Consider four straight lines: $(i) l_{1}; 3 y = 4 x + 5 (i i) l_{2}; 4 y = 3 x - 1 (i i i) l_{3}; 4 y + 3 x = 7 (i v) l_{4}; 4 x + 3 y = 2$

l_{1} a n d l_{4} a r e p a r a l l e l

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$l_{1} a n d l_{2} a r e p e r p e n d i c u l a r$

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l_{2} a n d l_{4} a r e p e r p e n d i c u l a r

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l_{2} a n d l_{3} a r e p a r a l l e l

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Solution

The correct option is C $l_{2} a n d l_{4} a r e p e r p e n d i c u l a r$
$F r o m t h s g i v e n e q u a t i o n s, W e c a n f i n d t h e s l o p e o f t h e l i n e s b y e q u a t i n g w i t h a x + b y + c = 0 S l o p e = \frac{- a}{b} S l o p e o f l i n e l_{1} = \frac{4}{3} S l o p e o f l i n e l_{2} = \frac{3}{4} S l o p e o f l i n e l_{3} = - \frac{3}{4} S l o p e o f l i n e l_{4} = - \frac{4}{3} A p p l y i n g t h e s l o p e s f o r c o n d i t i o n s o f p e r p e n d i c u l a r a n d p a r a l l e l l i n e s (i) l_{1} \times l_{2} = \frac{4}{3} \times \frac{3}{4} = 1 \neq - 1 (F a l s e) (i i) l_{1} = \frac{4}{3}, l_{4} = - \frac{4}{3} l_{1} \neq l_{4} (F a l s e) (i i i) l_{2} \times l_{4} = \frac{3}{4} \times \frac{- 4}{3} = - 1 (T r u e) (i v) l_{2} = \frac{3}{4}, l_{3} = - \frac{3}{4} l_{2} \neq l_{3} (F a l s e) H e n c e O p t i o n C i s t h e c o r r e c t a n s w e r .$

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Similar questions

Draw the graphs of the following straight −lines:

(i) y = 3 − x; (ii) y = x − 3; (iii) y = 3x − 2; (iv) y = 5 − 3x;

(v) 4y = − x + 3; (vi) 3y = 4x + 1; (vii) x = 4; (viii) 3y = 1.

Q. The straight lines

L_{1} \equiv 4 x - 3 y + 2 = 0, L_{2} \equiv 3 x + 4 y - 4 = 0

and

L_{3} \equiv x - 7 y + 6 = 0

Q. Find the angles between each of the following pairs of straight lines:
(i) 3x + y + 12 = 0 and x + 2y − 1 = 0
(ii) 3x − y + 5 = 0 and x − 3y + 1 = 0
(iii) 3x + 4y − 7 = 0 and 4x − 3y + 5 = 0
(iv) x − 4y = 3 and 6x − y = 11
(v) (m² − mn) y = (mn + n²) x + n³ and (mn + m²) y = (mn − n²) x + m³.

Q. Consider the lines given by

L_{1} : x + 3 y - 5 = 0

L_{2} : 3 x - k y - 1 = 0

L_{3} : 5 x + 2 y - 12 = 0

Find the equation of circle circumscribing the quadrilateral formed by four lines $3 x + 4 y - 5 = 0$ , $4 x - 3 y - 5 = 0$ , $3 x + 4 y + 5 = 0$ and $4 x - 3 y + 5 = 0$

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Consider four straight lines: (i) l1; 3y=4x+5 (ii) l2; 4y=3x−1 (iii) l3;4y+3x=7 (iv) l4;4x+3y=2

Consider four straight lines: $(i) l_{1}; 3 y = 4 x + 5 (i i) l_{2}; 4 y = 3 x - 1 (i i i) l_{3}; 4 y + 3 x = 7 (i v) l_{4}; 4 x + 3 y = 2$