Question

Find the equation of straight line passes through the point (3,-2)and cuts off positive intercepts on the x-axis and y-axis which sre in the ratio 4:3

Answer 1

Step-by-step explanation:

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Answer 2

Given:

• The line passes through the points (3,-2)
• The intercepts are in the ratio of 4:3

To Find:

• The equation of the line passing through points (3,-2)

Solution:

Let the ratios 4:3 be denoted as 4a and 3a.

∴ The intercepts are given as,

⇒ x/4a + y/3a = 1 → {equation 1}

It is given that the line passes through the points (3,-2) which implies that "x = 3 and y = -2". On substituting the values of "x" and "y" in equation 1 we get,

⇒ 3/4a - 2/3a = 1

On finding the LCM of the above equation we get,

⇒ [(3/4a)×12a - (2/3a)×12a]/12a = 1 {LCM = 12a}

On canceling the divisible terms from the above equation we get,

⇒ (9-8)/12a = 1

⇒ 9-8 = 12a {subtracting the terms in LHS}

⇒ 1 = 12a

⇒ a = 1/12

Now we need to find the values of 4a and 3a. Let us consider,

4a = 4 × (1/12) = 1/3

3a = 3×(1/12) = 1/4

On substituting the values of 4a and 3a in equation 1 we get,

⇒ x/(1/3) + y/(1/4) = 1

⇒ 3x + 4y = 1 or 3x + 4y - 1 = 0

∴ The equation of the line passing through the point (3,-2) = 3x + 4y = 1 or 3x + 4y -1 = 0.