Unit-7 Coordinate Geometry
Explore subjects and practice MCQs
11
Topics
1,100
Total MCQs
Unit-7 Coordinate Geometry
Answer: The origin is the point (0, 0) where the two coordinate axes intersect.
Answer: d = sqrt((4-1)^2 + (6-2)^2) = sqrt(9 + 16) = sqrt(25) = 5.
Answer: Distances: sqrt(16+9)=5, sqrt(16+9)=5, and 8; two equal sides, so it is isosceles.
Answer: The horizontal line is the x-axis and the vertical line is the y-axis.
Answer: d = sqrt((-3-0)^2 + (4-0)^2) = sqrt(9 + 16) = sqrt(25) = 5.
Answer: Each side has length 2, so the triangle is equilateral.
Answer: The vertical line is the y-axis.
Answer: d = sqrt((-1-3)^2 + (1-(-2))^2) = sqrt(16 + 9) = sqrt(25) = 5.
Answer: Area = 1/2 |1(5-1) + 3(1-2) + 6(2-5)| = 1/2 |4 - 3 - 18| = 17/2 = 8.5.
Answer: The origin is the point where both coordinates are zero, i.e. (0, 0).
Answer: d = sqrt((-1-2)^2 + (3-(-1))^2) = sqrt(9 + 16) = sqrt(25) = 5.
Answer: Area = 1/2 |2(7-2) + 5(2-3) + 8(3-7)| = 1/2 |10 - 5 - 32| = 27/2 = 13.5.
Answer: The first coordinate x is the abscissa and the second coordinate y is the ordinate.
Answer: d = sqrt((-5-(-2))^2 + (-7-(-3))^2) = sqrt(9 + 16) = sqrt(25) = 5.
Answer: Area = 1/2 |0 + 2(2-0) + 5(0-5)| = 1/2 |4 - 25| = 21/2 = 10.5.
Answer: The second coordinate y is the ordinate.
Answer: d = sqrt((2-5)^2 + (5-1)^2) = sqrt(9 + 16) = sqrt(25) = 5.
Answer: The line through (2, 3) and (4, 7) has equation y = 2x - 1; putting y = 11 gives x = 6.
Answer: It is named after the French mathematician Rene Descartes.
Answer: d = sqrt((4-1)^2 + (5-1)^2) = sqrt(9 + 16) = sqrt(25) = 5.