Point of Intersection Calculator

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**Step 1 :**

**Step 2 :**

Intersection of two lines is a point as shown below.

Here, P is the point of intersection of lines 1 and 2.

Deriving the formula for finding the**point of intersection** of two lines:

Let a_{1}x + b_{1}y + c_{1} = 0 ......... (1) and a_{2}x + b_{2}y + c_{2} = 0 ......... (2) be two intersecting lines.

Let (x_{}, y_{}) be the intersecting point of lines (1) and (2), then the point (x_{}, y_{}) should lie on both the lines represented by (1) and (2). i.e. the point (x, y)satisfies both the equations (1) and (2).

Solving the equations (1) and (2), we get the intersection point of two line (x_{}, y)

$\frac{x}{(b_{1}c_{2} - b_{2}c_{1})}$ = $\frac{y}{(a_{2}c_{1} - a_{1}c_{2})}$ = $\frac{1}{(a_{1}b_{2} - a_{2}b_{1})}$

So, x = $\frac{(b_{1}c_{2} - b_{2}c_{1})}{(a_{1}b_{2} - a_{2}b_{1})}$

y = $\frac{(a_{2}c_{1} - a_{1}c_{2})}{(a_{1}b_{2} - a_{2}b_{1})}$

Therefore, point of intersection (x, y) = [ $\frac{(b_{1}c_{2} - b_{2}c_{1})}{(a_{1}b_{2} - a_{2}b_{1})}$, $\frac{(a_{2}c_{1} - a_{1}c_{2})}{(a_{1}b_{2} - a_{2}b_{1})}$ ]

**Point of Intersection Calculator** will help us to find the point the point of intersection for a given system of equations.

Here, P is the point of intersection of lines 1 and 2.

Deriving the formula for finding the

Let a

Let (x

Solving the equations (1) and (2), we get the intersection point of two line (x

$\frac{x}{(b_{1}c_{2} - b_{2}c_{1})}$ = $\frac{y}{(a_{2}c_{1} - a_{1}c_{2})}$ = $\frac{1}{(a_{1}b_{2} - a_{2}b_{1})}$

So, x = $\frac{(b_{1}c_{2} - b_{2}c_{1})}{(a_{1}b_{2} - a_{2}b_{1})}$

y = $\frac{(a_{2}c_{1} - a_{1}c_{2})}{(a_{1}b_{2} - a_{2}b_{1})}$

Therefore, point of intersection (x, y) = [ $\frac{(b_{1}c_{2} - b_{2}c_{1})}{(a_{1}b_{2} - a_{2}b_{1})}$, $\frac{(a_{2}c_{1} - a_{1}c_{2})}{(a_{1}b_{2} - a_{2}b_{1})}$ ]

Observe the given system of equations.

Apply the formula:

point of intersection (x, y) = [ $\frac{(b_{1}c_{2} - b_{2}c_{1})}{(a_{1}b_{2} - a_{2}b_{1})}$, $\frac{(a_{2}c_{1} - a_{1}c_{2})}{(a_{1}b_{2} - a_{2}b_{1})}$ ]

Find the point of intersection of the lines:

x + y = 1

2x + 3y = 5

**Step 1 :**Given system of equations:

x + y = 1

2x + 3y = 5

So, a

_{1}= 1, b_{1}= 1 and c_{1}= -1

a

_{2}= 2, b_{2}= 3 and c_{2}= -5**Step 2 :**point of intersection (x, y) = [ $\frac{(b_{1}c_{2} - b_{2}c_{1})}{(a_{1}b_{2} - a_{2}b_{1})}$, $\frac{(a_{2}c_{1} - a_{1}c_{2})}{(a_{1}b_{2} - a_{2}b_{1})}$ ]

point of intersection (x, y) = [ $\frac{((1)(-5) - (3)(-1))}{((1)(3) - (2)(1))}$, $\frac{((-2)(1) - (-1)(5))}{((1)(3) - (2)(1))}$]

point of intersection (x, y) = (-2, 3)

**Answer :**(2, -3)

Find the point of intersection of the lines:

x + y = -1

2x + 3y = -5

**Step 1 :**Given system of equations:

x + y = -1

2x + 3y = -5

So, a

_{1}= 1, b_{1}= 1 and c_{1}= 1

a

_{2}= 2, b_{2}= 3 and c_{2}= 5**Step 2 :**point of intersection (x, y) = [ $\frac{(b_{1}c_{2} - b_{2}c_{1})}{(a_{1}b_{2} - a_{2}b_{1})}$, $\frac{(a_{2}c_{1} - a_{1}c_{2})}{(a_{1}b_{2} - a_{2}b_{1})}$ ]

point of intersection (x, y) = [ $\frac{((1)(5) - (3)(1))}{((1)(3) - (2)(1))}$, $\frac{((2)(1) - (1)(5))}{((1)(3) - (2)(1))}$]

point of intersection (x, y) = (2, -3)

**Answer :**(2, -3)