1. Take an arbitrary point that lies on the first line y=3x+10. Let x=0, then y=10 and point has coordinates (0,10).
2. Use formula
to find the distance from point
to the line Ax+By+C=0.
The second line has equation y=3x-20, that is 3x-y-20=0. By the previous formula the distance from the point (0,10) to the line 3x-y-20=0 is:
.
3. Since lines y=3x+10 and y=3x-20 are parallel, then the distance between these lines are the same as the distance from an arbitrary point from the first line to the second line.
Answer:
.
Answer:
D
Step-by-step explanation:
Question is Incomplete, Complete question is given below.
Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.
Answer:
∆ABC is right angled triangle with right angle at B.
Step-by-step explanation:
Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.
We need to prove that triangle is the right angled triangle.
Let the triangle be denoted by Δ ABC with side as;
AB = (a - 1) cm
BC = (2√ a) cm
CA = (a + 1) cm
Hence,
Now We know that

So;


Now;

Also;

Now We know that




[By Pythagoras theorem]

Hence, 
Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.
This proves that ∆ABC is right angled triangle with right angle at B.
Answer:
64 is the sum of the numbers
58 + 6 = 64
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11.2 * 1.4 = 15.68
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