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प्रश्न
The distance of the point (–6, 8) from x-axis is ______.
विकल्प
6 units
– 6 units
8 units
10 units
उत्तर
The distance of the point (–6, 8) from x-axis is 8 units.
Explanation:
The distance of the point (x, y) from x-axis is y-coordinate
∴ The distance of the point (–6, 8) from x-axis is 8 units.
संबंधित प्रश्न
On which axis do the following points lie?
P(5, 0)
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If the point ( x,y ) is equidistant form the points ( a+b,b-a ) and (a-b ,a+b ) , prove that bx = ay
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Find the ratio in which the line segment joining the points A(3, 8) and B(–9, 3) is divided by the Y– axis.
Find the possible pairs of coordinates of the fourth vertex D of the parallelogram, if three of its vertices are A(5, 6), B(1, –2) and C(3, –2).
If the vertices of a triangle are (1, −3), (4, p) and (−9, 7) and its area is 15 sq. units, find the value(s) of p.
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If \[D\left( - \frac{1}{5}, \frac{5}{2} \right), E(7, 3) \text{ and } F\left( \frac{7}{2}, \frac{7}{2} \right)\] are the mid-points of sides of \[∆ ABC\] , find the area of \[∆ ABC\] .
If P (2, p) is the mid-point of the line segment joining the points A (6, −5) and B (−2, 11). find the value of p.
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The points whose abscissa and ordinate have different signs will lie in ______.
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`
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Hence, the relation between x and y is `square`.
