Question

Find the value
of a so that (3 ; a) and
(4, 1) is √10



please help me ​

Answer 1

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 $\sf (x_1 , y_1) \:\:\&\:\:(x_2 , y_2)$ is

$\sf \large{ \sqrt{ {(x _{2} - x _{1} })^{2} + (y _{2} - y _{1}) ^{2} }}$

Let,

• $\sf x _{1}$ = 3

• $\sf x_{2}$ = 4

• $\sf y _{1}$ = a

• $\sf y_{2}$ = 1

Hence,

$\sf \implies \: \sqrt{( {4 - 3)}^{2} + (1 - a) ^{2} } = \sqrt{10} \\ \\ \sf \: on \: squaring \: both \: sides \: we \: get \\ \\ \sf \implies \: {1}^{2} + {(1 - a)}^{2} = 10 \\ \\ \sf \implies \: {(1 - a)}^{2} = 10 - 1 \\ \\ \sf \implies \: {(1 - a)} = \sqrt{9} \\ \\ \sf \implies \: {1 - a} = \pm \: 3 \\ \\ \sf \implies \: {1 - a} = 3 \\ \\ \sf \implies \: {1 - 3} = a \\ \\ \sf \implies \large{ a = - 2} \\ \\ \: \: \: \: \: \: \: \: \: \sf \: (or) \\ \\ \sf \implies \: 1 - a = - 3 \\ \\ \sf \implies \:1 + 3 = a \\ \\ \sf \implies \large{a = 4}$

Hence, the value of a is - 2 or 4.

Answer 2

Answer:

$\mapsto$ Given

$\leadsto$ Distance between (3,a) and (4,1) is √10 units.

$\mapsto$ To Find

$\leadsto$ What is the value of a.

$\mapsto$ Solution

✏ Let the point be A (3,a) and B (4,1)

✏ The distance AB = √10

➡ So AB²,

$\implies$ 10 = (3-4)² + (a-1)²

$\implies$ 10 = 1 + a² - 2a + 1

$\implies$ 10 = a² - 2a + 2

$\implies$ a² - 2a + 2 - 10 = 0

$\implies$ a² - 2a - 8 = 0

$\implies$ a² - (4-2)a - 8 = 0

$\implies$ a² - 4a + 2a - 8 = 0

$\implies$ a(a-4) + 2(a-4) = 0

$\implies$ (a-4) (a+2) = 0

$\implies$ (a-4) = 0 , (a-2) = 0

$\implies$ a = 4 , a = -2

Hence, the value of a = 4 or -2

$\mapsto$ 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:

HOPE IT HELP YOU