## 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 (a)  =  3 (y-intercept)

a  =  3b

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

So, the required equation is 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,

Coordinate of A is (a, 0) and coordinate of B is (0, b).

Midpoint of AB  =  (x1+x2)/2 , (y1+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 :

(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 :

(x/a)+(y/b)  =  1

(22/(b+5))+(-6/b)  =  1

(22b-6(b+5))/b(b+5)  =  1

(22b-6b-30)/(b2+5b)  =  1

(16b-30)/(b2+5b)  =  1

16b-30  =  b2+5b

b2+5b-16b-30  =  0

b2-11b-30  =  0

(b-6) (b-5)  =  0

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

The required equations are

x+2y-10  =  0 or 6x+11y-66  =  0. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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