Chapter 11 · Kerala SSLC Class 10 Maths
Geometry and Algebra
Two ways to learn — a crisp formula reference, or a story that builds from a parallelogram's missing corner to the equation of a circle.
Six scenes from a parallelogram's missing corner to a circle's equation.
How this works
Six scenes build from a parallelogram's missing corner all the way to the equation of a circle. Read the story, then tap "The Maths" for the formal result.
Completing the Parallelogram
Three vertices of a parallelogram are known: (3,2), (6,3), and (4,5). The fourth vertex is missing — but a parallelogram's opposite sides being equal and parallel is enough to pin it down exactly.
Moving from (3,2) to (6,3) — bottom-left to bottom-right — the x-coordinate increases by 6−3=3 and the y-coordinate increases by 3−2=1. This same "change in coordinates" must apply again moving along the top side, from (4,5) to the missing vertex — since opposite sides of a parallelogram are equal length AND parallel.
Bottom side: x-change = 3, y-change = 1
Top side (same change): from (4,5), add 3 to x and 1 to y
Missing vertex = (4+3, 5+1) = (7, 6)
The two triangles formed (one under each side) have equal hypotenuses — they're opposite sides of the parallelogram — and equal angles (matching angles from parallel lines). So the triangles are congruent, and their legs match exactly.
Splitting the Difference
Two points, (2,3) and (8,5), mark the ends of a line segment. Find its midpoint.
The change from (2,3) to (8,5) is 6 in x and 2 in y. The midpoint is exactly HALFWAY along this change — half of 6 is 3, half of 2 is 1.
Midpoint = (2+3, 3+1) = (5, 4)
This halving works because the small triangle (start point to midpoint) and the large triangle (start point to end point) have identical angles, so their sides scale by the same factor — here, exactly 1/2.
Suppose instead of the midpoint, we want the point dividing the segment from (2,4) to (8,7) in ratio 1:2 — meaning it's 1/3 of the way along.
Δx = 8−2=6, Δy=7−4=3
1/3 of these: Δx'=2, Δy'=1
Point = (2+2, 4+1) = (4, 5)
Same idea, different fraction: instead of halving (for the midpoint), take m/(m+n) of the total change.
The Constant Ratio
Look at the line through (2,5) and (6,7). Moving between these two points, x increases by 4 and y increases by 2 — so the y-change is HALF the x-change.
Pick any other two points on this same line, and the same thing happens: the y-change is always half the x-change, no matter which two points you choose.
On this line: (change in y) = (1/2) × (change in x), always
This fixed multiplying number is called the slope of the line.
slope = (change in y) / (change in x) = (y₂−y₁)/(x₂−x₁)
Given that a line through (1,6) has slope 3/4, what is the y-coordinate of the point on this line with x-coordinate 4?
x-change from (1,6) to (4,?) = 4−1 = 3
y-change = (3/4) × 3 = 9/4
y = 6 + 9/4 = 33/4
So (4, 33/4) is on this line — found using only ONE known point and the slope, no second point needed.
Naming the Line
Every point (x,y) on the line through (2,4) with slope 2/3 satisfies one equation: the y-change from (2,4) to (x,y) is 2/3 of the x-change.
y − 4 = (2/3)(x − 2)
Clear the fraction and collect terms:
3(y−4) = 2(x−2)
3y−12 = 2x−4
2x − 3y + 8 = 0
This single equation is satisfied by EVERY point on the line, and no point off the line — it is called the equation of the line.
Perpendicular Slopes
Two lines cross at point (a,b), with slopes p and q. Is there a relationship between p and q if the lines are perpendicular?
Moving 1 unit along x from (a,b) on the first line lands at (a+1, b+p) — since slope p means the y-change per unit x-change is p. On the second line, the matching point is (a+1, b+q).
These three points — (a,b), (a+1,b+p), (a+1,b+q) — form a right triangle, with the right angle at (a,b) since the two lines meet perpendicularly there.
Two perpendicular legs: lengths √(1+p²) and √(1+q²)
Hypotenuse: from (a+1,b+p) to (a+1,b+q), length |p−q|
By Pythagoras: (1+p²) + (1+q²) = (p−q)²
Expand and simplify:
2 + p² + q² = p² − 2pq + q²
2 = −2pq
pq = −1
The Circle's Own Equation
A line has an equation connecting the x and y of every point ON it. Does a circle have one too?
Take the circle with centre (1,4) and radius 2. Every point (x,y) on this circle is at distance exactly 2 from (1,4).
distance from (1,4) to (x,y) = √((x−1)²+(y−4)²) = 2
Squaring both sides:
(x−1)² + (y−4)² = 4
Any point satisfying this equation is automatically at distance 2 from (1,4) — so it's on the circle. Any point NOT satisfying it is at some other distance, and isn't on the circle. This equation captures the circle exactly.
The Big Picture
Every idea in this chapter uses the same coordinate-change logic from Chapter 7 — applied to shapes instead of single distances.
Fourth vertex of a parallelogram?
D = B + C − A
Midpoint of two points?
M = ((x₁+x₂)/2, (y₁+y₂)/2)
Point dividing a segment m:n?
add [m/(m+n)] of the coordinate changes
Steepness of a line?
slope = (y₂−y₁)/(x₂−x₁)
Equation of a line?
y−y₁=m(x−x₁) → ax+by+c=0
Perpendicular lines?
m₁ × m₂ = −1
Equation of a circle?
(x−h)² + (y−k)² = r²
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