Coordinate Geometry — Telangana (SSC) Class 9 Mathematics Solutions (Free)

The coordinate plane is split into four quadrants by the sign of each coordinate: both positive puts a point top-right (Quadrant I), a negative x with positive y puts it top-left (Quadrant II), both negative puts it bottom-left (Quadrant III), and positive x with negative y puts it bottom-right (Quadrant IV). A(6, 3): both coordinates positive → Quadrant I B(-5, 2): negative x, positive y → Quadrant II C(-2, -6): both coordinates negative → Quadrant III D(3, -4): positive x, negative y → Quadrant IV

The distance between any two points is found by treating the horizontal and vertical gaps between them as the two legs of a right triangle, and applying the Pythagorean theorem to find the straight-line (hypotenuse) distance: distance = √[(horizontal gap)² + (vertical gap)²] Horizontal gap: 5 − 2 = 3 Vertical gap: 7 − 3 = 4 distance = √(3² + 4²) = √(9 + 16) = √25 = 5 The distance between P and Q is 5 units.

The midpoint of a segment is found by simply averaging the x-coordinates of its two endpoints, and separately averaging their y-coordinates. Midpoint x-coordinate: (-3 + 9)/2 = 6/2 = 3 Midpoint y-coordinate: (7 + (-1))/2 = 6/2 = 3 The midpoint of AB is (3, 3).

Points lying on the same straight line must show a constant rate of rise between each consecutive pair — in other words, the slope calculated between any two of them must match. Rate of rise from A to B: (7 − 3)/(3 − 1) = 4/2 = 2 Rate of rise from B to C: (11 − 7)/(5 − 3) = 4/2 = 2 Both segments rise at the identical rate of 2, which is only possible if A, B, and C all sit on one continuous straight line — so the three points are collinear.

For a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃), the enclosed area can be computed directly from the coordinates without needing to first find the base and height separately: Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| Substituting A(2, 1), B(6, 1), C(6, 7): Area = ½ |2(1 − 7) + 6(7 − 1) + 6(1 − 1)| = ½ |2(-6) + 6(6) + 6(0)| = ½ |-12 + 36 + 0| = ½ (24) = 12 square units

Any point sitting on the x-axis always has a y-coordinate of 0, so we can represent it as (x, 0) and solve for the unknown x. Applying the distance relationship between (x, 0) and (5, 4), and setting the result equal to 5: 5² = (x − 5)² + (0 − 4)² 25 = (x − 5)² + 16 (x − 5)² = 9 x − 5 = ±3 This gives x = 8 or x = 2. Checking both: the distance from (8, 0) to (5, 4) is √(9+16) = 5 ✓, and from (2, 0) to (5, 4) is also √(9+16) = 5 ✓. The point is either (8, 0) or (2, 0).

The slope of a line through two points is the ratio of how much y changes to how much x changes between them: slope = (change in y) / (change in x) = (1 − 7)/(8 − 4) = -6/4 = -1.5 A negative slope means y decreases as x increases, so the line is falling as it moves from left to right.

A quadrilateral qualifies as a rectangle when its opposite sides come out equal in length AND its two diagonals also come out equal in length. Side lengths: AB, from (0,0) to (7,0): length 7 BC, from (7,0) to (7,5): length 5 CD, from (7,5) to (0,5): length 7 DA, from (0,5) to (0,0): length 5 AB matches CD at 7 units, and BC matches DA at 5 units — the opposite-sides condition holds. Diagonal lengths: AC, from (0,0) to (7,5): √(7² + 5²) = √74 BD, from (7,0) to (0,5): √(7² + 5²) = √74 Both diagonals measure √74, so they're equal too. With both conditions satisfied, ABCD is confirmed to be a rectangle.