Find the equation of a line parallel to 5x +6y = 0 at distance sqrt 61 from (-3,7)
I need help please. I don't have any idea
To find the equation of a line parallel to the line 5x + 6y = 0, we need to determine its slope.
First, let's rearrange the given equation into slope-intercept form (y = mx + b), where m represents the slope:
5x + 6y = 0
6y = -5x
y = -(5/6)x
Since the given line is parallel to the line we are trying to find the equation for, they will have the same slope.
The next step is to find the equation of a line at a distance sqrt(61) from the point (-3, 7). To do this, we can use the point-slope form of a linear equation (y - y1 = m(x - x1)), where (x1, y1) represent the coordinates of a point on the line we are trying to find, and m is the slope we found earlier.
Using (-3, 7) as the point and the slope -(5/6), we have:
y - 7 = -(5/6)(x - (-3))
y - 7 = -(5/6)(x + 3)
y - 7 = -(5/6)x - (5/6)(3)
y - 7 = -(5/6)x - 5/2
Finally, we can rearrange the equation to the standard form (Ax + By = C) by multiplying every term by 6 to eliminate the fractions:
6(y-7) = -5x - (15/2)
6y - 42 = -5x - 15/2
10x + 6y = -45
Therefore, the equation of the line parallel to 5x + 6y = 0 at a distance sqrt(61) from (-3, 7) is 10x + 6y = -45.
To find the equation of a line that is parallel to a given line, we need to understand the relationship between the coefficients of the equations.
The given equation is 5x + 6y = 0. To find a line parallel to this equation, the new line must have the same slope. In this case, we can rearrange the equation into slope-intercept form (y = mx + b), where m is the slope of the line:
6y = -5x
y = (-5/6)x
So, the slope of the given line is -5/6.
Now, we know the slope of the new line we are trying to find and a point that lies on the line, which is (-3, 7).
Let's use the point-slope form of a linear equation to find the equation of the new line:
y - y1 = m(x - x1)
In this equation, (x1, y1) represents the coordinates of the given point, and m represents the slope.
Plugging in the values, we get:
y - 7 = (-5/6)(x - (-3))
y - 7 = (-5/6)(x + 3)
y - 7 = (-5/6)x - (5/6)(3)
y - 7 = (-5/6)x - 5/2
y = (-5/6)x + (7 - 5/2)
y = (-5/6)x + 9/2
Therefore, the equation of the line parallel to 5x + 6y = 0 and at a distance sqrt(61) from (-3,7) is:
y = (-5/6)x + 9/2
y = -(5/6)x for the first line, so the slope of your second (parallel) line is also -5/6. Its equation is
y' = -(5/6)x + C
C is the y-intercept of the second line.
You have to choose C so that y and y' are sqrt61 apart.
C is the hypotenuse of a right triangle with sides sqrt61 and (5/6)*sqrt61. Its value is sqrt[61 + (25/36)*61]
= sqrt[(61/36)*61] = 61/6
The line you want is
y' = (-5/6)x + 61/6
There is a second line that is also a solution, below the first line
y' = (-5/6)x - 61/6