Find the equation of the line which cuts off an intercept 4 on negative direction of x-axis and an intercept of 2 on positive direction of y-axis.

x - 2y + 4 = 0

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x + 2y - 4 = 0

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2x - y + 4 = 0

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2x + y - 4 = 0

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Solution

The correct option is A x - 2y + 4 = 0
x intercept = -4
y intercept = 2

The required equation is

$\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow \frac{x}{(- 4)} + \frac{y}{2} = 1$

$\Rightarrow \frac{- x + 2 y}{4} = 1$

$\Rightarrow - x + 2 y = 4 \Rightarrow - x + 2 y - 4 = 0$

$\Rightarrow - 1 (x - 2 y + 4) = 0$

$\Rightarrow x - 2 y + 4 = 0$

So, option b is correct.

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Find the equation of the line which cuts off an intercept 4 on negative direction of x-axis and an intercept of 2 on positive direction of y-axis.

The correct option is A x - 2y + 4 = 0x intercept = -4 y intercept = 2 The required equation is xa+yb=1⇒x(−4)+y2=1 ⇒ −x+2y4=1 ⇒ −x+2y=4⇒−x+2y−4=0 ⇒ −1(x−2y+4)=0 ⇒ x−2y+4=0 So, option b is correct.