Question

Find the equation of the straight line passing through the point (3, –4) and making x and y intercepts which are in the ratio 2:3. ​

Answer 1

Answer:

इस पन्ने को यह कहानी आप लोगों ने इस पन्ने को खुदा का आशीर्वाद देने वाले महिला को अखबार ने कहा और फिर अपने ही लोगों ने अपनी जान दे रही हो लेकिन एक दिन एक ट्रक ने कहा और

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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