If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t.
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Answers
Given :
Given three points :
- A = (5,2)
- B = (2,-2)
- C = (-2,t)
These points form a right-angled triangle with <B=90°.
To find :
We've to find the value of t.
Solution :
The main concept that lies here is the distance formula.
First, we should find the distance between these points and then find the value of t using the pythagoras theorem.
- Distance between A and B :
→ √{(2-5)²+(-2-2)²}
→ √{(-3)²+(-4)²}
→ √{9+16}
→ √25
→ AB =5units
- Distance between B and C :
→ √{(-2-2)²+(t+2)²}
→ √{(-4)²+t²+4+4t}
→ √{16+4+4t+t²}
→ BC=√{20+4t+t²}units
- Distance between C and A :
→ √{(5+2)²+(2-t)²}
→ √{49+5+t²-4t}
→ CA=√{53+t²-4t}units
Now, from pythagoras theorem :
→ (AB)²+(BC)²=(AC)²
→ (5)²+(√{20+4t+t²})²=(√{53+t²-4t})²
→ 25+20+4t+t²=53+t²-4t
→ 45+4t+t²-53-t²+4t=0
→ 45-53+4t+4t+t²-t2=0
→ -8+8t=0
→ t=1
Verification :
Now, we'll verify whether t=1 is the correct value or not by using pythagoras theorem and substituting t=1 in the distances :
- BC = √{20+4t+t²}units
→ √{20+4(1)+(1)²}
→ √{20+4+1}
→ √{25}
→ 5units
- CA = √{53+t²-4t}units
→ √{53+(1)²-4(1)}
→ √{53+1-4}
→ √50units
Using pythagoras theorem :
→ (AB)²+(BC)²+(CA)²
→ (5)²+(5)²=(√50)²
→ 25+25=50
→ 50=50
→ LHS = RHS
→Henceforth, t=1.
_____________________
Small correction : We don't use the word diagram in math! We call them figures. And in science, we say them diagrams.
Hey! Hope uh get me right :D
from Pythagoras theorem :
(AB)²+(BC)²=(AC)²
(5)²+(√{20+4t+t²})²=(√{53+t²-4t})²
25+20+4t+t²=53+t²-4t
45+4t+t²-53-t²+4t=0
45-53+4t+4t+t²-t2=0
-8+8t=0
t=1
see the attachment for diagram..
