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The point of intersection of both the axes is called:
The horizontal axis in the coordinate plane is called the xaxis. The vertical axis is called the yaxis. The point at which the two axes intersect is called the origin.
If two points P and Q have same abscissae and different ordinates, then points P and Q will definitely lie on
If P and Q have same abscissa but different ordinates then P and Q have coordinates, as P (a, c ), Q (a, b)
∵ xcoordinate is constant.
∴ P and Q will lie on line parallel to yaxis.
Minimum distance of point (4, 6) from x axis will be
Minimum distance of point (α, b) from xaxis = Perpendicular distance between point and xaxis = ycoordinate (ordinate) of the point = 6 = 6
Point (4, 3) have (+, ) sign convention, that belongs to IV^{th} quadrant.
If two points M and N lie on yaxis, and have same absolute value of abscissa but different signs. If the abscissa of point M is K, then the distance between M and N is equal to:
Let the coordinates of point M be (0, k)
{∵ M is on yaxis}
∴ Coordinates of point N = (0,  K)
∴ Distance between point M and N
= K  (K) = 2K
∵ (3, 2) has (, ) sign convention.
∴ (3, 2)/0 Belongs to 3^{rd} quadrant.
The mirror image of point (4, 2) about xaxis will lie in:
Point A is the reflected point.
Point A will have coordinates, as, abscissa will not change and ordinate will change sign
∴ Reflected point will lie in 2^{nd} quadrant.
The distance between (12, 5) and origin is
In ΔOAB
OA^{2} = OB^{2} + AB^{2}
= (12)^{2} + (OC)^{2} = (12)^{2} + (5)^{2}
= 169
⇒ OA = √169 = 13 Units
On every point of yaxis, abscissa = 0
∴ x = 0 as the equation of yaxis
The area of triangle formed by points A (3, 0), B (0, 4), C (0, 4) will be (in sq. units)
Coordinates of point Q ≡ (3, 5).
After plotting ΔPOQ on graph, it can be clearly viewed that ΔPOQ is isosceles.
∴ Area of ΔPOQ = 1/2 × OR × PQ
= 1/2 × 5 × [3 (3)]
= 1/2 × 5 × 6 = 15 sq. units
Distance between points (24, 10) and (48, 10) will be
Distance between (24, 10) and (48,10) will be equal to the absolute value of difference between the abscissa of the points as, the points have same ordinate.
∴ Distance = 48  (24) = 72 units
The difference between ordinates of point P (3, 6) and Q (6, 3) is
Ordinates of points are 6 and 3
∴ Difference = 6  3 = 9
The point of intersection of lines having equations x + y = 6 and x  y = 2, is
Let the coordinates of point of intersection be (x_{1}, y_{1})
∴ x_{1 }+ y_{1} = 6
x_{1}  y_{1 }= 2
Solving these two equations, we get
x_{1} = 4, and , y_{1} = 2
∴ point of intersection ≡ (4, 2)
Distance of point ( 24 ,10) from origin will be
In ΔOBC
BC^{2} + OC^{2} = OB^{2}
⇒ (OA)^{2} + (OC)^{2} = OB^{2}
⇒ (10)^{2} + (24)^{2} = OB^{2}
⇒ OB = √676 = 26 units
The area of triangle formed by the points A (2, 0), B (6, 0), C (4, 6) is
After plotting figure, it can be clearly seen that ABC is an isosceles triangle in which,
AB = (6  2) = 4 units
CD = (6  0) = 6 units
∴ Ares of ΔABC = 1/2 × AB × CD
= 1/2 × 4 × 6 = 12 sq. units
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