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Practice each question, then open the answer or worked solution.
The distance of the point P (2, 3) from the x-axis is
(A) 2 (B) 3 (C) 1 (D) 5
(B) 3
We know that,
(x, y) is a point on the Cartesian plane in first quadrant.
Then,
x = Perpendicular distance from Y – axis and
y = Perpendicular distance from X – axis
Therefore, the perpendicular distance from X-axis = y coordinate = 3
The distance between the points A (0, 6) and B (0, –2) is
(A) 6 (B) 8 (C) 4 (D) 2
(B) 8
Distance formula: d2 = (x2 – x1)2 + (y2 – y1)2
According to the question,
We have,
x1 = 0, x2 = 0
y1 = 6, y2 = – 2
d2 = (0 – 0)2 + ( – 2 – 6)2
d= √((0)2+ (-8)2)
d = √64
d = 8 units
Therefore, the distance between A (0, 6) and B (0, 2) is 8
The distance of the point P (–6, 8) from the origin is
(A) 8 (B) 2√7 (C) 10 (D) 6
(C) 10
Distance formula: d2 = (x2 – x1)2 + (y2 – y1)2
According to the question,
We have;
x1 = – 6, x2 = 0
y2 = 8, y2 = 0
d2 = [0 – ( – 6)]2 + [0 – 8]2
d= √((0-(-6))2+ (0-8)2
d= √((6)2+ (-8)2)
d = √(36 + 64)
d = √100
d = 10
Therefore, the distance between P ( – 6, 8) and origin O (0, 0) is 10
The distance between the points (0, 5) and (–5, 0) is
(A) 5 (B) 5√2 (C) 2√5 (D) 10
(B) 5√ 2
Distance formula: d2 = (x2 – x1)2 + (y2 – y1)2
According to the question,
We have;
x1 = 0, x2 = – 5
y2 = 5, y2 = 0
d2 = (( – 5) – 0)2 + (0 – 5)2
d= √(-5-0)2+ (0-5)2
d= √((-5)2+ (-5)2)
d = √(25 + 25)
d= √50= 5√2
So the distance between (0, 5) and ( – 5, 0) = 5√ 2
AOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is
(A) 5 (B) 3 (C) √34 (D) 4
(C) √34
The three vertices are: A = (0, 3), O = (0, 0), B = (5, 0)
We know that, the diagonals of a rectangle are of equal length,
Length of the diagonal AB = Distance between the points A and B
Distance formula: d2 = (x2 – x1)2 + (y2 – y1)2
According to the question,
We have;
x1 = 0, x2 = 5
y2 = 3, y2 = 0
d2 = (5 – 0)2 + (0 – 3)2
d= √((5-0)2+ (0-3)2)
d = √(25 + 9)= √34
Distance between A (0, 3) and B (5, 0) is √34
Therefore, the length of its diagonal is √34
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(A) 5 (B) 12 (C) 11 (D) 7+ √5
(B) 12
The vertices of a triangle are (0, 4), (0, 0) and (3, 0).
Now, perimeter of ΔAOB = Sum of the length of all its sides:
= distance between (OA+OB+AB)
Distance between the points (x1,y1) and (x2, y2) is given by,
d = √ ((x2 – x1)2 + (y2 – y1)2)
To find:
Distance between A(0, 4) and O(0, 0) + Distance between O(0, 0) and B(3, 0) + Distance between A(0, 4) and B(3, 0)
Therefore, the required perimeter of triangle is 12
The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is
(A) 14 (B) 28 (C) 8 (D) 6
(C) 8
Vertices of the triangle are,
A (x1, y1) = (3, 0)
B (x2, y2) = (7, 0)
C (x3, y3) = (8, 4)
Therefore, the area of ΔABC is 8.