : Any point on the X-axis is of the form -Maths 9th

Last Answer : (c) Every point on the X-axis has its y-coordinate equal to zero. i.e., y = 0 Hence, the general form of every point on X-axis is (x, 0).

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

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).

The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Description : The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Last Answer : (b) Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

Any point on the X-axis is of the form -Maths 9th

Description : Any point on the X-axis is of the form -Maths 9th

Last Answer : (c) Every point on the X-axis has its y-coordinate equal to zero. i.e., y = 0 Hence, the general form of every point on X-axis is (x, 0).

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

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).

Is (x, 0) a point on the x–axis? Give reason. -Maths 9th

Description : Is (x, 0) a point on the x–axis? Give reason. -Maths 9th

Last Answer : Solution :-

Write the equation of a line parallel to x-axis and passing through the point (–3, –4). -Maths 9th

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

Write the coordinates of a point on x-axis at a distance of 6 units from the origin in the positive direction of x-axis and then justify your answer. -Maths 9th

Description : Write the coordinates of a point on x-axis at a distance of 6 units from the origin in the positive direction of x-axis and then justify your answer. -Maths 9th

Last Answer : Solution :- As, any point on x-axis has coordinates (,)x0 where x is the distance from origin, so required coordinates are (6, 0).

Find the co-ordinates of the points on the x-axis which are at a distance of 10 units from the point (– 4, 8)? -Maths 9th

Description : Find the co-ordinates of the points on the x-axis which are at a distance of 10 units from the point (– 4, 8)? -Maths 9th

Last Answer : (a) Internal division: If P(x, y) divides the line segment formed by the joining of the points A (x1, y1) and B (x2, y2) internally in the ratio m1 : m2. Then\(x=rac{m_1x_2+m_2x_1}{m_1+m_2}\) and \(y=rac ... : 1, the co-ordinates of the mid-point are \(\bigg(rac{x_1+x_2}{2},rac{y_1+y_2}{2}\bigg)\).