TWO POINT FORM EQUATION OF A LINE

Let A(x1, y1) and B(x2, y2) be two given distinct points.

Slope of the straight line passing through these points is given by

m = (y2 - y1)/(x2 - x1)

Using point-slope form equation of a line, 

(y - y1) = [(y2 - y1)/(x2 - x1)](x - x1)

Hence, the equation of line in two-point form : 

(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)

Example 1 :

Find the general equation of a line passing through the points (-2, 1) and (4, -7).

Solution :

Given : Two points on the straight line : (-2, 1) and (4, -7).

Equation of line in two-point form :

(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)

Substitute (x1 , y1) = (-2, 1) and (x2, y2) = (4, -7).

(y - 1)/(-7 - 1) = (x + 2)/(4 + 2)

(y - 1)/(-8) = (x + 2)/6

6(y - 1) = -8(x + 2)

Distribute.

6y - 6 = -8x - 16

Simplify. 

8x + 6y + 10 = 0

Divide each side by 2. 

4x + 3y + 5 = 0

Example 2 :

Find the equation of a line in slope-intercept form which passing through the points (-2, 5) and (3, 6).

Solution :

Given : Two points on the straight line : (-2, 5) and (3, 6).

Equation of line in two-point form :

(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)

Substitute (x1 , y1) = (-2, 5) and (x2, y2) = (3, 6).

(y - 5)/(6 - 5) = (x + 2)/(3 + 2)

(y - 1)/1 = (x + 2)/5

5(y - 1) = x + 2

Distribute.

5y - 5 = x + 2

Add 5 to each side.

5y = x + 7

Divide each side by 5.

y = x/5 + 7/5

y = (1/5)x + 7/5

Example 3 :

The vertices of a triangle ABC are A(2, 1), B(-2, 3) and C(4, 5). Find the equation of the median through the vertex A.

Solution :

Median is a straight line joining a vertex and the midpoint of the opposite side.

In ΔABC above, midpoint of BC :

= D((-2 + 4)/2, (3 + 5)/2)

= D(1, 4)

The median through A is the line joining two points A (2, 1) and D(1, 4).

Equation of the median through A :

(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)

Substitute (x1 , y1) = (2, 1) and (x2, y2) = (1, 4).

(y - 1)/(4 - 1) = (x - 2)/(1 - 2)

(y - 1)/3 = (x - 2)/(-1)

-1(y - 1) = 3(x - 1)

-y + 1 = 3x - 3

3x + y - 4 = 0

Example 4 :

Two buildings of different heights are located at opposite sides of each other. If a heavy rod is attached joining the terrace of the buildings from (6, 10) to (14, 12), find the equation of the rod joining the buildings ?

Solution :

The equation of the rod is the equation of the line passing through the two points (6, 10) and (14, 12).

Equation of the line in two-point form :

(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)

Substitute (x1 , y1) = (6, 10) and (x2, y2) = (14, 12).

(y - 10)/(12 - 10) = (x - 6)/(14 - 6)

(y - 10)/2 = (x - 6)/8

Least common multiple of the denominators 2 and 8 is 8. 

Multiply each side by 8.

4(y - 10) = x - 6

4y - 40 = x - 6

x - 4y + 34 = 0

Hence, equation of the rod is x - 4y + 34 = 0.

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