Solved Examples

How to Use This Section

This is your practice powerhouse for coordinate geometry. Below are 30+ fully solved problems covering the whole chapter — distance, section and midpoint formulas, area of a triangle, collinearity, and shape identification — roughly easy to hard.

How to read: Cover the solution, attempt each yourself, then check the steps. In coordinate geometry, marks come from writing the formula first and substituting carefully.

Keep these handy:

• Distance: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Section ($m:n$): $\left(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\right)$.
• Midpoint: $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$.
• Area: $\tfrac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$.
• Centroid: $\left(\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}\right)$.

Solved Examples

Example 1: Distance

Find the distance between $(2, 3)$ and $(5, 7)$.

Solution:

1. $d = \sqrt{(5-2)^2 + (7-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

Final Answer: 5 units.

Takeaway: A $(3,4,5)$ triple again.

Example 2: Distance from origin

Find the distance of $(7, 24)$ from the origin.

Solution:

1. $d = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.

Final Answer: 25 units.

Takeaway: $(7, 24, 25)$ is a Pythagorean triple.

Example 3: Midpoint

Find the midpoint of $(-3, 5)$ and $(7, -1)$.

Solution:

1. $M = \left(\dfrac{-3+7}{2}, \dfrac{5-1}{2}\right) = (2, 2)$.

Final Answer: $(2, 2)$.

Takeaway: Average the coordinates.

Example 4: Section formula

Find the point dividing $(2, 3)$ and $(7, 8)$ in ratio $2 : 3$.

Solution:

1. $x = \dfrac{2(7)+3(2)}{5} = \dfrac{20}{5} = 4$.
2. $y = \dfrac{2(8)+3(3)}{5} = \dfrac{25}{5} = 5$.

Final Answer: $(4, 5)$.

Takeaway: Substitute the ratio carefully.

Example 5: Area of a triangle

Find the area of the triangle $(0, 0)$, $(5, 0)$, $(0, 4)$.

Solution:

1. Area $= \tfrac12|0(0-4) + 5(4-0) + 0(0-0)| = \tfrac12|20| = 10$.

Final Answer: 10 square units.

Takeaway: For axis-aligned legs, $\tfrac12 \times 5 \times 4 = 10$ confirms it.

Example 6: Collinearity

Are $(1, -1)$, $(2, 1)$, $(4, 5)$ collinear?

Solution:

1. Expression $= 1(1-5) + 2(5+1) + 4(-1-1) = -4 + 12 - 8 = 0$.

Final Answer: Yes, collinear.

Takeaway: Zero area ⇒ collinear.

Example 7: Centroid

Find the centroid of $(2, 4)$, $(6, 4)$, $(4, 10)$.

Solution:

1. $G = \left(\dfrac{2+6+4}{3}, \dfrac{4+4+10}{3}\right) = (4, 6)$.

Final Answer: $(4, 6)$.

Takeaway: Divide the coordinate sums by 3.

Example 8: Unknown coordinate from distance

The distance between $(3, y)$ and $(3, 7)$ is 5. Find $y$.

Solution:

1. Same x, so $|y - 7| = 5 \Rightarrow y - 7 = \pm 5$.
2. $y = 12$ or $y = 2$.

Final Answer: $y = 12$ or $y = 2$.

Takeaway: Equal x-coordinates ⇒ distance is just the y-difference.

Example 9: Fourth vertex of a parallelogram

$A(1, 1)$, $B(2, 3)$, $C(5, 4)$ are three vertices of parallelogram $ABCD$. Find $D$.

Solution:

1. Midpoint of $AC$ = midpoint of $BD$.
2. Midpoint of $AC = (3, 2.5)$.
3. $\dfrac{2+x}{2} = 3 \Rightarrow x = 4$; $\dfrac{3+y}{2} = 2.5 \Rightarrow y = 2$.

Final Answer: $D = (4, 2)$.

Takeaway: Equate diagonal midpoints.

Example 10: Ratio of division by a point

In what ratio does $(4, 5)$ divide the segment joining $(2, 3)$ and $(7, 8)$?

Solution:

1. Let ratio $k : 1$ on x: $4 = \dfrac{7k + 2}{k+1} \Rightarrow 4k + 4 = 7k + 2 \Rightarrow 3k = 2 \Rightarrow k = \dfrac{2}{3}$.
2. Ratio $= 2 : 3$.

Final Answer: $2 : 3$.

Takeaway: Use $k:1$ on one coordinate, then read the ratio.

Example 11: Isosceles triangle

Show that $(0, 5)$, $(-3, 1)$, $(3, 1)$ form an isosceles triangle.

Solution:

1. $AB = \sqrt{9 + 16} = 5$; $AC = \sqrt{9 + 16} = 5$; $BC = \sqrt{36 + 0} = 6$.
2. $AB = AC = 5 \neq BC$.

Final Answer: Isosceles.

Takeaway: Two equal sides ⇒ isosceles.

Example 12: Points of trisection

Find the points of trisection of the segment joining $(3, -2)$ and $(-3, -4)$.

Solution:

1. $P$ ($1:2$): $\left(\dfrac{1(-3)+2(3)}{3}, \dfrac{1(-4)+2(-2)}{3}\right) = (1, -\tfrac83)$.
2. $Q$ ($2:1$): $\left(\dfrac{2(-3)+1(3)}{3}, \dfrac{2(-4)+1(-2)}{3}\right) = (-1, -\tfrac{10}{3})$.

Final Answer: $(1, -\tfrac83)$ and $(-1, -\tfrac{10}{3})$.

Takeaway: Ratios $1:2$ and $2:1$.

Example 13: Right triangle test

Do $(2, 1)$, $(5, 1)$, $(2, 5)$ form a right triangle?

Solution:

1. $AB = 3$, $AC = 4$, $BC = \sqrt{9 + 16} = 5$.
2. $3^2 + 4^2 = 25 = 5^2$.

Final Answer: Yes, right-angled at $(2, 1)$.

Takeaway: Check Pythagoras with the longest side.

Example 14: Find k for collinearity

Find $k$ if $(7, -2)$, $(5, 1)$, $(3, k)$ are collinear.

Solution:

1. $7(1-k) + 5(k+2) + 3(-2-1) = 0$.
2. $7 - 7k + 5k + 10 - 9 = 0 \Rightarrow -2k + 8 = 0 \Rightarrow k = 4$.

Final Answer: $k = 4$.

Takeaway: Set the area expression to 0 and solve.

Example 15: Equidistant point on y-axis

Find the point on the y-axis equidistant from $(5, -2)$ and $(-3, 2)$.

Solution:

1. Let $P(0, y)$: $25 + (y+2)^2 = 9 + (y-2)^2$.
2. $25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4$.
3. $29 + 4y = 13 - 4y \Rightarrow 8y = -16 \Rightarrow y = -2$.

Final Answer: $(0, -2)$.

Takeaway: Set squared distances equal with $x = 0$.

Example 16: Diameter centre

The endpoints of a diameter are $(3, 4)$ and $(-1, -2)$. Find the centre and radius.

Solution:

1. Centre = midpoint $= (1, 1)$.
2. Radius = distance from centre to $(3, 4) = \sqrt{4 + 9} = \sqrt{13}$.

Final Answer: Centre $(1, 1)$, radius $\sqrt{13}$.

Takeaway: Centre is the midpoint; radius is centre-to-endpoint distance.

Example 17: Section by x-axis

In what ratio does the x-axis divide the segment joining $(2, -3)$ and $(5, 6)$?

Solution:

1. On x-axis $y = 0$; ratio $k:1$: $0 = \dfrac{6k - 3}{k+1} \Rightarrow 6k = 3 \Rightarrow k = \tfrac12$.
2. Ratio $= 1 : 2$.

Final Answer: $1 : 2$.

Takeaway: Set $y = 0$ to find where the x-axis cuts the segment.

Example 18: Square verification

Show that $(0, 0)$, $(3, 4)$, $(7, 1)$, $(4, -3)$ form a square.

Solution:

1. $AB = \sqrt{9+16} = 5$; $BC = \sqrt{16+9} = 5$; $CD = \sqrt{9+16} = 5$; $DA = \sqrt{16+9} = 5$.
2. Diagonals: $AC = \sqrt{49+1} = \sqrt{50}$; $BD = \sqrt{1+49} = \sqrt{50}$ — equal.
3. All sides equal + equal diagonals ⇒ square.

Final Answer: Square.

Takeaway: Sides equal and diagonals equal characterise a square.

Example 19: Distance between two general points

Find the distance between $(a, b)$ and $(-a, -b)$.

Solution:

1. $d = \sqrt{(2a)^2 + (2b)^2} = 2\sqrt{a^2 + b^2}$.

Final Answer: $2\sqrt{a^2 + b^2}$.

Takeaway: Opposite points about the origin are $2\sqrt{a^2+b^2}$ apart.

Example 20: Third vertex from centroid

The centroid of a triangle is $(0, 0)$ and two vertices are $(2, 3)$ and $(-4, 1)$. Find the third.

Solution:

1. $\dfrac{2 - 4 + x}{3} = 0 \Rightarrow x = 2$.
2. $\dfrac{3 + 1 + y}{3} = 0 \Rightarrow y = -4$.

Final Answer: $(2, -4)$.

Takeaway: Sum of coordinates must be 0 when the centroid is the origin.

Example 21: Area of a quadrilateral

Find the area of quadrilateral $A(1, 1)$, $B(7, -3)$, $C(12, 2)$, $D(7, 21)$.

Solution:

1. $\triangle ABC = \tfrac12|1(-3-2) + 7(2-1) + 12(1+3)| = \tfrac12|{-5 + 7 + 48}| = \tfrac{50}{2} = 25$.
2. $\triangle ACD = \tfrac12|1(2-21) + 12(21-1) + 7(1-2)| = \tfrac12|{-19 + 240 - 7}| = \tfrac{214}{2} = 107$.
3. Total $= 25 + 107 = 132$.

Final Answer: 132 square units.

Takeaway: Split along $AC$ and add the two triangle areas.

Example 22: Collinear by section formula

If $(x, 2)$ divides the segment joining $(1, 4)$ and $(5, -4)$, find $x$ and the ratio.

Solution:

1. On y: $2 = \dfrac{-4k + 4}{k+1} \Rightarrow 2k + 2 = -4k + 4 \Rightarrow 6k = 2 \Rightarrow k = \tfrac13$.
2. Ratio $1:3$; $x = \dfrac{1(5)+3(1)}{4} = \dfrac{8}{4} = 2$.

Final Answer: $x = 2$, ratio $1 : 3$.

Takeaway: Use the known coordinate to find $k$, then the other coordinate.

Example 23: Equilateral triangle side

Find the relation so that $(0, 0)$, $(3, \sqrt3)$, and $(a, b)$ form an equilateral triangle with the first two as a side.

Solution:

1. Side $AB = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt3$.
2. The third vertex is equidistant ($2\sqrt3$) from both $A$ and $B$.
3. One such point: rotate; e.g. $(0, 2\sqrt3)$ gives $OA' = 2\sqrt3$ and check $BA' = \sqrt{9 + 3} = 2\sqrt3$. Valid.

Final Answer: $(0, 2\sqrt3)$ is one valid third vertex.

Takeaway: The third vertex of an equilateral triangle is equidistant from both endpoints by the side length.

Example 24: Midpoint to find endpoint

If $(2, -1)$ is the midpoint of the segment joining $(x, 3)$ and $(6, y)$, find $x$ and $y$.

Solution:

1. $\dfrac{x + 6}{2} = 2 \Rightarrow x = -2$.
2. $\dfrac{3 + y}{2} = -1 \Rightarrow y = -5$.

Final Answer: $x = -2$, $y = -5$.

Takeaway: Solve each midpoint equation separately.

Example 25: Distance between points on a line

Find the distance between $(1, 2)$ and $(4, 6)$, and the coordinates of the midpoint.

Solution:

1. Distance $= \sqrt{9 + 16} = 5$.
2. Midpoint $= \left(\dfrac{5}{2}, 4\right)$.

Final Answer: Distance 5; midpoint $(2.5, 4)$.

Takeaway: Two formulas, one pair of points.

Example 26: Area to test a degenerate triangle

Find the area of the triangle $(a, b+c)$, $(b, c+a)$, $(c, a+b)$.

Solution:

1. Area $= \tfrac12|a((c+a)-(a+b)) + b((a+b)-(b+c)) + c((b+c)-(c+a))|$.
2. $= \tfrac12|a(c-b) + b(a-c) + c(b-a)| = \tfrac12|ac - ab + ab - bc + bc - ac| = 0$.

Final Answer: 0 — the points are collinear.

Takeaway: A classic identity: these three points are always collinear.

Example 27: Rectangle verification

Show that $(2, -2)$, $(8, 4)$, $(5, 7)$, $(-1, 1)$ form a rectangle.

Solution:

1. $AB = \sqrt{36+36} = \sqrt{72}$; $BC = \sqrt{9+9} = \sqrt{18}$; $CD = \sqrt{36+36} = \sqrt{72}$; $DA = \sqrt{9+9} = \sqrt{18}$ ⇒ opposite sides equal.
2. Diagonals: $AC = \sqrt{9+81} = \sqrt{90}$; $BD = \sqrt{81+9} = \sqrt{90}$ ⇒ equal.
3. Parallelogram + equal diagonals ⇒ rectangle.

Final Answer: Rectangle.

Takeaway: Opposite sides equal and equal diagonals ⇒ rectangle.

Example 28: Distance using a parameter

If the point $(x, y)$ is equidistant from $(7, 1)$ and $(3, 5)$, find the relation between $x$ and $y$.

Solution:

1. $(x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2$.
2. $-14x + 49 - 2y + 1 = -6x + 9 - 10y + 25$.
3. $-14x - 2y + 50 = -6x - 10y + 34 \Rightarrow -8x + 8y + 16 = 0 \Rightarrow x - y = 2$.

Final Answer: $x - y = 2$.

Takeaway: Equidistance gives the equation of the perpendicular bisector.

Example 29: Centroid and area together

Find the centroid and area of the triangle $(1, 1)$, $(4, 1)$, $(1, 5)$.

Solution:

1. Centroid $= \left(\dfrac{1+4+1}{3}, \dfrac{1+1+5}{3}\right) = (2, \tfrac73)$.
2. Area $= \tfrac12|1(1-5) + 4(5-1) + 1(1-1)| = \tfrac12|{-4 + 16 + 0}| = 6$.

Final Answer: Centroid $(2, \tfrac73)$; area 6 square units.

Takeaway: Two independent computations from the same vertices.

Example 30: Ratio in which y-axis divides

Find the ratio in which the y-axis divides the segment joining $(-4, 7)$ and $(8, -5)$, and the point of division.

Solution:

1. On y-axis $x = 0$; ratio $k:1$: $0 = \dfrac{8k - 4}{k+1} \Rightarrow 8k = 4 \Rightarrow k = \tfrac12$.
2. Ratio $1 : 2$; $y = \dfrac{1(-5) + 2(7)}{3} = \dfrac{9}{3} = 3$.

Final Answer: Ratio $1 : 2$; point $(0, 3)$.

Takeaway: Solve for $k$ at $x = 0$, then get the y-coordinate.

Example 31: Larger area problem

Find the area of the triangle with vertices $(-5, -1)$, $(3, -5)$, $(5, 2)$.

Solution:

1. Area $= \tfrac12|{-5(-5-2) + 3(2+1) + 5(-1+5)}|$.
2. $= \tfrac12|35 + 9 + 20| = \tfrac12(64) = 32$.

Final Answer: 32 square units.

Takeaway: Substitute carefully; the modulus keeps the area positive.