I have math HW and these are the questions i wasn't aware of how to answer... please help.
a)Prove that 4x-y-1=0 is parallel to y=4x-5
b)Show that the points A(3,1), B(5,2) and C(11,5) are collinear
c)Show that the lines y=2x-4 and x+2y-10=0 are perpendicular to one other.
d) Prove that triangle ABC is isoscles given A(3,1), B(-3,7) and C(-1,3)
Thank you!
a) change the first into the form y = mx + b
4x - y - 1 = 0
-y = -4x + 1
y = 4x - 1
the value of m is the same for both equations, so they are parallel
b) find the slope between 2 different line segments.
If the slope is the same, then they are collinear.
c) slope of first is 2
After changing the second to the form y = mx + b, the value of m should be -1/2 to be perpendicular.
d) find the length of the 3 sides using your distance formula
Two of the lines should be the same
a) To prove that the equation 4x - y - 1 = 0 is parallel to the equation y = 4x - 5, we need to show that their slopes are equal. The slope of a linear equation in the form y = mx + b, where m is the slope, is given by the coefficient of x.
First, let's rearrange the equation 4x - y - 1 = 0 to put it in slope-intercept form:
- y = -4x + 1
Divide both sides by -1:
y = 4x - 1
Now we can compare the equations:
The equation y = 4x - 5 has a slope of 4.
The equation y = 4x - 1 also has a slope of 4.
Since both equations have the same slope, they are parallel.
b) To show that the points A(3,1), B(5,2), and C(11,5) are collinear, we can calculate the slopes of the lines formed by any two pairs of these points. If the slopes are equal, the points are collinear.
Let's find the slope of the line passing through points A and B:
slope_AB = (y2 - y1) / (x2 - x1)
= (2 - 1) / (5 - 3)
= 1 / 2
Now, let's find the slope of the line passing through points B and C:
slope_BC = (y3 - y2) / (x3 - x2)
= (5 - 2) / (11 - 5)
= 3 / 6
= 1 / 2
Since the slopes of both lines are equal (slope_AB = slope_BC = 1/2), the points A, B, and C are collinear.
c) To show that the lines y = 2x - 4 and x + 2y - 10 = 0 are perpendicular, we need to check if the product of their slopes is -1.
First, let's put the equation x + 2y - 10 = 0 in slope-intercept form:
2y = -x + 10
y = -(1/2)x + 5
Now we can compare the equations:
The equation y = 2x - 4 has a slope of 2.
The equation y = -(1/2)x + 5 has a slope of -(1/2).
The product of the slopes is 2 * -(1/2) = -1.
Since the product of the slopes is -1, the lines y = 2x - 4 and x + 2y - 10 = 0 are perpendicular.
d) To prove that triangle ABC, with vertices A(3,1), B(-3,7), and C(-1,3), is isosceles, we need to show that at least two of the sides have equal lengths.
Let's calculate the distances between each pair of points:
Distance between points A and B:
d_ab = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(-3 - 3)^2 + (7 - 1)^2]
= √[(-6)^2 + (6)^2]
= √[36 + 36]
= √72
= 6√2
Distance between points B and C:
d_bc = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(-1 - (-3))^2 + (3 - 7)^2]
= √[2^2 + (-4)^2]
= √[4 + 16]
= √20
= 2√5
Distance between points C and A:
d_ca = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(3 - (-1))^2 + (1 - 3)^2]
= √[4^2 + (-2)^2]
= √[16 + 4]
= √20
= 2√5
Since d_bc = d_ca, triangle ABC is isosceles because two of its sides have equal lengths.