Coordinate Geometry Concepts

1. Distance Between Two Points

• Formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

• Example: Question: Find the distance between ((2,3)) and ((8,11)).

  Solution:

  $d=\sqrt{(8-2)^2+(11-3)^2}$

  $=\sqrt{6^2+8^2}$

  $=\sqrt{36+64}$

  $=\boxed{10}$

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2. Midpoint of a Line Segment

• Formula: $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

• Example: Question: Find the midpoint of ((2,4)) and ((8,10)).

  Solution:

  $M=\left(\frac{2+8}{2},\frac{4+10}{2}\right)$

  $=\boxed{(5,7)}$

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3. Section Formula — Internal Division

• Formula:

  If (P) divides (A(x_1,y_1)) and (B(x_2,y_2)) in ratio (m:n),

  $P=\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)$

• Example: Question: Find the point dividing ((2,4)) and ((8,10)) internally in the ratio (1:2).

  Solution:

  $x=\frac{1(8)+2(2)}{1+2}=4$

  $y=\frac{1(10)+2(4)}{3}=6$

  Therefore,

  $\boxed{P=(4,6)}$

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4. Section Formula — External Division

• Formula:

  $P=\left(\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n}\right)$

• Example: Question: Find the point dividing ((2,4)) and ((8,10)) externally in the ratio (2:1).

  Solution:

  $x=\frac{2(8)-1(2)}{2-1}=14$

  $y=\frac{2(10)-1(4)}{2-1}=16$

  Therefore,

  $\boxed{(14,16)}$

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5. Slope of a Line

• Formula:

  $m=\frac{y_2-y_1}{x_2-x_1}$

• Example: Question: Find the slope of the line joining ((2,3)) and ((6,11)).

  Solution:

  $m=\frac{11-3}{6-2}$

  $=\frac84$

  $=\boxed{2}$

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6. Parallel Lines

• Formula:

  Two non-vertical lines are parallel if:

  $m_1=m_2$

• Example: Question: Are the lines joining ((1,2),(3,6)) and ((4,5),(6,9)) parallel?

  Solution:

  First slope:

  $m_1=\frac{6-2}{3-1}=2$

  Second slope:

  $m_2=\frac{9-5}{6-4}=2$

  Since:

  $m_1=m_2$

  Answer: Yes, the lines are parallel.

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7. Perpendicular Lines

• Formula:

  $m_1m_2=-1$

• Example: Question: Two lines have slopes (2) and (-\frac12). Are they perpendicular?

  Solution:

  $m_1m_2=2\left(-\frac12\right)=-1$

  Therefore,

  $\boxed{\text{Yes, they are perpendicular.}}$

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8. Equation of a Line Using a Point and Slope

• Formula:

  $y-y_1=m(x-x_1)$

• Example: Question: Find the equation of the line passing through ((2,3)) with slope (4).

  Solution:

  $y-3=4(x-2)$

  $y-3=4x-8$

  $\boxed{y=4x-5}$

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9. Equation of a Line Through Two Points

• Formula:

  $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

• Example: Question: Find the equation of the line passing through ((1,2)) and ((3,6)).

  Solution:

  First find slope:

  $m=\frac{6-2}{3-1}=2$

  Using ((1,2)):

  $y-2=2(x-1)$

  $\boxed{y=2x}$

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10. Collinearity of Three Points

• Formula:

  Three points are collinear if their slopes are equal:

  $m_{AB}=m_{BC}$

• Example: Question: Determine whether ((1,2),(3,6),(5,10)) are collinear.

  Solution:

  $m_{AB}=\frac{6-2}{3-1}=2$

  $m_{BC}=\frac{10-6}{5-3}=2$

  Since the slopes are equal:

  $\boxed{\text{The points are collinear.}}$

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11. Collinearity Using Area

• Formula:

  Three points are collinear if the area of the triangle formed by them is zero.

  \frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$$

• Example: Question: Check whether ((1,2),(3,4),(5,6)) are collinear.

  Solution:

  $\text{Area}=\frac12|1(4-6)+3(6-2)+5(2-4)|$

  $=\frac12|-2+12-10|$

  $=0$

  Therefore,

  $\boxed{\text{The points are collinear.}}$

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12. Area of a Triangle Using Coordinates

• Formula:

  $A=\frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

• Example: Question: Find the area of the triangle with vertices ((0,0),(4,0),(4,5)).

  Solution:

  $A=\frac12|0(0-5)+4(5-0)+4(0-0)|$

  $=\frac12(20)$

  $=\boxed{10\text{ square units}}$

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13. Centroid of a Triangle

• Formula:

  \frac{y_1+y_2+y_3}{3}\right)$$

• Example: Question: Find the centroid of a triangle with vertices ((1,2),(3,4),(5,0)).

  Solution:

  $G=\left(\frac{1+3+5}{3},\frac{2+4+0}{3}\right)$

  $=\boxed{(3,2)}$

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14. Distance of a Point From the Origin

• Formula:

  For (P(x,y)):

  $OP=\sqrt{x^2+y^2}$

• Example: Question: Find the distance of ((6,8)) from the origin.

  Solution:

  $OP=\sqrt{6^2+8^2}$

  $=\sqrt{100}$

  $=\boxed{10}$

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15. Distance Between a Point and a Line

• Formula:

  For line:

  $ax+by+c=0$

  and point ((x_1,y_1)):

  $d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$

• Example: Question: Find the distance of ((2,3)) from the line (3x+4y-10=0).

  Solution:

  $d=\frac{|3(2)+4(3)-10|}{\sqrt{3^2+4^2}}$

  $=\frac{|6+12-10|}{5}$

  $=\frac85$

  $\boxed{\frac85}$

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16. Area of a Quadrilateral Using Coordinates

• Formula:

  x_1y_2+x_2y_3+x_3y_4+x_4y_1 -(y_1x_2+y_2x_3+y_3x_4+y_4x_1) \right|$$

• Example: Question: Find the area of the quadrilateral with vertices ((0,0),(4,0),(4,3),(0,3)).

  Solution:

  This is a rectangle:

  $A=4\times3$

  $\boxed{12\text{ square units}}$

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Advanced Variants

17. Internal Division With an Unknown Ratio

• Formula:

  \frac{my_2+ny_1}{m+n}\right)$$

• Example: Question: A point (P) divides ((2,3)) and ((8,9)) internally in the ratio (m:n). If its (x)-coordinate is (6), find the ratio.

  Solution:

  $6=\frac{8m+2n}{m+n}$

  $6m+6n=8m+2n$

  $4n=2m$

  $m:n=2:1$

  Answer:

  $\boxed{2:1}$

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18. Finding a Missing Coordinate Using Distance

• Formula:

  $d^2=(x_2-x_1)^2+(y_2-y_1)^2$

• Example: Question: The distance between ((2,3)) and ((x,7)) is (5). Find (x).

  Solution:

  $5^2=(x-2)^2+(7-3)^2$

  $25=(x-2)^2+16$

  $9=(x-2)^2$

  $x-2=\pm3$

  Therefore,

  $\boxed{x=5\text{ or }-1}$

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19. Point Equidistant From Two Points

• Formula:

  If (P(x,y)) is equally distant from (A) and (B):

  $PA=PB$

  Compare their squared distances to avoid square roots.

• Example: Question: Find the points on the (x)-axis that are equidistant from ((2,3)) and ((6,1)).

  Solution:

  Let the point be ((x,0)).

  $PA^2=(x-2)^2+9$

  $PB^2=(x-6)^2+1$

  Equate:

  $(x-2)^2+9=(x-6)^2+1$

  Solving:

  $x=3$

  Therefore,

  $\boxed{(3,0)}$

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20. Coordinate Geometry + Ratio + Distance

• Example: Question: A point divides the line joining (A(1,2)) and (B(7,8)) in the ratio (1:2). Find its distance from (A).

  Solution:

  First find the point:

  $P=\left(\frac{1(7)+2(1)}3,\frac{1(8)+2(2)}3\right)$

  $P=(3,4)$

  Now:

  $AP=\sqrt{(3-1)^2+(4-2)^2}$

  $=\sqrt{4+4}$

  $=\boxed{2\sqrt2}$