Class 10 CBSE · Mathematics · Chapter 7
Mastering Coordinate Geometry:
Step-by-Step Problem Sets
Three pillars, fifteen problems, zero skipped steps. Every solution uses a rigorous Chain-of-Thought approach — the same logic you need in your board exam.
Distance: d = √[(x₂−x₁)² + (y₂−y₁)²]
Section: x = (m₁x₂ + m₂x₁) / (m₁+m₂)
Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2)
§1 The Distance Formula
Rooted in Pythagoras' theorem, this formula calculates the exact distance between any two points on the Cartesian plane — no ruler, no protractor required.
\[ d \;=\; \sqrt{\,(x_2 - x_1)^2 + (y_2 - y_1)^2\,} \]
Foundational
Core
Intermediate
Applied
Challenge
1
Find the distance between the points \(A(2, 3)\) and \(B(4, 1)\).
Foundational
Identify
Let \(x_1=2,\; y_1=3,\; x_2=4,\; y_2=1\).
Substitute
\[ d = \sqrt{(4-2)^2 + (1-3)^2} \]
Compute
\[ d = \sqrt{(2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2} \]
Verification: Swap the points: \(\sqrt{(2-4)^2+(3-1)^2} = \sqrt{4+4} = 2\sqrt{2}\). Same result — distance is always symmetric. ✓
2
Find the distance of point \(P(6, 8)\) from the origin \(O(0, 0)\).
Core
Recognise
The origin is \(O(0,0)\). Distance simplifies to: \(\text{OP} = \sqrt{x^2 + y^2}\).
Apply
\[ \text{OP} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]
Insight
\((6, 8, 10)\) is a Pythagorean triple — \(6^2 + 8^2 = 10^2\). Always spot these in exams!
Verification: \(36 + 64 = 100 = 10^2\). Perfect square — no rounding needed. ✓
3
Find the distance between the points \((a, b)\) and \((-a, -b)\).
Intermediate
Setup
Let \(P = (a, b)\) and \(Q = (-a, -b)\). Notice Q is the reflection of P through the origin.
Differences
\[ x_2 - x_1 = -a - a = -2a \qquad y_2 - y_1 = -b - b = -2b \]
Apply
\[ d = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} \]
Factorise
\[ d = \sqrt{4(a^2 + b^2)} = 2\sqrt{a^2 + b^2} \]
Professor's Note: The distance from P to its origin-reflection is exactly twice the distance from P to the origin itself. Elegant!
4
Find the value(s) of \(y\) if the distance between \(P(2, -3)\) and \(Q(10, y)\) is 10 units.
Applied
Write
\[ \sqrt{(10-2)^2 + (y+3)^2} = 10 \]
Square
\[ 64 + (y+3)^2 = 100 \implies (y+3)^2 = 36 \]
Solve
\[ y + 3 = \pm 6 \implies y = 3 \quad \text{or} \quad y = -9 \]
Verification: For \(y=3\): \(\sqrt{64+36} = \sqrt{100} = 10\) ✓. For \(y=-9\): \(\sqrt{64+36} = 10\) ✓
5
Find the point on the \(x\)-axis which is equidistant from \(A(2, -5)\) and \(B(-2, 9)\).
Challenge
Assume
Any point on the \(x\)-axis has \(y = 0\). Let the required point be \(P(x, 0)\). We need \(PA = PB\), so \(PA^2 = PB^2\).
Set up
\[ (x-2)^2 + 25 = (x+2)^2 + 81 \]
Expand
\[ x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 \implies -4x + 29 = 4x + 85 \]
Solve
\[ -8x = 56 \implies x = -7 \]
Professor's Note: This is where the perpendicular bisector of AB crosses the \(x\)-axis — a beautiful connection between distance and geometry.
§2 The Section Formula
When point P divides segment AB in ratio \(m_1 : m_2\), similar triangles give us P's exact coordinates.
\[ P(x,y) = \left(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2},\;\; \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right) \]
Special case — Midpoint: \(\displaystyle M = \left(\frac{x_1+x_2}{2},\;\frac{y_1+y_2}{2}\right)\)
Foundational
Core
Intermediate
Applied
Challenge
1
Find the coordinates of the point dividing the join of \((-1, 7)\) and \((4, -3)\) in the ratio \(2:3\).
Foundational
Label
\(A(-1,7)\), \(B(4,-3)\), \(m_1=2\), \(m_2=3\).
x-coord
\[ x = \frac{2(4) + 3(-1)}{2+3} = \frac{8-3}{5} = \frac{5}{5} = 1 \]
y-coord
\[ y = \frac{2(-3) + 3(7)}{5} = \frac{-6+21}{5} = \frac{15}{5} = 3 \]
Verification: \(AP = \sqrt{(1+1)^2+(3-7)^2} = \sqrt{4+16} = \sqrt{20}\). \(PB = \sqrt{(4-1)^2+(-3-3)^2} = \sqrt{9+36} = \sqrt{45}\). Ratio \(\sqrt{20}:\sqrt{45} = 2:3\) ✓
2
Find the midpoint of the segment joining \((2, 2)\) and \((4, 8)\).
Core
Recall
Midpoint is the section formula with \(m_1 : m_2 = 1 : 1\), which reduces to averaging both coordinates.
Apply
\[ M = \left(\frac{2+4}{2},\;\frac{2+8}{2}\right) = \left(3,\; 5\right) \]
Verification: Distance from \((2,2)\) to \((3,5)\) = \(\sqrt{1+9}=\sqrt{10}\). Distance from \((3,5)\) to \((4,8)\) = \(\sqrt{1+9}=\sqrt{10}\). Equal — confirmed midpoint. ✓
3
Find the ratio in which \((-1, 6)\) divides the segment joining \((-3, 10)\) and \((6, -8)\).
Intermediate
Assume
Let ratio be \(k:1\). The \(x\)-coordinate of the dividing point is:
\[ x = \frac{6k + (-3)}{k+1} = \frac{6k-3}{k+1} \]
Solve
Set \(x = -1\):
\[ -1(k+1) = 6k-3 \implies -k-1 = 6k-3 \implies 7k = 2 \implies k = \tfrac{2}{7} \]
Verify y
\[ y = \frac{\tfrac{2}{7}(-8) + 1(10)}{\tfrac{2}{7}+1} = \frac{\tfrac{54}{7}}{\tfrac{9}{7}} = 6 \quad \checkmark \]
4
Find the coordinates of the points of trisection of the segment joining \(A(4, -1)\) and \(B(-2, -3)\).
Applied
Understand
Trisection: \(AP = PQ = QB\). So P divides AB in ratio \(1:2\) and Q divides AB in ratio \(2:1\).
Point P
Ratio \(1:2\):
\[ P_x = \frac{1(-2)+2(4)}{3} = \frac{6}{3} = 2, \qquad P_y = \frac{1(-3)+2(-1)}{3} = \frac{-5}{3} \]
Point Q
Ratio \(2:1\):
\[ Q_x = \frac{2(-2)+1(4)}{3} = \frac{0}{3} = 0, \qquad Q_y = \frac{2(-3)+1(-1)}{3} = \frac{-7}{3} \]
5
Find the ratio in which the \(y\)-axis divides the segment joining \((5, -6)\) and \((-1, -4)\). Also find the point of intersection.
Challenge
Key insight
Any point on the \(y\)-axis has \(x = 0\). Let ratio be \(k:1\) and apply section formula for \(x\) only.
Solve x
\[ 0 = \frac{k(-1) + 1(5)}{k+1} \implies -k + 5 = 0 \implies k = 5 \]
Find y
\[ y = \frac{5(-4)+1(-6)}{5+1} = \frac{-26}{6} = -\frac{13}{3} \]
§3 Geometry Verification
Coordinate geometry is the ultimate lie-detector for geometric claims. We don't say “it looks like a square” — we prove it with algebra.
Collinear: \(AB + BC = AC\) ·
Square: all 4 sides equal AND diagonals equal ·
Parallelogram: midpoints of both diagonals coincide
Foundational
Core
Intermediate
Applied
Challenge
1
Are the points \((1, 5)\), \((2, 3)\) and \((-2, -11)\) collinear?
Foundational
Strategy
Three points A, B, C are collinear if and only if \(AB + BC = AC\). Let \(A(1,5)\), \(B(2,3)\), \(C(-2,-11)\).
AB
\[ AB = \sqrt{(2-1)^2+(3-5)^2} = \sqrt{1+4} = \sqrt{5} \]
BC
\[ BC = \sqrt{(-2-2)^2+(-11-3)^2} = \sqrt{16+196} = \sqrt{212} = 2\sqrt{53} \]
AC
\[ AC = \sqrt{(-2-1)^2+(-11-5)^2} = \sqrt{9+256} = \sqrt{265} \]
Check
\(\sqrt{5} \approx 2.24\), \(2\sqrt{53} \approx 14.56\). Sum \(\approx 16.80\). But \(\sqrt{265} \approx 16.28\). Since \(AB + BC \neq AC\), the points are not collinear.
Verification (Slope method): Slope \(AB = \dfrac{3-5}{2-1} = -2\). Slope \(BC = \dfrac{-11-3}{-2-2} = 3.5\). Slopes differ → not collinear. ✓
2
Show that \((5, -2)\), \((6, 4)\) and \((7, -2)\) form an isosceles triangle.
Core
Label
Let \(A(5,-2)\), \(B(6,4)\), \(C(7,-2)\).
Sides
\[ AB = \sqrt{1^2+6^2} = \sqrt{37} \]
\[ BC = \sqrt{1^2+(-6)^2} = \sqrt{37} \]
\[ CA = \sqrt{(-2)^2+0^2} = 2 \]
Conclude
\(AB = BC = \sqrt{37}\) but \(CA = 2\). Two sides are equal — the triangle is isosceles by definition.
Verification: A and C share \(y = -2\), so AC is horizontal. B is directly above the midpoint of AC at \(x=6\), confirming the axis of symmetry. ✓
3
Prove that \((1, 7)\), \((4, 2)\), \((-1, -1)\) and \((-4, 4)\) are the vertices of a square.
Intermediate
Plan
A quadrilateral is a square iff all four sides are equal AND both diagonals are equal. Let \(A(1,7)\), \(B(4,2)\), \(C(-1,-1)\), \(D(-4,4)\).
Four sides
\[ AB = \sqrt{9+25} = \sqrt{34}, \quad BC = \sqrt{25+9} = \sqrt{34} \]
\[ CD = \sqrt{9+25} = \sqrt{34}, \quad DA = \sqrt{25+9} = \sqrt{34} \]
Diagonals
\[ AC = \sqrt{4+64} = \sqrt{68}, \qquad BD = \sqrt{64+4} = \sqrt{68} \]
Conclude
All four sides \(= \sqrt{34}\) ✓. Both diagonals \(= \sqrt{68}\) ✓. Also \(68 = 2 \times 34\), confirming right angles at every vertex.
4
If \(A(6, 1)\), \(B(8, 2)\), \(C(9, 4)\) and \(D(p, 3)\) are vertices of a parallelogram, find \(p\).
Applied
Key fact
Diagonals of a parallelogram bisect each other, so midpoint of AC = midpoint of BD.
Midpoint AC
\[ \text{Mid}(AC) = \left(\frac{6+9}{2},\;\frac{1+4}{2}\right) = \left(\frac{15}{2},\;\frac{5}{2}\right) \]
Midpoint BD
\[ \text{Mid}(BD) = \left(\frac{8+p}{2},\;\frac{5}{2}\right) \]
Equate
\[ \frac{8+p}{2} = \frac{15}{2} \implies p = 7 \]
Professor's Note: The midpoint strategy avoids all four side-length calculations. In vector algebra this same property becomes the algebraic definition of a parallelogram.
5
Find the area of a rhombus with vertices \((3, 0)\), \((4, 5)\), \((-1, 4)\) and \((-2, -1)\).
Challenge
Formula
Area of a rhombus \(= \dfrac{1}{2} \times d_1 \times d_2\). Diagonals: \(A(3,0)\leftrightarrow C(-1,4)\) and \(B(4,5)\leftrightarrow D(-2,-1)\).
d₁ = AC
\[ d_1 = \sqrt{(-4)^2+4^2} = \sqrt{32} = 4\sqrt{2} \]
d₂ = BD
\[ d_2 = \sqrt{(-6)^2+(-6)^2} = \sqrt{72} = 6\sqrt{2} \]
Area
\[ \text{Area} = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 48 = 24 \]
Verification: Mid(AC) \(= (1,2)\) and Mid(BD) \(= (1,2)\). Same midpoint confirms it is a parallelogram, and equal sides confirm it is a rhombus. ✓
