Find the point of intersection of the lines x-3 = 0 and y-5 = 0.

Mathematics

1 answer:

Vikki [24]3 years ago

3 0

Answer:

x = 3 and y = 5

Step-by-step explanation:

Given equations are :

x-3 = 0 ....(1)

y-5 = 0 ....(2)

We need to find the point of intersection of the lines.

Add 3 both sides in equation (1).

x-3+3=3

x = 3

Now, add 5 to both sides of the equation.

y-5+5=0+5

y = 5

Hence, the required point of intersection of the lines is x = 3 and y = 5.

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ABC is a right-angled triangle.

CaHeK987 [17]

Answer:

A= 10 units squared

Step-by-step explanation:

Because we are fully given points A and B, as well as the knowledge that ABC is a right triangle, we can assume that point C is the last needed point to form the right angle of this triangle. Meaning, point c has to have <em>one</em> identical coordinate out of (x,y) that is the same from <em>each</em> of our pre-existing coordinates in order to successfully form a right angle.

Due to the fact that we are told that Point C lies on y=2, we know that it already has an identical 'y' coordinate as Point A.

We can now decode that the x coordinate of Point C must be identical to the x coordinate of Point B, In this case, -2

We can conclude that point C lies on (-2,2)

If we look at the length of the sides of the triangle, ignoring the hypotenuse, we can see that one is 4 units long and the other is 5 units long.

To find the area of a square the formula is A= l x w x 0.5

Since 4x 5 = 20 and half of 20 is ten, we know that the final answer will be 10 units^2

to help you visualize this problem I've added an attachment :)