## EQUATION OF LINE WITH A POINT AND INTERCEPTS

Example 1 :

Find the equation of the line passing through the point (9, -1) and having its x-intercept thrice and its y-intercept.

Solution :

Let a and b are x and y-intercepts respectively.

x-intercept = 3(y-intercept)

a = 3b

Intercept form equation of a line :

x/a + y/b = 1

Here, and are and y-intercepts respectively.

The required line is passing through the point (9 , -1)

x/a  + y/b = 1

9/3b - 1/b = 1

(9 - 3)/3b = 1

6/3b = 1

2/b = 1

2 = b

y-intercept (b) = 2

x -intercept (a) = 3(2) = 6

x/6 + y/2 = 1

x/6 + 3y/6 = 1

x + 3y = 6

x + 3y - 6 = 0

Example 2 :

A straight line cuts coordinate axes at A and B. If the midpoint of AB is (3, 2), then find the equation of AB. Solution :

x-intercept = a and y-intercept = b

From the picture, coordinates of A are (a, 0) and coordinates of B are (0, b).

Midpoint of AB = (x+ x2)/2 , (y+ y2)/2

(3, 2) = (a + 0)/2, (0 + b)/2

By equating x and y coordinates, we get

3 = a/2       2 = b/2

a = 6          b = 4

Intercept form equation of a line :

x/a + y/b = 1

x/6 + y/4 = 1

(2x + 3y)/12 = 1

2x + 3y = 12

2x + 3y - 12 = 0

Example 3 :

Find the equation of the line passing through (22, -6) and having intercept on x-axis exceeds the intercept on y-axis by 5.

Solution :

Let x -intercept be a and y-intercept be b.

x -intercept (a) = b + 5

Intercept form equation of a line :

x/a + y/b = 1

(22/(b+5)) + (-6/b) = 1

[22b - 6(b + 5)]/b(b + 5) = 1

(22b - 6b - 30)/(b+ 5b) = 1

(16b - 30)/(b+ 5b) = 1

16b - 30 = b+ 5b

b+ 5b - 16b - 30 = 0

b- 11b - 30 = 0

(b - 6)(b - 5) = 0

b = 6  or  b = 5

 When b = 6,a = 6 + 5a = 11 When b = 5,a = 5 + 5a = 10

Equation of the line when a = 11 and b = 6 :

x/11 + y/6 = 1

6x + 11y = 66

6x + 11y - 66 = 0

Equation of the line when a = 10 and b = 5 :

x/10 + y/5 = 1

x + 2y = 10

x + 2y - 10 = 0

Equation of the required line is

x + 2y - 10 = 0  or  6x + 11y - 66 = 0

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