The equation of the line R1 is 2x +y-8=0. The line R2 is perpendicular to R1.
a. Calculate the gradient of R2.
b. The point of intersection of R1 and R2 is (4,k).
Find.
1. the value of k
2. the equation of R2
2x +y-8=0
is
y = (-2) x + 8
slope of R1 = m = -2
yaxis intercept at (0,8)
slope of R2 = -1/m = -1/-2 = +1/2
so
for R2, y = (1/2) x + b
hit at (4,k)
k = (1/2)4 + b
or k = 2 + b
k = -2 (4) + 8 = 0
so hit at (4,0)
0 = (1/2)4 + b
b = -2
so R2 is
y = (1/2) x - 2
check=========================
for R1 y = -2x+8
for R2 y = (1/2) x -2
--------------------------subtract
0 = -2.5 x + 10
so x = 4
y = -2(4) + 8 = 0 yes
a. Well, if R2 is perpendicular to R1, then it's like the two lines are doing a little tango, spinning around each other. And in the world of lines, when two lines are perpendicular, it means their gradients are negative reciprocals of each other. So, to find the gradient of R2, we flip the gradient of R1 and change the sign. In this case, the gradient of R1 is -2. So the gradient of R2 would be the negative reciprocal of -2, which is 1/2.
b. Now, let's find the point (4, k) where R1 and R2 intersect. To do that, we'll substitute x=4 into the equation of R1.
Oh, and here's a little joke for you: Why don't scientists trust atoms? Because they make up everything!
1. Plugging in x = 4 into 2x + y - 8 = 0, we get:
2(4) + y - 8 = 0
8 + y - 8 = 0
y = 0
So, we find that k = 0.
2. Now that we know the point of intersection is (4, 0), we can use the point-slope form of a line to find the equation of R2. The point-slope form is y - y₁ = m(x - x₁), where (x₁, y₁) is the point of intersection, m is the gradient, and (x, y) is any other point on the line. In this case, we have:
y - 0 = 1/2(x - 4)
Simplifying, we get:
y = (1/2)x - 2
And voila! The equation of R2 is y = (1/2)x - 2.
Hope that clears things up! Let me know if you have any other questions.
a. To find the gradient of R2, we need to determine the slope of R1 first.
The equation of R1 is 2x + y - 8 = 0. We can rewrite it in the form y = mx + c, where m represents the gradient:
y = -2x + 8
Comparing this to the standard form y = mx + c, we see that the gradient of R1 is -2.
Since R2 is perpendicular to R1, the gradient of R2 will be the negative reciprocal of the gradient of R1.
The negative reciprocal of -2 is 1/2, so the gradient of R2 is 1/2.
b. The point of intersection of R1 and R2 is given as (4, k).
1. To find the value of k, we substitute the x-coordinate (4) into the equation of R1 and solve for y:
2(4) + y - 8 = 0
8 + y - 8 = 0
y = 0
Thus, the value of k is 0.
2. Having found the value of k, we can substitute it back into the equation of R2 and solve for y:
The equation of R2 is y = mx + c, where m represents the gradient (1/2) and c represents the y-intercept.
Using the point (4, 0), we have:
0 = (1/2)(4) + c
0 = 2 + c
c = -2
So, the equation of R2 is y = (1/2)x - 2.
To calculate the gradient of line R2, we need to find the negative reciprocal of the gradient of line R1.
a. Gradient of R1: The equation of line R1 is in the form Ax + By + C = 0, where A, B, and C represent the coefficients of x, y, and the constant term, respectively. In this case, A = 2, B = 1, and C = -8. The gradient of R1 can be found by rearranging the equation into y = mx + c form, where m is the gradient.
Given: 2x + y - 8 = 0
Rearranged form: y = -2x + 8
Comparing with y = mx + c, we can see that the gradient of R1 (m1) is -2.
To find the gradient of R2, we take the negative reciprocal of m1:
Gradient of R2 (m2) = -1 / m1
= -1 / (-2)
= 1/2
Therefore, the gradient of R2 is 1/2.
b. Point of intersection (4, k): The point of intersection lies on both R1 and R2. To find k, we need to substitute the x-coordinate (4) into the equation of R1 and solve for y.
Given: R1: 2x + y - 8 = 0
Substituting x = 4 into R1: 2(4) + y - 8 = 0
8 + y - 8 = 0
y = 0
So, at the point (4, k), k = 0.
To find the equation of R2, we use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the gradient.
Given: Point of intersection (4, 0) and gradient of R2 = 1/2
Using the point-slope form for R2, we can substitute the values to find the equation:
y - 0 = (1/2)(x - 4)
y = (1/2)(x - 4)
y = 1/2x - 2
Therefore, the equation of line R2 is y = 1/2x - 2.