Find the equation of the straight line passing through the point (2, 1

1.	Find the equation of the straight line passing through the point (2, 1) and bisecting the portion of the straight line 3x – 5y = 15 lying between the axes.
Answer» Concept Used: The equation of a line in intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) Given: The line passes through (2, 1) \(\frac{2}{a}+\frac{1}{b}=1\) ……(i) Assuming: The line 3x – 5y = 15 intercept the x-axis and the y-axis at A and B, respectively. Explanation: At x = 0 we have, 0– 5y = 15 ⇒ 5y = -15 ⇒ y = -3 At y = 0 we have, 3x – 0 =15 ⇒ x = 5 A= (0, -3) and B = (5, 0) The midpoint of AB is (\(\frac{5}{2},-\frac{3}{2}\)) lies on the line \(\frac{x}{a}+\frac{y}{b}=1\) \(\frac{5}{2a}-\frac{3}{2b}=1\) ……(ii) Using\(\frac{3}{2}\times\) eq(i) + eq(ii) we get, \(\frac{3}{a}+\frac{5}{2a}=\frac{3}{2}+1\) ⇒ a = \(\frac{11}{5}\) For a = \(\frac{11}{5}\) we have, \(\frac{10}{11}+\frac{1}{b}=1\) ⇒ b = 11 Therefore, the equation of the required line is: \(\frac{x}{\frac{11}{5}}+\frac{y}{11}=1\) \(\frac{5x}{11}+\frac{y}{11}=1\) ⇒ 5x + y = 11 Hence, the equation of line is 5x + y = 11

Find the equation of the straight line passing through the point (2, 1) and bisecting the portion of the straight line 3x – 5y = 15 lying between the axes.

Answer»

Concept Used:

The equation of a line in intercept form is \(\frac{x}{a}+\frac{y}{b}=1\)

Given:

The line passes through (2, 1)

\(\frac{2}{a}+\frac{1}{b}=1\) ……(i)

Assuming:

The line 3x – 5y = 15 intercept the x-axis and the y-axis at A and B, respectively.

Explanation:

At x = 0 we have,

0– 5y = 15

⇒ 5y = -15

⇒ y = -3

At y = 0 we have,

3x – 0 =15

⇒ x = 5

A= (0, -3) and B = (5, 0)

The midpoint of AB is (\(\frac{5}{2},-\frac{3}{2}\)) lies on the line \(\frac{x}{a}+\frac{y}{b}=1\)

\(\frac{5}{2a}-\frac{3}{2b}=1\) ……(ii)

Using\(\frac{3}{2}\times\) eq(i) + eq(ii) we get,

\(\frac{3}{a}+\frac{5}{2a}=\frac{3}{2}+1\)

⇒ a = \(\frac{11}{5}\)

For a = \(\frac{11}{5}\) we have,

\(\frac{10}{11}+\frac{1}{b}=1\)

⇒ b = 11

Therefore, the equation of the required line is:

\(\frac{x}{\frac{11}{5}}+\frac{y}{11}=1\)

\(\frac{5x}{11}+\frac{y}{11}=1\)

⇒ 5x + y = 11

Hence, the equation of line is 5x + y = 11