Question
A square has vertices (0, 0), (3, 0), and (0, 3). Find the fourth Vertex.
The correct answer is: So we get that D = (3,3)
SOLUTION:
HINT: Use the properties of a square.
Complete step by step solution:
Here we have ABCD as the square.
Now, AB = 3
⇒ CD = 3 (Since ABCD is a square)
And, AC = 3
⇒BD=3 (Since ABCD is a square)
So we get that D = (3,3)
Related Questions to study
In the equation above, p and t are constants. Which of the following could be the value of p ?
In the equation above, p and t are constants. Which of the following could be the value of p ?
What is the sum of the complex numbers 2 +3i and 4 +8i , where ?
A) 17
B) 17i
C) 6 +11i
D) 8 +24i
What is the sum of the complex numbers 2 +3i and 4 +8i , where ?
A) 17
B) 17i
C) 6 +11i
D) 8 +24i
A square has vertices (0, 0), (1, 0), and (0, 1). Find the fourth vertex.
A square has vertices (0, 0), (1, 0), and (0, 1). Find the fourth vertex.
Find AB if MN is the midsegment of △ ABC.
Find AB if MN is the midsegment of △ ABC.
Solve each inequality and graph the solution:
-0.5x>-10.
Note:
When we have an inequality with the symbols < or > , we use dotted lines to graph them. This is because we do not include the end point of the inequality.
Solve each inequality and graph the solution:
-0.5x>-10.
Note:
When we have an inequality with the symbols < or > , we use dotted lines to graph them. This is because we do not include the end point of the inequality.
Find out the simple interest on Rs. 7500 for 2 ½ years at 15 % per annum. Also find the amount
Find out the simple interest on Rs. 7500 for 2 ½ years at 15 % per annum. Also find the amount
Find x if MN is the midsegment of △ ABC.
Find x if MN is the midsegment of △ ABC.
QUESTION 12: A square has vertices (0, 0), (m, 0), and (0, m). Find the fourth
vertex.
QUESTION 12: A square has vertices (0, 0), (m, 0), and (0, m). Find the fourth
vertex.
Solve each inequality and graph the solution :
x+9>15.
Note:
When we have an inequality with the symbols , we use dotted lines to graph them. This is because we do not include the end point of the inequality.
Solve each inequality and graph the solution :
x+9>15.
Note:
When we have an inequality with the symbols , we use dotted lines to graph them. This is because we do not include the end point of the inequality.
How is the solution to -4(2x-3)>36 similar to and different from the solution-4(2x-3)=36 .
Note:
The number of solutions for an inequality can be infinity. Whereas, a linear equation has only a maximum of one solution.
How is the solution to -4(2x-3)>36 similar to and different from the solution-4(2x-3)=36 .
Note:
The number of solutions for an inequality can be infinity. Whereas, a linear equation has only a maximum of one solution.