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प्रश्न
Two points having same abscissae but different ordinate lie on
विकल्प
x-axis
y-axis
a line parallel to y-axis
a line parallel to x-axis
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उत्तर
Let the points p (3,5)and Q (3,2) having the same abscissa but different ordinates be shown in the graph given below:
Fig: (location of two considered points)
And these points lie on a line parallel to y−axis
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संबंधित प्रश्न
Let ABCD be a square of side 2a. Find the coordinates of the vertices of this square when A coincides with the origin and AB and AD are along OX and OY respectively.
Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.
Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.
Find the coordinates of the point which divides the line segment joining (−1,3) and (4, −7) internally in the ratio 3 : 4
The distance of the point P (4, 3) from the origin is
If the centroid of the triangle formed by points P (a, b), Q(b, c) and R (c, a) is at the origin, what is the value of a + b + c?
Write the ratio in which the line segment doining the points A (3, −6), and B (5, 3) is divided by X-axis.
The points (–5, 2) and (2, –5) lie in the ______.
The coordinates of a point whose ordinate is `-1/2` and abscissa is 1 are `-1/2, 1`.
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`.
