Question

A straight line makes on the co-ordinate axes positive intercepts whose sum is 5.If the line passes through the  point  P(-3,4).Find its equation.

Answer 1

sum of intercepts is 5
i.e., a + b = 5
 ⇒  b = 5 - a
the line passes through P(-3,4)
w.k.t eq of line is x/a + y/b = 1
substitute P(-3,4) in above eq
we get -3/a + 4/b = 1
          4/b = 1 + 3/a
        a (4/5-a) = a + 3
         4a = 5a-a²+15-3a 
        a²+2a-15 = 0
        a = 3 or a = -5
if a =3 , b= 5-3 = 2
then eq of required line is x/3 + y/2 = 1⇒ 2x+3y-6 = 0
if a =-5 , b = 5+5 = 10
then eq of required line is x/-5 + y/10 = 1⇒ -2x+y-10 = 0
       

Answer 2

There could be multiple answers to this question.

         x/a + y/b = 1
If we write the equation of a straight line in this form, then the x axis intercept is a and y axis intercept is b.       Given a+b = 5

bx + a y = ab
(5 - a) x + a y = a (5 - a)

Point P (-3,4) lies on the straight line.

-3 (5-a) + 4 a = 5a - a²
-15 + 3a + 4a = 5a - a²
a² + 2a - 15 = 0
(a + 5)(a-3) = 0          a = -5 or 3
b = 10 or 2

So equations of the two straight lines are :

10 x - 5y = -50  or  2 x - y + 10 = 0  and          2 x + 3 y = 6

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we can also solve this question by using  y = m x + c form where c is the y axis intercept and x axis intercept is -c/m.
c - c/m = 5
m c - c = 5m        c = 5 m /(m-1)

4 = -3m+c
4 = - 3 m  + 5 m /(m-1)
4 (m-1) = -3m(m-1) + 5 m
4 m - 4 = - 3 m² + 3m + 5m
3m² - 4m - 4 = 0
m = [4 +- √(16+48)]/6 = (4+- 8 )/ 6 = 2 or -4/6 or -2/3
c = 5 * 2/1 = 10  or  (5*-2/3) / (-1-2/3)  = 2

y = 2 x + 10      or    y = - 2x /3   +    2