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Standard XII

Mathematics

Condition for Two Lines to Be Parallel

If the lines ...

Question

If the lines $a x - b y + c = 0$ and $- 5 x + 3 y + 7 = 0$ are parallel then value of $\frac{a + b}{a - b}$ is

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Solution

If $a x - b y + c = 0$ and $- 5 x + 3 y + 7 = 0$ are parallel
Then $m_{1} = m_{2}$
$\Rightarrow - \frac{a}{(- b)} = - \frac{- 5}{3}$
$\Rightarrow \frac{a}{b} = \frac{5}{3}$
Applying componendo and dividendo, we get
$\Rightarrow \frac{a + b}{a - b} = \frac{5 + 3}{5 - 3} = 4$

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(i) Lines 2x - by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.

(ii) Lines mx + 3y + 7 = 0 and 5x - ny - 3 = 0 are perpendicular to each other. Find the relation connecting m and n.

if the equation of tangent to the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ which are parallel to the line 4x+3y+5=0 is ax+by+c=0,find the value of $- (\frac{a}{b} + \frac{c}{b})$ , where $∣ ∣ \frac{c}{a} ∣ ∣ > 6$

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Q. If the line

a x + b y + c = 0; a \neq 0, b \neq 0

is perpendicular to line

3 y + 4 x + 5 = 0

, then the value of

\frac{3}{a} + \frac{4}{b}

Q. If the lines

2 x - b y + 5 = 0

and

a x + 3 y = 2

are parallel, find the relation connecting

a

and

b

(i) Lines 2x - by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.

(ii) Lines mx + 3y + 7 = 0 and 5x - ny - 3 = 0 are perpendicular to each other. Find the relation connecting m and n.

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Condition for Two Lines to Be Parallel

Standard XII Mathematics

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If the lines ax−by+c=0 and −5x+3y+7=0 are parallel then value of a+ba−b is

If ax−by+c=0 and −5x+3y+7=0 are parallel Then m1=m2 ⇒−a(−b)=−−53 ⇒ab=53 Applying componendo and dividendo, we get ⇒a+ba−b=5+35−3=4

If the lines $a x - b y + c = 0$ and $- 5 x + 3 y + 7 = 0$ are parallel then value of $\frac{a + b}{a - b}$ is

If $a x - b y + c = 0$ and $- 5 x + 3 y + 7 = 0$ are parallel
Then $m_{1} = m_{2}$
$\Rightarrow - \frac{a}{(- b)} = - \frac{- 5}{3}$
$\Rightarrow \frac{a}{b} = \frac{5}{3}$
Applying componendo and dividendo, we get
$\Rightarrow \frac{a + b}{a - b} = \frac{5 + 3}{5 - 3} = 4$