find the equation of the line that is parallel to the line y=15x and contains the point (1,-4)
y=
find the distance d (p_1, and p_2) between the points p_1 and p_2
p_1=(-2,1)
p_2=(1,4)
(1) The slope must be 15. To pass through (1,-4), it must be the equation
y + 4 = 15 (x - 1)
That can be rewritten in standard form as
y = 15x -19
(2) (distance)^2 = (y2-y1)^2 +(x2-x1)^2
=(4-1)^2 + [1 - (-3)]^2 = 9 + 16 = 25
distance = 5
Write the equation in slope intercept form of a line that is perpendicular to y=-15x-7 but goes through the point (0,2).
To find the equation of the line that is parallel to y = 15x and passes through the point (1, -4), we can use the point-slope formula.
The slope of the given line y = 15x is 15 (since the equation is in the form y = mx + b, where m is the slope).
Since the line we are looking for is parallel to y = 15x, it will also have a slope of 15.
Using the point-slope formula: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation:
y - (-4) = 15(x - 1)
y + 4 = 15x - 15
y = 15x - 19
Therefore, the equation of the line that is parallel to y = 15x and contains the point (1, -4) is y = 15x - 19.
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To find the distance between points p1 (-2,1) and p2 (1,4), we can use the distance formula.
The distance formula is given by d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
Substituting the given coordinates:
d = sqrt((1 - (-2))^2 + (4 - 1)^2)
= sqrt(3^2 + 3^2)
= sqrt(9 + 9)
= sqrt(18)
= 3√2
Therefore, the distance d between points p1 (-2,1) and p2 (1,4) is 3√2.
To find the equation of a line that is parallel to the line y = 15x and passes through the point (1, -4), we need to use the slope-intercept form of a line, which is y = mx + b.
First, let's find the slope of the given line y = 15x. The slope, represented by m, is the coefficient of x. In this case, the slope (m) is 15.
Since the line we're looking for is parallel to the given line, it will have the same slope of 15. Now we can plug this slope and the coordinates (1, -4) into the slope-intercept form to find the y-intercept (b).
Using the formula y = mx + b and substituting the values, we have:
-4 = 15(1) + b
Simplifying, we get:
-4 = 15 + b
To isolate b, we subtract 15 from both sides:
-4 - 15 = b
-19 = b
Now we have the slope (m = 15) and the y-intercept (b = -19). We can substitute these values into the slope-intercept form to get the equation of the parallel line:
y = 15x - 19
Therefore, the equation of the line that is parallel to y = 15x and passes through the point (1, -4) is y = 15x - 19.
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To find the distance (d) between two points, p1 and p2, we can use the distance formula in a Cartesian coordinate system.
The distance formula is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Given the points p1 (-2, 1) and p2 (1, 4), we can substitute the coordinates into the formula:
d = √((1 - (-2))^2 + (4 - 1)^2)
Simplifying, we have:
d = √((3)^2 + (3)^2)
Calculating further:
d = √(9 + 9)
d = √(18)
d ≈ 4.242
Therefore, the distance (d) between the points p1 and p2 is approximately 4.242.