Question

Find the value of a if the distance between the points A(-3,-14) and B(a,-5) is 9 units.​

Answer 1

Answer:

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Answer 2

Given Question :-

• Find the value of a if the distance between the points A(-3,-14) and B(a,-5) is 9 units.

Answer

Given :-

• The distance between the points A(-3,-14) and B(a,-5) is 9 units.

To Find :-

• The value of a

Concept Used :-

Concept Used :- Distance Formula :-

Let us consider a line segment joining the points A and B, then distance between A and B is given by

$\bf\implies \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

where,

$\tt \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) and (x_2,y_2).$

$\large\underline{\bold{Solution :- }}$

Given that

• Two points A (- 3, - 14) and B (a, - 5)

and

• Distance between A and B is 9 units

We know,

• Distance between two points A and B is calculated by using the formula given below,

$\rm\implies \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

Here,

• x₁ = - 3

• x₂ = a

• y₁ = - 5

• y₂ = - 14

Since,

• distance between these points is 9 units.

$\rm :\implies\:9 = \sqrt{ {(a + 3)}^{2} + {( - 5 + 14)}^{2} }$

• On squaring both sides, we get

$\rm :\longmapsto\:81 = {(a + 3)}^{2} + {9}^{2}$

$\rm :\longmapsto\: \cancel{81} = {(a + 3)}^{2} + \cancel{81}$

$\rm :\longmapsto\: {(a + 3)}^{2} = 0$

$\rm :\longmapsto\:a + 3 = 0$

$\rm :\implies\: \boxed{ \bf \: a \: = - \: 3}$

Additional Information :-

$\underline{\:\textsf{Section Formula\; :}}$

• Section Formula is used to find the co ordinates of the point (x, y) which divides the line segment joining the points (A) and (B) internally in the ratio m : n

${\underline{\boxed{ \bf{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} , \: \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$

where,

$\tt \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) and (x_2,y_2).$

$\underline{\:\textsf{Mid Point Formula\; :}}$

• Mid Point formula is used to find the mid points of any line segment.

• Let us assume a line segment joining the points A and B, then midpoint (x, y) of AB is given by

${\underline{\boxed{\bf{\quad \Big(x, y \Big) = \Bigg(\dfrac{x_2 + x_1}{2} , \: \dfrac{y_2 + y_1}{2}\Bigg) \quad}}}}$