Two points on the grid hold all the information you need. The gap between them is found with Pythagoras, and the point halfway between them is just the average of the coordinates.
Two formulas, nothing more
• Distance. Make a right triangle: the horizontal leg is x2−x1, the vertical leg is y2−y1, and the segment is the hypotenuse: d=(x2−x1)2+(y2−y1)2.
• Midpoint. Average each coordinate: (2x1+x2,2y1+y2).
Spot the easy triples — a (3,4,5) or (5,12,13) pattern in the legs gives the distance with no surd at all.
📐Coordinate Plane ExplorerLive calculations as you move points
P1 (x₁, y₁)
x₁
,
y₁
P2 (x₂, y₂)
x₂
,
y₂
Distance P1P2√((4−1)² + (5−2)²) = √(9+9)4.243 units
Midpoint M((1+4)/2, (2+5)/2)(2.5, 3.5)
Slope (m)(5−2) / (4−1) = 3/31
Line Equationy = mx + c → y = 1x + (1)y = 1x + 1
The distance between (0, 0) and (6, 8) is 36+64= .
📝Practice Questions
Q1The distance between (0, 0) and (3, 4) is:
Q2The midpoint of the segment joining (2, 4) and (6, 8) is:
Q3The distance between (1, 2) and (4, 6) is:
📚
Real exam questions — Distance & Midpoint
2 question types · 8 solved examples from real SSC papers
Distance and midpoint are the two workhorse formulas of coordinate geometry. Distance is the Pythagorean theorem (spot a triplet and skip the arithmetic); midpoint is plain averaging (add the coordinates and halve). Most SSC questions reward recognising the shortcut over plugging into the full formula.
How to solve this type
The distance between (x1,y1) and (x2,y2) is (x2−x1)2+(y2−y1)2. From the origin it shrinks to x2+y2. First find the horizontal gap and vertical gap, then read them as the two legs of a right triangle. If the gaps form a known Pythagorean triplet (3-4-5, 5-12-13, 8-15-17, or any multiple like 6-8-10), write the hypotenuse straight away. If one coordinate is shared, the gap there is 0 and the distance is simply the difference of the other coordinate. Be ruthless with signs: 4−(−4)=8, not 0.
Find the distance of the point (5,12) from the origin.
