Find the value
of a so that (3 ; a) and
(4, 1) is √10
please help me
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Find the value
of a so that (3 ; a) and
(4, 1) is √10
please help me
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Question:-
Find the value of a so that distance between (3 ; a) and (4, 1) is √10 units.
Answer:-
Given:
Distance between (3 , a) & (4 , 1) is √10 units.
We know that,
Distance between two points with coordinates [tex] \sf (x_1 , y_1) \:\:\&\:\:(x_2 , y_2) [/tex] is
[tex] \sf \large{ \sqrt{ {(x _{2} - x _{1} })^{2} + (y _{2} - y _{1}) ^{2} }}[/tex]
Let,
Hence,
[tex] \sf \implies \: \sqrt{( {4 - 3)}^{2} + (1 - a) ^{2} } = \sqrt{10} \\ \\ \sf \: on \: squaring \: both \: sides \: we \: get \\ \\ \sf \implies \: {1}^{2} + {(1 - a)}^{2} = 10 \\ \\ \sf \implies \: {(1 - a)}^{2} = 10 - 1 \\ \\ \sf \implies \: {(1 - a)} = \sqrt{9} \\ \\ \sf \implies \: {1 - a} = \pm \: 3 \\ \\ \sf \implies \: {1 - a} = 3 \\ \\ \sf \implies \: {1 - 3} = a \\ \\ \sf \implies \large{ a = - 2} \\ \\ \: \: \: \: \: \: \: \: \: \sf \: (or) \\ \\ \sf \implies \: 1 - a = - 3 \\ \\ \sf \implies \:1 + 3 = a \\ \\ \sf \implies \large{a = 4}[/tex]
Hence, the value of a is - 2 or 4.
Answer:
[tex]\mapsto[/tex] Given
[tex]\leadsto[/tex] Distance between (3,a) and (4,1) is √10 units.
[tex]\mapsto[/tex] To Find
[tex]\leadsto[/tex] What is the value of a.
[tex]\mapsto[/tex] Solution
✏ Let the point be A (3,a) and B (4,1)
✏ The distance AB = √10
➡ So AB²,
[tex]\implies[/tex] 10 = (3-4)² + (a-1)²
[tex]\implies[/tex] 10 = 1 + a² - 2a + 1
[tex]\implies[/tex] 10 = a² - 2a + 2
[tex]\implies[/tex] a² - 2a + 2 - 10 = 0
[tex]\implies[/tex] a² - 2a - 8 = 0
[tex]\implies[/tex] a² - (4-2)a - 8 = 0
[tex]\implies[/tex] a² - 4a + 2a - 8 = 0
[tex]\implies[/tex] a(a-4) + 2(a-4) = 0
[tex]\implies[/tex] (a-4) (a+2) = 0
[tex]\implies[/tex] (a-4) = 0 , (a-2) = 0
[tex]\implies[/tex] a = 4 , a = -2
Hence, the value of a = 4 or -2
[tex]\mapsto[/tex] Verification
▶ AB² = (3-4)² + (1-4)² = 1+9 = 10
▶ AB² = (3-4)² + (1+2)² = 1+9 = 10
Step-by-step explanation:
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