Question

The equation of the line passing through (–4, 3) and having intercepts whose ratio is 5 : 3 is

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a) 9x + 20y – 96 = 0
b) 3x + 5y = 3
c) 9x + 20y + 96 = 0
d) 9x – 20y – 96 = 0​

Answer 1

Answer:

$\large\boxed{\sf{(b)\;3x+5y=3}}$

Step-by-step explanation:

Let the intercepts made on X and Y axes are 'a' and 'b' respectively.

Also, their ratio is 5:3

Therefore, we will get,

$= > \dfrac{a}{b} = \dfrac{5}{3} \\ \\ = > a = \dfrac{5}{3} b \: \: \: \: \: \: .........(i)$

Now, we know that, equation of line having intercepts 'a' and 'b' on X and Y axes respectively is given by,

$\dfrac{x}{a} + \dfrac{y}{b} = 1 \: \: \: \: \: \: ...........(ii)$

Also, the given line passes through (-4, 3).

Therefore, it will satisfy this equation (ii).

Therefore,substiting the values in (ii), we get,

$= > \dfrac{ - 4}{a} + \dfrac{3}{b} = 1$

Now, substituting the value of a from eqn (i), we get,

$= > \dfrac{ - 4}{ \frac{5b}{3} } + \dfrac{3}{b} = 1 \\ \\ = > \dfrac{ - 12}{5b} + \dfrac{3}{b} = 1 \\ \\ = > \dfrac{ - 12 + 15}{5b} = 1 \\ \\ = > 5b = 3 \\ \\ = > b = \dfrac{3}{5}$

Therefore, we will get,

$= > a = \dfrac{5}{3} \times \dfrac{3}{5} \\ \\ = > a = 1$

Now, substituting the value of 'a' and 'b' in eqn (ii), we will get the desired equation, i.e.,

$= > \dfrac{x}{1} + \dfrac{5y}{3} = 1 \\ \\ = > \frac{3x + 5y}{3} = 1 \\ \\ = > 3x + 5y = 3$

Hence, the correct option is (b) 3x + 5y = 3

Answer 2

Answer:

m:n=2:5

(x1,y1)=(4,3)

nx+my=nx1+my1

3x+5y=3*(-4)+5*(3)

3x+5y=-12+15

3x+5y=3

3x+5y-3=0