Question

Let A(-2,y) and B(3,2) are two points on the line segment. If the distance between these two points is 5 units, find y.

Answer 1

S O L U T I O N :

$\underline{ \bf{Given\::}}$

We have two points A(-2, y) & B(3,2) on the line segment. If the distance between these two points is 5 unit.

$\underline{ \bf{Explanation\::}}$

As we know that formula of the distance;

$\boxed{ \bf{D= \sqrt{( x_2- x_1)^{2} + (y_2 - y_1)^{2} } }}$

A/q

• x1 = -2
• x2 = 3
• y1 = y
• y2 = 2

↠ 5 = √[(3 - (-2)]² + (2- y)²

↠ 5 = √(3 + 2)² + (2)² + (y)² - 2 × 2 × y

[using (a-b)²]

↠ 5 = √(5)² + 4 + y² - 4y

[Squaring both sides : ]

↠ (5)² = [√(5)² + 4 + y² - 4y]²

↠ 25 = (5)² + 4 + y² -4y

↠ 25 = 25 + 4 + y² -4y

↠25 = 25 + (y - 2)²

↠ 25/25 = (y - 2)²

↠ 0 = (y - 2)²

↠ y - 2 = √0

↠ y - 2 = 0

↠ y = 2

Thus,

The value of y will be 2 .

Answer 2

$\huge{ \underline{ \tt{Question \ratio}}}$

Let A(-2,y) and B(3,2) are two points on the line segment. If the distance between these two points is 5 units, find y.

$\huge {\underline{ \tt{Answer \ratio}}}$

$\tt \: the \: value \: of \: \blue{ y = 2}$

$\huge{ \underline{ \tt{Solution \ratio - }}}$

As we know,

$\tt \: Distance = \sqrt{(x_2 - x_1) {}^{2} + (y_2 - y_1) {}^{2} }$

Now,

$\tt \: x_1 = - 2 \\ \tt \: x _2 = 3 \\ \tt y_1 = y \\ \tt y_2 = 2$

$\tt \to5 = \sqrt{ [(3 - ( - 2)] {}^{2} + (2 - y) {}^{2} }$

$\tt \to5 = \sqrt{(3 + 2 {)}^{2} + (2 {)}^{2 }+ (y {)}^{2} - (2 \times 2 \times y)} \: \: \: \: \: \: \: \: \{\because(a - b {)}^{2} \}$

$\tt \to5 = \sqrt{(5 {)}^{2} + 4 + (y {)}^{2} - 4y }$

Now on squaring both the side; we get;

$\tt \to(5 {)}^{2} = [ \sqrt{(5 {)}^{2} + 4 + {y}^{2} - 4y } ] {}^{2}$

$\tt \to25 = (5 {)}^{2} + 4 + {y}^{2} - 4y$

$\tt \to25 = 25 + 4 + {y}^{2} - 4y$

$\tt \to25 = 25 + {y}^{2} - 4y + 4$

$\tt \because(y - 2 {)}^{2} = {y}^{2} - 4y + 4$

$\tt \to25 = 25 + (y - 2 {)}^{2}$

$\tt \to {25 - 25} = (y - 2 {)}^{2}$

$\tt \to0 = (y - 2 {)}^{2}$

$\tt \to \sqrt{0} = y - 2$

$\tt \to0 = y - 2$

$\tt \to \: y - 2 = 0$

$\tt \to \: \red{y = 2}$