Write the parametric equations of a line perpendicular to 4x + 8y +7 =0 with the same x-intercept as [𝑥,𝑦]=[2,7]+𝑡[−10,3].
the line has slope -1/2
so a perpendicular line has slope 2
now find the x-intercept for the parametric line, and you have a point and a slope, so use the point-slope form for your new line.
Oh, perpendicular lines, they always like to do their own thing! So, we have the original line with the equation 4x + 8y + 7 = 0. To find a line perpendicular to this one, we need to find its slope first.
The slope of the given line can be found by rearranging the equation into slope-intercept form: y = (-1/2)x - 7/8. The slope is the coefficient of x, which is -1/2.
Now, for a line to be perpendicular to this one, its slope must be the negative reciprocal of (-1/2). So, the slope of our new line will be 2.
Since the line has the same x-intercept as [𝑥,𝑦] = [2,7] + t[−10,3], we can see that the x-coordinate of the x-intercept is 2. Therefore, our new line will pass through the point (2, 0) since the x-intercept has a y-coordinate of 0.
Using the point-slope form of a line, we can write the equation as:
y = 2(x - 2)
And that's it! We now have the parametric equation of a line that is perpendicular to 4x + 8y + 7 = 0 and has the same x-intercept as [𝑥,𝑦] = [2,7] + t[−10,3].
To find the equation of a line perpendicular to the line 4x + 8y + 7 = 0, we need to determine the slope of the given line. By rearranging the equation into slope-intercept form, y = mx + b, we can identify the slope:
4x + 8y + 7 = 0
8y = -4x - 7
y = -0.5x - 7/8
From this, we can observe that the slope of the given line is -0.5.
For a line to be perpendicular to this line, its slope has to be the negative reciprocal of -0.5, which is 2.
The given line [x, y] = [2, 7] + t[-10, 3] can be represented parametrically as:
x = 2 - 10t
y = 7 + 3t
To find a line perpendicular to the given line with the same x-intercept, we can keep the x-intercept constant and change the y-intercept while maintaining the same slope.
Let's assume the x-intercept is (a, 0), where 'a' is the constant value.
Since the line has a slope of 2, its equation can be written as y = 2x + b, where 'b' is the y-intercept we need to determine.
Plugging in the x-intercept into the equation, we get:
0 = 2a + b
Solving for b, we find that b = -2a.
Therefore, the parametric equations of the line perpendicular to 4x + 8y + 7 = 0 with the same x-intercept as [𝑥,𝑦]=[2,7]+𝑡[−10,3] can be written as:
x = a - 5t
y = 2x - 2a = 2(a - 5t) - 2a = -10t
where 'a' is the constant x-intercept value.
To find the parametric equations of a line perpendicular to the equation 4x + 8y + 7 = 0, we can start by finding the slope of the given line. The slope-intercept form of a line is given by y = mx + b, where m is the slope.
The equation 4x + 8y + 7 = 0 can be rearranged to the slope-intercept form: y = (-1/2)x - 7/8. So, the slope of this line is -1/2.
Since we want a line perpendicular to this, the slope of the new line will be the negative reciprocal of -1/2, which is 2.
Now, we need a point on the new line. The given line has a point [𝑥,𝑦] = [2,7] when t = 0. So, this point [2,7] will also lie on the new line.
We now have the slope (2) and a point (2,7) on the line. We can use the point-slope form of the line to find the parametric equations.
The point-slope form of a line is given by (y - y1) = m(x - x1), where (x1, y1) is the given point and m is the slope.
Plugging in the values, we have (y - 7) = 2(x - 2). Simplifying, we get y - 7 = 2x - 4.
To express this line in parametric form, we can let x = t and solve for y:
y = 2t + 3
Therefore, the parametric equations of the line perpendicular to 4x + 8y + 7 = 0 with the same x-intercept as [𝑥,𝑦]=[2,7]+𝑡[−10,3] are x = t and y = 2t + 3.