A straight line passes through the point (-1,2) and its distance from the origin is 1 unit. Find the equation
To find the equation of the straight line that passes through the point (-1,2) and has a distance of 1 unit from the origin, we can use the distance formula.
Let the equation of the line be y = mx + b, where m is the slope of the line and b is the y-intercept.
Since the line passes through (-1,2), we can substitute these values into the equation to get 2 = m(-1) + b.
The distance between the origin (0,0) and any point (x,y) on the line can be calculated using the distance formula:
d = sqrt((x - 0)^2 + (y - 0)^2)
For the line to be 1 unit away from the origin, we can set the distance formula equal to 1 and substitute the point (-1,2) on the line:
1 = sqrt((-1 - 0)^2 + (2 - 0)^2)
1 = sqrt(1 + 4)
1 = sqrt(5)
Squaring both sides of the equation, we get:
1 = 5
This is not true, so the equation y = mx + b does not satisfy the condition of having a distance of 1 unit from the origin.
Therefore, there is no straight line that passes through the point (-1,2) and has a distance of 1 unit from the origin.
To find the equation of a straight line passing through the point (-1,2) and having a distance of 1 unit from the origin, we can use the formula for the distance between a point and a line.
The equation of a straight line can be written in the form y = mx + c, where m represents the slope of the line and c represents the y-intercept.
Since the line passes through the point (-1,2), we can substitute these values into the equation to get:
2 = m(-1) + c
Next, we need to find the equation of the line that is 1 unit away from the origin. The distance between a point (x1, y1) and the origin (0, 0) is given by the formula:
√[(x1 - 0)^2 + (y1 - 0)^2] = 1
In this case, since the point on the line closest to the origin is unknown, we can assign it coordinates (x, y). Substituting this point into the distance formula, we get:
√[(x - 0)^2 + (y - 0)^2] = 1
Simplifying this equation yields:
√[x^2 + y^2] = 1
Now, we have a system of two equations:
2 = -m + c
√[x^2 + y^2] = 1
To solve this system, we can eliminate one of the variables. Solving the first equation for c, we get:
c = 2 + m
Next, we substitute this value of c into the second equation:
√[x^2 + y^2] = 1
Now, we can square both sides to eliminate the square root:
x^2 + y^2 = 1
This equation represents a circle with radius 1 centered at the origin (0, 0). To find the equation of the line that is 1 unit away from the origin, we need to find the equation of the circle and then find the equation of the line perpendicular to the radius, passing through the given point.
Since the line is tangent to the circle, the radius drawn to the point of tangency will be perpendicular to the line. The slope of the tangent line will be equal to the negative reciprocal of the slope of the radius.
Let's assume the equation of the line is y = mx + b.
Since the line is perpendicular to the radius, the product of the slopes will be -1:
(m) * (-1/m) = -1
Simplifying this expression, we get:
m^2 = -1
This equation has no real solutions, which means no line can satisfy the given conditions.
Therefore, there is no straight line that passes through the point (-1,2) and is 1 unit away from the origin.