Question

find the equation of the straight line passing through the point (3,-4) and making X and Y intercepts which are in ratio 2:3​

Answer 1

Answer:

Step-by-step explanation:

X    Y

2  :  3

X=2/5

Y=3/5

y=mx-c

-4=m(3)_-3/5

-4-3/5=3m

-4/3÷3=3m/3

    5

-1/8=m

  15

-4= (-1/8)3_-3/5

         15

-4=-4/3+3

         5  5

Answer 2

Given data:

The equation of the straight line is passing through the point (3,-4) and making intercept which is in the ratio 2:3.

To find:

The equation of the straight line as detailed above.

Solution:

Given, the ratio of the intercepts a, b as a:b = 2:3

Then, x-intercept = 2m

          y-intercept = 3m

As we know that, the equation of the line is:  $\frac{x}{a}+\frac{y}{b}=1$

On substituting the x, y-intercepts, the line of equation will be

$\frac{x}{2m}+\frac{y}{3m}=1$

As this line is passing through the point (3, -4), substitute them in the above equation to get the value of m.

⇒ $\frac{3}{2m}+\frac{-4}{3m}=1$

⇒ $\frac{3}{2}+\frac{-4}{3}=m$

⇒ $\frac{9-8}{6}=m$

⇒ $m = \frac{1}{6}$

Substitute the value of:

x-intercept = 2($\frac{1}{6}$) = $\frac{1}{3}$

y-intercept = 3($\frac{1}{6}$) = $\frac{1}{2}$

Then, the equation will be:

⇒ $\frac{x}{\frac{1}{3} } + \frac{y}{\frac{1}{2} }=1$

⇒ 3x + 2y = 1

⇒ 3x + 2y - 1 = 0

Therefore, the equation of the straight line passing through the point (3,-4), making intercepts which are in the ratio 2:3 is

3x + 2y - 1 = 0

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