If A ( 7, 10), B( -2, k) and C(3, -4) are vertices of a right angled triangle at B, then k=
Answers
Answer:
1
Step-by-step explanation:
Given that, In right triangle, ABC, right angled at B,
Coordinates of A is (7, 10)
Coordinates B is (- 2, k)
Coordinates of C is (3, - 4)
So, using Pythagoras Theorem, we have
\begin{gathered}\rm \: {AB}^{2} + {BC}^{2} = {AC}^{2} \\ \end{gathered}
AB
2
+BC
2
=AC
2
So, using Distance formula, we get
\begin{gathered}\rm \: {(7 + 2)}^{2} + {(10 - k)}^{2} + {( - 2 - 3)}^{2} + {(k + 4)}^{2} = {(7 - 3)}^{2} + {(10 + 4)}^{2} \\ \end{gathered}
(7+2)
2
+(10−k)
2
+(−2−3)
2
+(k+4)
2
=(7−3)
2
+(10+4)
2
\begin{gathered}\rm \: 81 + {(10 - k)}^{2} + 25 + {(k + 4)}^{2} = 16 + 196 \\ \end{gathered}
81+(10−k)
2
+25+(k+4)
2
=16+196
\begin{gathered}\rm \: 106 + {(10 - k)}^{2} + {(k + 4)}^{2} = 212 \\ \end{gathered}
106+(10−k)
2
+(k+4)
2
=212
\begin{gathered}\rm \: {(10 - k)}^{2} + {(k + 4)}^{2} = 106 \\ \end{gathered}
(10−k)
2
+(k+4)
2
=106
\begin{gathered}\rm \: 100 + {k}^{2} - 20k + {k}^{2} + 16 + 8k = 106 \\ \end{gathered}
100+k
2
−20k+k
2
+16+8k=106
\begin{gathered}\rm \: 2{k}^{2} + 116 - 12k = 106 \\ \end{gathered}
2k
2
+116−12k=106
\begin{gathered}\rm \: 2{k}^{2} + 116 - 12k - 106 = 0\\ \end{gathered}
2k
2
+116−12k−106=0
\begin{gathered}\rm \: 2{k}^{2} - 12k + 10 = 0\\ \end{gathered}
2k
2
−12k+10=0
\begin{gathered}\rm \: 2({k}^{2} - 6k + 5) = 0\\ \end{gathered}
2(k
2
−6k+5)=0
\begin{gathered}\rm \:{k}^{2} - 6k + 5 = 0\\ \end{gathered}
k
2
−6k+5=0
\begin{gathered}\rm \:{k}^{2} - 5k - k + 5 = 0\\ \end{gathered}
k
2
−5k−k+5=0
\begin{gathered}\rm \: k(k - 5) - 1(k - 5) = 0 \\ \end{gathered}
k(k−5)−1(k−5)=0
\begin{gathered}\rm \: (k - 5)(k - 1) = 0 \\ \end{gathered}
(k−5)(k−1)=0
\begin{gathered}\rm\implies \:k \: = \: 5 \: \: or \: \: k \: = \: 1 \\ \end{gathered}
⟹k=5ork=1