Distance Formula
ICSE Class 9 · Mathematics · Chapter 25
What this chapter covers
- Why a formula for distance is needed
- Distance between points on a line parallel to an axis
- Deriving the distance formula
- Distance of a point from the origin
- Using the distance formula
- Finding an unknown coordinate
- Points equidistant from given points
- Collinearity of three points
- Classifying triangles using distances
- Classifying quadrilaterals using distances
- Common mistakes to avoid
Key points
- The distance between two points is the length of the straight segment joining them; it is never negative, and AB = BA.
- For points on a line parallel to the x-axis, AB = |x2 - x1|; for points on a line parallel to the y-axis, AB = |y2 - y1|.
- Distance formula: AB = √((x2 - x1)2 + (y2 - y1)2), which is Pythagoras theorem written in coordinates.
- Distance of P(x, y) from the origin: OP = √(x2 + y2).
- To find an unknown coordinate, square both sides and solve; expect two values unless a condition removes one.
- PA = PB is the same as PA2 = PB2; take a point on the x-axis as (x, 0) and on the y-axis as (0, y).
- Three points are collinear when the largest of the three distances equals the sum of the other two.
- Triangles: compare the three sides for equilateral, isosceles or scalene, and use the converse of Pythagoras theorem for a right angle.
- Quadrilaterals: four equal sides with equal diagonals make a square; four equal sides with unequal diagonals, a rhombus; equal opposite sides with equal diagonals, a rectangle; equal opposite sides with unequal diagonals, a parallelogram.
- Leave answers in simplest surd form unless a decimal is asked for, and always state the unit.
Questions answered
State the distance formula for the points (x1, y1) and (x2, y2), and name the theorem it comes from.
The distance is √((x2 - x1)2 + (y2 - y1)2) units. It comes from Pythagoras theorem applied to the right-angled triangle whose legs are the horizontal step |x2 - x1| and the vertical step |y2 - y1|.
Find the distance between (-3, -7) and (-3, 5) without using the full formula.
The abscissae are equal, so the segment is parallel to the y-axis. Distance = |5 - (-7)| = 12 units.
Find the distance between (2, -3) and (-4, 5).
√((-4 - 2)2 + (5 + 3)2) = √(36 + 64) = √(100) = 10 units.
Find the distance of the point (-8, 15) from the origin.
√(64 + 225) = √(289) = 17 units.
Find the distance between (7, -4) and (-5, 1).
√((-12)2 + 52) = √(144 + 25) = √(169) = 13 units.
The distance between (a, 2) and (3, -6) is 10 units. Find the possible values of a.
(3 - a)2 + 64 = 100, so (3 - a)2 = 36 and 3 - a = ± 6. Hence a = -3 or a = 9.
Find the point on the x-axis that is equidistant from (7, 6) and (-3, 4).
Let the point be (x, 0). Then (x - 7)2 + 36 = (x + 3)2 + 16, which gives -14x + 85 = 6x + 25, so x = 3. The point is (3, 0), at a distance of √(52) = 2√(13) units from each.
Show that (-1, -1), (2, 3) and (8, 11) are collinear.
The distances are √(9 + 16) = 5, √(36 + 64) = 10 and √(81 + 144) = 15 units. Since 5 + 10 = 15, the points are collinear, with (2, 3) between the other two.
Past board exam questions
ICSE 2019 sample paper · 3 marks
Find the coordinates of the points on Y-axis which are at a distance of 5√(2) units from the point (5,8).
Show answer
Let the point on the Y-axis be (0, y).
√((5-0)2 + (8-y)2) = 5√(2)
25 + (8-y)2 = 50 ⇒ (8-y)2 = 25 ⇒ 8 - y = ± 5
y = 3 or y = 13.
The points are (0, 3) and (0, 13).