find an equation for the line with the given properties. express the equation in slope-intercept form.
containing the pointsP=(-1,1)& Q= (1,-2)
What is the equation of the line?
y=
Find the the equation of a line that is perpendicular to the line y=1/2x+4 and contains the point (-3,0)
please show work
To find the equation of a line in slope-intercept form (y = mx + b), we need to find the slope (m) and the y-intercept (b).
For the first question, to find the slope (m), we can use the formula: m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are the coordinates of two points on the line. Let's use the points P = (-1, 1) and Q = (1, -2):
m = (-2 - 1) / (1 - (-1))
= -3 / 2
Now, we need to find the y-intercept (b). We can use the formula: b = y - mx, where (x, y) is any point on the line. Let's use point P = (-1, 1):
b = 1 - (-3/2) * (-1)
= 1 + 3/2
= 5/2
Therefore, the equation of the line in slope-intercept form is y = (-3/2)x + (5/2).
For the second question, we need to find the slope of the given line first. The equation of the given line is already in slope-intercept form, y = (1/2)x + 4. The slope is the coefficient of x, which is 1/2.
A line that is perpendicular to the given line will have a slope that is the negative reciprocal of the given slope. The negative reciprocal of 1/2 is -2.
Now, we have the slope (-2) and a point (-3, 0). We can use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope:
y - 0 = -2(x - (-3))
y = -2(x + 3)
y = -2x - 6
Therefore, the equation of the line that is perpendicular to y = (1/2)x + 4 and contains the point (-3, 0) is y = -2x - 6.