Question

Find the equation of a straight line which passes through the point (3, 4) and sum of its intercepts on the x and y axis is 14.

• A. x + y = 7
• B. 4x + 3y = 24
• C. x + y = 14
• D. 3x + 4y = 24

Answer 1

Answer: (A), (B)

4x + 3y = 24

Let intercept on x and y-axis be a & b.

Equation of the time be $\frac{x}{a}+\frac{y}{b}=1$

This equation passes through (3,4)

Therefore $\frac{3}{a}+\frac{4}{b}=1$-------------(1)

Given, a + b =14

b = 14 - a -----------------(2)

Substituting the value of b in equation (1)

$\frac{3}{a}+\frac{4}{(14-a)}=1$

$\frac{42-3a+4a}{a(14-a)}=1$

$42+a=14a-a^2$

$a^2-13a+42=0$

(a - 7) (a - 6) = 0

a = 7,6

For a = 7, b = 14 - 7 = 7

For a = 6, b = 14 - 6 = 8

Substituting the value of a and b in equation (1)

We get the equations of lines

$\frac{x}{7}+\frac{y}{7}=1$ and $\frac{x}{6}+\frac{y}{8}=1$

x + y = 7 and 4x + 3y = 24