CBSE X Maths — Coordinate Geometry: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit III: Coordinate Geometry

Coordinate Geometry

Two formulas do almost everything in this topic: one measures the straight-line distance between two points, the other finds a point that divides a segment in a given ratio.

1. Distance Formula

d = √[(x₂−x₁)² + (y₂−y₁)²]

Worked example — distance between two points Find the distance between (3, 4) and (7, 1).
d = √[(7−3)² + (1−4)²] = √[16 + 9] = √25 = 5

2. Section Formula

The point P(x, y) dividing the segment joining A(x₁,y₁) and B(x₂,y₂) internally in the ratio m:n is:

x = (mx₂ + nx₁)/(m+n), y = (my₂ + ny₁)/(m+n)

Worked example — section formula Find the point dividing the join of (4,−3) and (8,5) in the ratio 3:1.
x = (3×8 + 1×4)/4 = 28/4 = 7 y = (3×5 + 1×(−3))/4 = 12/4 = 3
Point: (7, 3)

Worked example — midpoint (ratio 1:1) Find the midpoint of the segment joining (2,−3) and (−6,7).
Midpoint = ((2+(−6))/2, (−3+7)/2) = (−2, 2)

Worked example — using distances to check collinearity Show that (1,1), (−2,7), and (3,−3) are collinear.
Distance (1,1)-(−2,7) = √(9+36) = 3√5. Distance (1,1)-(3,−3) = √(4+16) = 2√5. Distance (−2,7)-(3,−3) = √(25+100) = 5√5.
Since 3√5 + 2√5 = 5√5, the point (1,1) lies exactly between the other two — the three points are collinear.

Common mistakes to watch for

Mixing up which point is (x₁,y₁) and which is (x₂,y₂) in the section formula — the ratio m:n must match the order A:B you chose.
Forgetting to square root at the end of the distance formula.
When finding a ratio (rather than being given one), remembering to write it as k:1 and solve for k using either coordinate.

Key Vocabulary Recap

Collinear points — points lying on the same straight line

Internal division — a point lying between the two given points

Syllabus Reference

This note covers CBSE Class X Mathematics Unit III, Coordinate Geometry: review of coordinate geometry concepts, the distance formula, and the section formula (internal division).

Practice questions

CBSE Class X Mathematics · Unit III: Coordinate Geometry — Practice Question Bank (no answers)

Coordinate Geometry — Practice

Covers: the distance formula and the section formula. 10 questions.

Find the distance between (3, 4) and (7, 1).

Find the distance between (−2, 3) and (6, −3).

Show that the points (1, 1), (−2, 7), and (3, −3) are collinear.

Find the coordinates of the point which divides the join of (4, −3) and (8, 5) in the ratio 3:1.

Find the midpoint of the line segment joining (2, −3) and (−6, 7).

Find the ratio in which the point (−4, 6) divides the line segment joining A(−6, 10) and B(3, −8).

Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).

If the distance between the points (x, 7) and (1, 3) is 5, find the value of x.

Find the coordinates of the points of trisection of the line segment joining (2, −2) and (−7, 4).

Q10

If A(2, 3) and B(4, k) are such that the distance AB = 2√2, find the value of k.

More Practice

Q11

Find a relation between x and y such that the point (x, y) is equidistant from (3, 6) and (−3, 4).

Q12

AB is the diameter of a circle with centre (2, −3). If B is (1, 4), find the coordinates of A.

Q13

If the point P(x, y) is equidistant from A(5, 1) and B(−1, 5), prove that 3x = 2y.

Q14

Find the value of y for which the distance between P(2, −3) and Q(10, y) is 10 units.

Q15

Show that the points (4,4), (3,5), and (−1,−1) form a right-angled triangle, and identify the vertex at which the right angle lies.

Q16

Find the coordinates of the point which divides the join of A(−1, 7) and B(4, −3) in the ratio 2:3.

Q17

Find the ratio in which the y-axis divides the line segment joining (5, −6) and (−1, −4).

Q18

Points A(2, −2) and B(−7, 4) are given. Find the point P on AB such that AP = (2/5)AB.

Q19

Determine whether the points (3, −2), (5, 2), and (8, 8) are collinear, using the distance method.

Q20

Find the value of k if the point P(2, 4) is equidistant from A(5, k) and B(k, 7).

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit III: Coordinate Geometry — Solutions

Coordinate Geometry — Practice: Solutions

Distance (3,4)-(7,1).

d=√[(x₂−x₁)²+(y₂−y₁)²]

√[16+9]=√25

✓5

Distance (−2,3)-(6,−3).

Apply the distance formula

√[64+36]=√100

✓10

Show (1,1),(−2,7),(3,−3) collinear.

Find all three pairwise distances

d₁=√45=3√5, d₂=√20=2√5, d₃=√125=5√5

✓3√5+2√5=5√5, so (1,1) lies between the other two — the points are collinear

Divides (4,−3),(8,5) in ratio 3:1.

x=(mx₂+nx₁)/(m+n), y similarly

x=(3×8+4)/4=7, y=(3×5−3)/4=3

✓(7, 3)

Midpoint of (2,−3),(−6,7).

Midpoint = average of coordinates

((2−6)/2, (−3+7)/2)

✓(−2, 2)

Ratio for (−4,6) dividing A(−6,10),B(3,−8).

Let ratio be k:1. Use the x-coordinate

(3k−6)/(k+1)=−4 → 7k=2

✓k = 2/7, so the ratio is 2:7

Point on x-axis equidistant from (2,−5),(−2,9).

Let point be (x,0). Set the two distances² equal

(x−2)²+25=(x+2)²+81

Expand and solve

−4x+29=4x+85 → x=−7

✓(−7, 0)

Distance (x,7)-(1,3)=5. Find x.

(x−1)²+16=25

(x−1)²=9 → x−1=±3

✓x = 4 or x = −2

Trisection points of (2,−2),(−7,4).

First point divides in ratio 1:2

((−7+4)/3, (4−4)/3)=(−1,0)

Second point divides in ratio 2:1

((−14+2)/3, (8−2)/3)=(−4,2)

✓(−1, 0) and (−4, 2)

Q10

A(2,3),B(4,k), AB=2√2. Find k.

Square both sides of the distance formula

4+(k−3)²=8 → (k−3)²=4

✓k = 5 or k = 1

More Practice — Solutions

Q11

Relation for (x,y) equidistant from (3,6),(−3,4).

Set the two squared distances equal

(x−3)²+(y−6)²=(x+3)²+(y−4)²

Expand and simplify

−12x−4y+20=0

✓3x + y = 5

Q12

Centre (2,−3), B=(1,4). Find A.

Centre is the midpoint of AB

(x+1)/2=2 → x=3; (y+4)/2=−3 → y=−10

✓A = (3, −10)

Q13

P equidistant from A(5,1),B(−1,5). Prove 3x=2y.

Set squared distances equal and expand

(x−5)²+(y−1)²=(x+1)²+(y−5)² → −12x+8y=0

✓−3x+2y=0, i.e. 3x=2y — proved

Q14

Distance P(2,−3)Q(10,y)=10. Find y.

(10−2)²+(y+3)²=100

(y+3)²=36 → y+3=±6

✓y = 3 or y = −9

Q15

Show (4,4),(3,5),(−1,−1) is right-angled.

Find all three squared side lengths

d₁²(4,4)-(3,5)=2; d₂²(3,5)-(−1,−1)=52; d₃²(4,4)-(−1,−1)=50

✓Since d₁²+d₃²=2+50=52=d₂², the triangle is right-angled, with the right angle at (4,4) (the vertex common to the two shorter sides)

Q16

Divides A(−1,7),B(4,−3) in ratio 2:3.

x=(2×4+3×−1)/5, y=(2×−3+3×7)/5

x=1, y=3

✓(1, 3)

Q17

Ratio y-axis divides (5,−6),(−1,−4).

Let ratio be k:1. Set the x-coordinate to 0

(k×−1+5)/(k+1)=0 → −k+5=0

✓k = 5, so the ratio is 5:1

Q18

AP=(2/5)AB for A(2,−2),B(−7,4). Find P.

AP:PB = 2:3

x=(2×−7+3×2)/5=−8/5; y=(2×4+3×−2)/5=2/5

✓P = (−8/5, 2/5)

Q19

Collinearity of (3,−2),(5,2),(8,8).

Find all three pairwise distances

d₁=√20=2√5, d₂=√45=3√5, d₃=√125=5√5

✓2√5+3√5=5√5, so the points are collinear

Q20

P(2,4) equidistant from A(5,k),B(k,7).

Set squared distances equal

(2−5)²+(4−k)²=(2−k)²+(4−7)²

Expand and simplify

16−8k=4−4k → 12=4k

✓k = 3

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