Find the equation of the straight line which passes through the point

1.	Find the equation of the straight line which passes through the point (-3, 8) and cuts off positive intercepts on the coordinate axes whose sum is 7
Answer» Concept Used: The equation of the line with intercepts a and b is \(\frac{x}{a}+\frac{y}{b}=1\) Given: Here a + b = 7, b = 7 – a Explanation: The line is passing through (-3, 8). \(\frac{-3}{a}+\frac{8}{b}=1\) Substituting b =7 – a, we get \(\frac{x}{a}+\frac{y}{7-b}=1\) ⇒ -3(7 - a) + = 7a - a² ⇒ a² + 4a – 21 = 0 ⇒ (a – 3)(a + 7)= 0 ⇒ a = 3 ( since, a can only be positive) Substituting a = 3 in equation (i) we get, b = 7 – 3 = 4 Hence, the equation of the line is \(\frac{x}{3}+\frac{y}{4}=1\)

Find the equation of the straight line which passes through the point (-3, 8) and cuts off positive intercepts on the coordinate axes whose sum is 7

Answer»

Concept Used:

The equation of the line with intercepts a and b is \(\frac{x}{a}+\frac{y}{b}=1\)

Given:

Here a + b = 7, b = 7 – a

Explanation:

The line is passing through (-3, 8).

\(\frac{-3}{a}+\frac{8}{b}=1\)

Substituting b =7 – a, we get

\(\frac{x}{a}+\frac{y}{7-b}=1\)

⇒ -3(7 - a) + = 7a - a²

⇒ a² + 4a – 21 = 0

⇒ (a – 3)(a + 7)= 0

⇒ a = 3 ( since, a can only be positive)

Substituting a = 3 in equation (i) we get, b = 7 – 3 = 4

Hence, the equation of the line is \(\frac{x}{3}+\frac{y}{4}=1\)