मराठी

Ordinate of all points on the x-axis is ______.

Advertisements
Advertisements

प्रश्न

Ordinate of all points on the x-axis is ______.

पर्याय

  • 0

  • 1

  • –1

  • any number

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

Ordinate of all points on the x-axis is 0.

Explanation:

Ordinate of all points on the x-axis is zero. Because ordinate (or y-coordinate) of a point is perpendicular distance of this point from the x-axis measured along the y-axis.

If point lies on x-axis, then the perpendicular distance of point from x-axis will be zero, so ordinate will be zero.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Coordinate Geometry - Exercise 3.1 [पृष्ठ २५]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
पाठ 3 Coordinate Geometry
Exercise 3.1 | Q 6. | पृष्ठ २५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

If G be the centroid of a triangle ABC, prove that:

AB2 + BC2 + CA2 = 3 (GA2 + GB2 + GC2)


Determine the ratio in which the point (-6, a) divides the join of A (-3, 1)  and B (-8, 9). Also, find the value of a.


If the points P(x, y) is equidistant from the points A(5, 1)and B(–1, 5), prove that 3x = 2y.


Show that the following points are the vertices of a square:

A(3, 2), B(0, 5), C(–3, 2) and D(0, –1)


Find the ratio in which the line segment joining the points A(3, −3) and B(−2, 7) is divided by the x-axis. Also, find the coordinates of the point of division.   


If Points (1, 2) (−5, 6) and (a, −2) are collinear, then a =


The line segment joining the points A(2, 1) and B (5, - 8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by  2x - y + k= 0  find the value of k.


The line segment joining the points (3, -1) and (-6, 5) is trisected. The coordinates of point of trisection are ______.


If the points P(1, 2), Q(0, 0) and R(x, y) are collinear, then find the relation between x and y.

Given points are P(1, 2), Q(0, 0) and R(x, y).

The given points are collinear, so the area of the triangle formed by them is `square`.

∴ `1/2 |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = square`

`1/2 |1(square) + 0(square) + x(square)| = square`

`square + square + square` = 0

`square + square` = 0

`square = square`

Hence, the relation between x and y is `square`.


The distance of the point (–4, 3) from y-axis is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×