A quadirateral has vertices at (-2, 1), (0, 4), (5, 2) and (1, -1). Do the diagonals have the same midpoint? Justify your answer
1 answer:
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Answer:
Step-by-step explanation:
Given
Required [Missing from the question]
G(T(x))
We have:
This implies that:
Substitute:
Open bracket
Ok so linear equations come in the form
y = mx + b
B is the y-intercept. In this equation the y intercept is 40. A y- intercept is where the graph crosses the y axis so at the point (0,40) the graph crosses the y axis.
M is the slope which is rise over run so if the slope was 10 (which it is) 10 is equivalent to 10/1 so you move up 10 units for every 1 unit you move across.
So to graph this equation, you would draw your first point at (0,40). For your next point, you would move right one unit, and up 10 units. Draw a point there which would be (1, 50). Hopefully you understand. For going left from the y intercept point, you would move left 1 and down 10.
Actual graph above.
Hope this helps C:
y = mx + b
B is the y-intercept. In this equation the y intercept is 40. A y- intercept is where the graph crosses the y axis so at the point (0,40) the graph crosses the y axis.
M is the slope which is rise over run so if the slope was 10 (which it is) 10 is equivalent to 10/1 so you move up 10 units for every 1 unit you move across.
So to graph this equation, you would draw your first point at (0,40). For your next point, you would move right one unit, and up 10 units. Draw a point there which would be (1, 50). Hopefully you understand. For going left from the y intercept point, you would move left 1 and down 10.
Actual graph above.
Hope this helps C: