1a.A line L1 passes through the points (3,3) and (5,7).Find the equation of L1 in the form Y=Mx+c where m and c are constants.
b.Another line L2 is perpendicular to L1 and passes through (-2,3).Find :
1.The equation of L2
2.The X-intercept of L2
L1 has slope 2, so
L1: y = 2x-3
L2 has slope -1/2, so
L2: y = -1/2 x + 2
Now just find where L2 has y=0.
a. To find the equation of line L1, we can use the formula for the slope (m) between two points (x1, y1) and (x2, y2):
m = (y2 - y1) / (x2 - x1)
Using the given points (3, 3) and (5, 7), we have:
m = (7 - 3) / (5 - 3) = 4 / 2 = 2
Now, we can plug in the slope value (m = 2) and one of the points (3, 3) into the equation of a line (y = mx + c) to solve for the y-intercept (c):
3 = 2 * 3 + c
3 = 6 + c
c = 3 - 6
c = -3
So, the equation of line L1 is:
y = 2x - 3
b. To find the equation of line L2, we know that it is perpendicular to line L1, which means the slopes of the two lines are negative reciprocals of each other.
The slope of L1 is 2, so the slope of L2 would be -1/2.
Now, to find the equation of L2 using the point (-2, 3), we can again use the equation of a line (y = mx + c):
3 = (-1/2) * (-2) + c
3 = 1 + c
c = 3 - 1
c = 2
Therefore, the equation of line L2 is:
y = (-1/2)x + 2
To find the x-intercept of L2, we set y = 0 and solve for x:
0 = (-1/2)x + 2
(-1/2)x = -2
x = (-2) / (-1/2)
x = 4
So, the x-intercept of L2 is 4.