Question

A straight line passes through the point (- 1 2 and its distance from the origin is 1 unit

Answer 1

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Answer 2

Answer:

$3x+4y=5$

Step-by-step explanation:

Let the equation of line be $x\cos\alpha+y\sin\alpha=p$

where, p is distance of line from origin.

α is angle of normal to line from origin$0\leq \alpha\leq \dfrac{\pi}{2}$

p = 1 and passing point (-1,2)

Substitute into equation

$-1\cos\alpha+2\sin\alpha=1$3

$2\sin\alpha=1+\cos\alpha$

$\alpha=53^\circ$

$\sin53^\circ=\dfrac{4}{5}$

$\cos53^\circ=\dfrac{3}{5}$

$\dfrac{3}{5}x+\dfrac{4}{5}y=1$

$3x+4y=5$

Hence, the required equation is 3x + 4y = 5