Using ratios in the coordinate plane practice
Sure! Here are a few practice problems using ratios in the coordinate plane:
1) The point (-3, 2) is on a line. Find a point on the same line that is twice as far from the origin (0, 0) as (-3, 2).
Solution:
The distance between the origin and (-3, 2) can be found using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
d = sqrt((0 - (-3))^2 + (0 - 2)^2)
d = sqrt(9 + 4)
d = sqrt(13)
To find a point that is twice as far from the origin, we can multiply the coordinates of (-3, 2) by 2:
(2*(-3), 2*2)
(-6, 4)
So, the point (-6, 4) is twice as far from the origin as (-3, 2), and it lies on the same line.
2) A point (x, 4) is on the line with the equation y = 2x + 1. Find the value of x.
Solution:
Since the point lies on the line, we can substitute the coordinates of the point into the equation of the line:
4 = 2x + 1
Now, we can solve for x:
2x + 1 = 4
2x = 3
x = 3/2
x = 1.5
So, the value of x is 1.5.
3) The point (1, y) is on the line with the equation 3x = 2y - 4. Find the value of y.
Solution:
Similar to the previous problem, we can substitute the coordinates of the point into the equation of the line:
3(1) = 2y - 4
Simplifying the equation, we have:
3 = 2y - 4
Now, we can solve for y:
2y - 4 = 3
2y = 7
y = 7/2
y = 3.5
So, the value of y is 3.5.
I hope these practice problems help! Let me know if you have any more questions.