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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
If two opposite vertices of a square are (5, 4) and (1, −6), find the coordinates of its remaining two vertices.
Three consecutive vertices of a parallelogram are (-2,-1), (1, 0) and (4, 3). Find the fourth vertex.
Find the ratio in which the point (2, y) divides the line segment joining the points A (-2,2) and B (3, 7). Also, find the value of y.
Find the area of a quadrilateral ABCD whose vertices area A(3, -1), B(9, -5) C(14, 0) and D(9, 19).
The area of the triangle formed by the points P (0, 1), Q (0, 5) and R (3, 4) is
If the points A(−2, 1), B(a, b) and C(4, −1) ae collinear and a − b = 1, find the values of aand b.
What is the distance between the points A (c, 0) and B (0, −c)?
Signs of the abscissa and ordinate of a point in the second quadrant are respectively.
Find the coordinates of the point whose ordinate is – 4 and which lies on y-axis.
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`.
