Question

Find the equation of line which passes through the point (-1,2) and whose distance from the origin is 1 unit.
Please answer as soon as possible.

Answer 1

Answer:

equation of line passing through point(-1,2)is

y-2=m(X+1)

y-2=mx+m

mx-y+(m+2)=0

perpendicular distance form origin is|ax+bx+c|\√a²+b²

1=|m+2|/√m²+1

1=(m+2)²/m²+1

m²+1=(m+2)²

m²+1=m²+4m+4

m=-3/4

so eqn of line is:

-3x-4y+5=0

3x+4y-5=0

which is required eqn of a line.

Step-by-step explanation:

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Answer 2

EXPLANATION.

→ equation of line which passes through the

point (-1,2).

→ distance from the origin is 1 unit.

$\sf : \implies \: equation \: of \: line \\ \\ \sf : \implies \: (y - y_{1}) = m(x - x_{1}) \\ \\ \sf : \implies \: (2 - y_{1}) = m( - 1 - x_{1}) \\ \\ \sf : \implies \: 2 - y_{1} = - m - mx_{1} \\ \\ \sf : \implies \: m x_{1} - y_{1} + m + 2 = 0 \: \: \: .....(1)$

→ distance from origin = 1 unit

→ coordinate of origin = (0,0)

$\sf : \implies \: formula \: of \: perpendicular \: distance \\ \\ \sf : \implies \: d \: = | \frac{a x_{1} + b y_{1} + c }{ \sqrt{ {a}^{2} + {b}^{2} } } | \\ \\ \sf : \implies \: 1 = | \frac{m(0) + ( - 1)(0) + (m + 2)}{ \sqrt{ {m}^{2} + ( - 1) {}^{2} } } | \\ \\ \sf : \implies \: 1 = | \frac{m + 2}{ \sqrt{ {m}^{2} + 1} } | \\ \\ \sf : \implies \: \sqrt{ {m}^{2} + 1} = m + 2 \\ \\ \sf : \implies \: squaring \: on \: both \: sides \: \: we \: \: get$

$\sf : \implies \: {m}^{2} + 1 = (m + 2) {}^{2} \\ \\ \sf : \implies \: {m}^{2} + 1 = {m}^{2} + 4 + 4m \\ \\ \sf : \implies \: - 3 = 4m \\ \\ \sf : \implies \: m \: = \frac{ - 3}{4} = slope \:$

$\sf : \implies \: put \: the \: value \: in \: equation \: (1) \\ \\ \sf : \implies \: \frac{ - 3}{4} x_{1} - y_{1} \: + \: ( \frac{ - 3}{4} ) + 2 = 0 \\ \\ \sf : \implies \: \frac{ - 3x - 4y - 3 + 8}{4} = 0 \\ \\ \sf : \implies \: - 3x - 4y + 5 = 0 \\ \\ \sf : \implies \: - (3x + 4y - 5) = 0 \\ \\ \sf : \implies \: 3x + 4y - 5 = 0 \:$