Two-dimensional coordinate geometry deals about the coordinates which are represented in a coordinate plane. A coordinate plane has two axes, the one which is horizontal is known as \(X-axis\) and the one which is vertical is known as \(Y-axis\). A point \(p(x,y)\) is represented in the \(X-Y plane\) as shown below.

â€˜\(O\)â€™ is known as origin whose coordinate is (0,0).

The perpendicular distance of \(p(x,y)\) from \(X-axis\) and \(Y-axis\) is \(’y'\) and \(’x'\) respectively.

## Coordinate Geometry in Two Dimensional Plane

For example, the point \((2,3)\) is \(3 ~units\) away from the \(X-axis\) measured along the positive \(Y-axis\) and \(2~ units\) away from \(Y-axis\) measured along the positive \(X-axis\).

The points having \(x\)-coordinate as \(â€˜0â€™\) lie on the \(Y-axis\) and points having \(y\)-coordinate as \(â€˜0â€™\) lie on the \(X-axis\).

For example, the points \((2,0),(5,0)\) lie on \(X-axis\) and the points \((0,-3),(0,7)\) lie on \(Y-axis\).

**Distance between two points:**

Consider two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in an \(XY\) plane.

Then the distance between \(A\) and \(B\) is,

\(AB\) = \(âˆš(x_2 ~-~ x_1)^2 + (y_2~-~y_1)^2 \)

For example; distance between the points \(A(2,-3)\) and \(B(5,1)\) is,

\( AB\) = \(âˆš(5~-~2)^2 ~+~ (1-(-3))^2\) = \(âˆš3^2~ + ~4^2\) = \( âˆš25 \) = \( 5~units \)

Similarly, distance between a point \(P(x,y)\) from the origin is,

\( OP \) = \( âˆš(x-0)^2 ~+ ~(y-0)^2\) = \(âˆš(x^2~ +~ y^2)\)

**Reflection of a point across the X-axis**

Reflection of a point \(P(x,y)\) across the \(X-axis\) is \(Q(x,-y)\), which is found by changing the sign of the \(y\)-coordinate of \(P(x,y)\).

**Reflection of a point across the Y-axis**

Reflection of a point \(P(x,y)\) across the \(Y-axis\) is \(Q(-x,y)\), which is found by changing the sign of the

\(x\)-coordinate of \(P(x,y)\).

**Section Formula:**

Consider two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in an \(XY\) plane.

\(P(x,y)\) is a point which divides the line segment \(AB\) internally in the ratio \(m:n\),

Then, the \(x\)-coordinate of \(P\) is,

\(x\) = \(\frac{m~x_2~ + ~n~x_1}{m~ + ~n}\)

the \(y\)-coordinate of \(P\) is,

\(y\) = \(\frac{m~y_2~ + ~n~y_1}{m~ + ~n}\)

If \( m\) = \(n\), then \(P\) is the mid-point of the line segment \(AB\). Then coordinates of the point \(P\) is,

\((\frac{x_1 ~+~ x_2}{2}, \frac{y_1 ~+ ~y_2}{2}\))

If the point \(P(x,y)\) is dividing the line segment \(AB\) externally in the ratio \(m:n\),

Then, the \(x\)-coordinate of \(P\) is,

\(x \) = \(\frac{m~x_2~ – ~n~x_1}{m ~- ~n}\)

they-coordinate of \(P\) is,

\(y\) = \(\frac{m~y_2 ~- ~n~y_1}{m~ – ~n}\)

Example: Find the coordinates of the point which divides the line segment joining \(P(-3,-4)\) and

\(Q(6,8)\) in the ratio \(1:2\).

Let \(M(x,y)\) be the point which divides \(PQ\) in the ratio \(1:2\), then

\(x\) = \(\frac{1 ~Ã—~ 6 ~+~ 2~ Ã—~ (-3)}{1 ~+~ 2}\) = \( \frac{6~ -~ 6}{3}\) = \(0\)

\(y\) = \(\frac{1 ~Ã—~ 8 ~+~ 2~ Ã—~ (-4)}{1 ~+ ~2} \) = \(\frac{8 – 8}{3}\) = \(0\)

Therefore origin \((0,0)\) divides the point \(PQ\) in the ratio \(1:2\).

**Area of a triangle:**

The area of the triangle whose vertices are \((x_1,y_1 ),(x_2,y_2)\) and \((x_3,y_3)\) is

\(\frac{1}{2}|x_1 (y_2~ -~ y_3)~ + ~x_2(y_3~ – ~y_1)~ +~ x_3(y_1~ – ~y_2)|\)

If the area of a triangle whose vertices are \((x_1,y_1),(x_2,y_2)\) and \((x_3,y_3)\) is zero, then the three points are collinear.

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