I;m not sure if i am doing this right...I keep getting a negative intersection point, which doesn't seem possible.
Please help!
a) find the equatoin of line 1 which passes through (1,3) and (9,7)
I get y=3+1/2(x+1)or y=1/2x+2.5
b) the line 2 is perpedicular to line 1 and passes through 5,0. find the equation
for this one i get y=-2x+10
c) find where line 1 and line 2 intersect each other. use this to compute the distance between the point (5,0) and line 1
I get (-5,0) for the intersection, which doesnt make sense to me?
Thanks
line 1:
(y-3)/(x-1) = (7-3)/(9-1)
y-3 = 1/2 (x-1)
y = 1/2 x + 5/2
you are correct
line 2:
(y-0) = -2 (x-5)
y = -2x + 10
you are correct
1/2 x + 5/2 = -2x + 10
5/2 x = 15/2
x = 3 so y=4
The lines intersect at (3,4)
d^2 = (3-5)^2 + (4-0)^2
d^2 = 4+16 = 20
d = 2√5
To solve the problem, let's go through each step and check your calculations:
a) Finding the equation of line 1:
Using the two points (1,3) and (9,7), we can find the slope of the line using the formula:
m = (y2 - y1) / (x2 - x1)
Substituting the values, we get:
m = (7 - 3) / (9 - 1) = 4 / 8 = 1/2
Now, let's use the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept. We substitute the slope (1/2) and one of the given points (1,3) to find b:
3 = (1/2)(1) + b
3 = 1/2 + b
b = 3 - 1/2
b = 5/2 or 2.5
So, the equation of line 1 is y = (1/2)x + 2.5.
b) Finding the equation of line 2:
To find the equation of a line perpendicular to line 1, we need the negative reciprocal of the slope of line 1. The negative reciprocal of 1/2 is -2/1 or -2.
Using the point-slope form of a line, we substitute the slope (-2) and the given point (5,0) into the equation:
y - y1 = m(x - x1)
y - 0 = -2(x - 5)
y = -2x + 10
So, the equation of line 2 is y = -2x + 10.
c) Finding the intersection point and distance:
To find the intersection point between two lines, we need to set their equations equal to each other and solve for x and y. Let's set the equations of line 1 and line 2 equal to each other:
(1/2)x + 2.5 = -2x + 10
Multiply both sides by 2 to eliminate fractions:
x + 5 = -4x + 20
5x + 4x = 20 - 5
9x = 15
x = 15/9 or 5/3
Now, substitute the value of x back into either equation to find y. Let's use line 1:
y = (1/2)(5/3) + 2.5
y = 5/6 + 15/6
y = 20/6 or 10/3
So, the intersection point is (5/3, 10/3), which simplifies to (1.67, 3.33) approximately.
To compute the distance between the point (5,0) and line 1, we can use the formula for the distance between a point and a line. The formula is:
d = |Ax + By + C| / sqrt(A^2 + B^2)
Where A, B, and C are coefficients of the equation of the line, and (x, y) are the coordinates of the point. In line 1's equation (y = (1/2)x + 2.5), A = -1/2, B = -1, and C = -2.5.
Plugging in the values, we have:
d = |(-1/2)(5) + (-1)(0) + (-2.5)| / sqrt((-1/2)^2 + (-1)^2)
d = |-5/2 + 0 - 5/2| / sqrt(1/4 + 1) = |-10/2| / sqrt(1/4 + 4/4)
d = | -5 | / sqrt(5/4) = 5 / (sqrt(5)/2)
d = 5 * 2 / sqrt(5) = 10 / sqrt(5)
So, the distance between the point (5,0) and line 1 is 10 / sqrt(5), approximately 4.47.
Therefore, the correct intersection point is (1.67, 3.33), and the distance is 10 / sqrt(5), not (-5, 0).