A) y= 2/5x+9 write an equation that is parallel to the given line and passes through the point (0, 2)
B) Write an equation that is perpendicular to the same line but passes through point (0, 3)
A) y= 2/5x+9 write an equation that is parallel to the given line and passes through the point (0, 2)
B) Write an equation that is perpendicular to the same line but passes through point (0, 3)
A) So the line you are given is written in slope intercept form y = mx + b meaning that the slope of the line is m = 2/5 and the intercept b = 9 (so the line hits the y axis at 0,9). A line that is parallel has to have the same slope so we know our m is the same. So we have y = (2/5)x + b. We need b. It passes (0, 2) since the x coordinate is 0 you know the y intercept is 2 but you can verify this using the point slope formula:
y - y1 = m(x - x1) plug in values
y - 2 = (2/5)(x - 0) now simplify
y - 2 = (2.5)x last solve
y = (2/5)x + 2 see how when the x coordinate is 0 you can automatically use the point slope intercept formula? Neato!
B) Now we want a line perpendicular to y = (2/5)x + 2 and that passes through 0,3. What are the properties of a line perpendicular to a line of the form y = mx + b and passes through (0, 3)? We know that the slope of this new line will be the reciprocal and have the opposite sign of m. So the new slope we will call it m_{new }= -(1/m) so plug in for m and you get m_{new} = -(1/[2/5]) if you simplify you get that the m_{new} = -(5/2). So we have half of our equation for the perpendicular: y = -(5/2)x + b. If you look at (0, 3) you see that the y intercept is 3 so b = 3. So:
y = -(5/2)x + 3 is a line perpendicular to y = (2/5)x + 2. This line is also perpendicular to y = (2/5)x + 9 because it has the same slope.
You are given the equation for the line y = (2/5)x + 9 , which is in slope-intercept form.
Note: Slope-intercept form means that an equation for a line is in the form y = mx + b, where m is the slope (rise/run) of the line and b is the y-intercept (point at which x=0).
Thus, for the original equation of the line y = (2/5)x + 9: m = 2/5 , b = 9
a.) Given the equation for the line above, find an equation that is parallel to this line and passes thru the point (0, 2).
Here, you are given two pieces of information:
1) The line you are looking to find is parallel to the line y = (2/5)x + 9. Lines that are parallel to one another have the same slope. Therefore, the slope of the line that is parallel to this one is m= 2/5.
2) The line in question passes thru the point (0, 2). Since x=0 at this point, we know that this is the y-intercept. Therefore, the y-intercept of this line is b= 2.
Using slope-intercept form (y = mx + b) we can find the equation of the line parallel to the original line (y = (2/5) + 9) and which passes thru (0, 2) by substituting 2/5 for m and substituting 2 for b.
Thus, the equation for the line that passes thru the point (0, 2) and that is parallel to the original line is:
y = (2/5)x + 2
b.) Given the equation for the line y = (2/5)x + 9 , find the equation that is perpendicular to this line but passes thru the point (0, 3).
Here, you are given the following information:
1.) The line you are looking to find is perpendicular to the line y = (2/5)x + 9. A line that is perpendicular to another line has a slope that is the negative reciprocal of the slope of the original line. Therefore, the slope of the line that is perpendicular to the original line is m= -(5/2).
2.) The perpendicular line whose equation we are looking to find passes thru the point (0, 3). Since x=0 at this point, we know that this is the y-intercept. Therefore, the y-intercept of the perpendicular line is b= 3.
Using y = mx + b , we can find the equation of the line that passes thru the point (0, 3) and that is perpendicular to the original line by substituting -(5/2) for m and substituting 3 for b.
Thus, the equation of the line that passes thru the point (0, 3) and that is perpendicular to the original line is:
y = -(5/2)x + 3
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