Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

3 Views|Posted a year ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f l i n e h a v i n g a a n d b o n t h e a x i s i s \\ \frac{x}{a} + \frac{y}{b} = 1 \dots (i) \\ G i v e n t h a t a + b = 14 \Rightarrow b = 14 - a \\ \Rightarrow \frac{x}{a} + \frac{y}{14 - a} = 1 \dots (i i) \\ I f e q n . (i i) p a s s e s t h r o u g h t h e (3, 4) t h e n \\ \frac{3}{a} + \frac{4}{14 - a} = 1 \\ \Rightarrow \frac{3 (14 - a) + 4 a}{a (14 - a)} = 1 \\ \Rightarrow 42 + a = 14 a - a^{2} \\ \Rightarrow a^{2} + a - 14 a + 42 = 0 \\ \Rightarrow a^{2} - 13 a + 42 = 0 \\ \Rightarrow a^{2} - 7 a - 6 a + 42 = 0 \\ \Rightarrow a (a - 7) - 6 (a - 7) = 0 \\ \Rightarrow (a - 6) (a - 7) = 0 \Rightarrow a = 6, 7 \\ ∴ b = 14 - 6 = 8, b = 14 - 7 = 7 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f l i n e s a r e \\ \frac{x}{6} + \frac{y}{8} = 1 \Rightarrow 4 x + 3 y = 24 \\ a n d \frac{x}{7} + \frac{y}{7} = 1 \Rightarrow x + y = 7 \end{array}$

Upvote

Similar Questions for you

If |z + 1| = αz + β (i + 1) and  $z = \frac{1}{2}$ – 2i, find α + β.

Alok Kumar Singh

$| \frac{1}{2} - 2 i + 1 | = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\sqrt[]{\frac{9}{4} + 4} = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\frac{5}{2} = α (\frac{1}{2}) + β + i (- 2 α + β)$

$\frac{α}{2} + β = \frac{5}{2}$ ...(1)

–2α + β = 0 …(2)

Solving (1) and (2)

$\frac{α}{2} + 2 α = \frac{5}{2}$

$\frac{5}{2} α = \frac{5}{2}$

a =

Alok Kumar Singh

Variance = $\frac{\sum x^{2}}{n} - {(\bar{x})}^{2}$

$\frac{6 0^{2} + 6 0^{2} + 4 4^{2} + 5 8^{2} + 6 8^{2} + α^{2} + β^{2} + 5 6^{2}}{8} = {(58)}^{2} = 66.2$

$\frac{7200 + 1936 + 3364 + 4624 + 3136 + α^{2} + β^{2}}{8} = 3364 = 66.2$

$2532.5 + \frac{α^{2} + β^{2}}{8} - 3364 = 66.2$

α² + β² = 897.7 × 8

= 7181.6

Alok Kumar Singh

Start with

(1) $\bar{E} : \frac{6!}{2!} = 360$

(2) $\bar{G E} : \frac{5!}{2!},$ $\bar{G N} : \frac{5!}{2!}$

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

Alok Kumar Singh

${(1 + x)}^{11} =^{11} C_{0} +^{11} C_{1} x +^{11} C_{2} x^{2} + . . . . . +^{11} C_{11} x^{11}$

= \frac{2^{12} - 2 - 24}{12}

= \frac{2^{12} - 26}{12} = \frac{4070}{12} = \frac{2035}{6} = \frac{m}{n}

m + n = 2035 + 6 = 2041

Let $f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$ 

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ 

The range of fog(x) is

Alok Kumar Singh

$f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ ->g(x) = |x|, x Î (–3, 1)

$f (g (x)) = {\begin{matrix} 2 + 2 | x |, & | x | \in (- 1, 0) \Rightarrow x \in ? \\ 1 - \frac{| x |}{3}, & | x | \in [0, 3) \Rightarrow x \in (- 3, 1) \end{matrix}$