Question

If the midpoint of the line segment joining (4a,2b-3) and (-4,3b) is (4,-3a); then the value of a and b is .
a.-3 , -3
b.-3 , 3
c.3 , -3
d.3 , 3

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

• The midpoint of the line segment joining (4a,2b-3) and (-4,3b) is (4,-3a).

We know,

Mid-point formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R (x, y) will be:

$\sf\implies \boxed{\tt{ (x, \: y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}}$

Here,

$\red{\rm :\longmapsto\:x = 4}$

$\red{\rm :\longmapsto\:x_1 = 4a}$

$\red{\rm :\longmapsto\:x_2 = - 4}$

$\red{\rm :\longmapsto\:y = - 3a}$

$\red{\rm :\longmapsto\:y_1 = 2b - 3}$

$\red{\rm :\longmapsto\:y_2 = 3 b}$

So, on substituting the values in above formula, we get

$\rm :\longmapsto\:(4, \: - 3a) = \bigg(\dfrac{4a - 4}{2}, \: \dfrac{2b - 3 + 3b}{2} \bigg)$

$\rm :\longmapsto\:(4, \: - 3a) = \bigg(2a - 2, \: \dfrac{5b - 3}{2} \bigg)$

So, on comparing we get

$\rm :\longmapsto\:2a - 2 = 4$

$\rm :\longmapsto\:2a = 4 + 2$

$\rm :\longmapsto\:2a = 6$

$\rm \implies\:\boxed{\tt{ \: a \: = \: 3 \: }}$

Now,

$\rm :\longmapsto\:\dfrac{5b - 3}{2} = - 3a$

$\rm :\longmapsto\:\dfrac{5b - 3}{2} = - 3 \times 3$

$\rm :\longmapsto\:\dfrac{5b - 3}{2} = - 9$

$\rm :\longmapsto\:5b - 3 = - 18$

$\rm :\longmapsto\:5b = - 18 + 3$

$\rm :\longmapsto\:5b = - 15$

$\rm \implies\:\boxed{\tt{ \: b \: = \: - \: 3 \: }}$

So, we have

$\red{\begin{gathered}\begin{gathered}\rm :\longmapsto\:\begin{cases} &\sf{a \: = \: 3} \\ \\ &\sf{b \: = \: - \: 3} \end{cases}\end{gathered}\end{gathered}}$

Hence, option (c) is correct.

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More to Know :-

1. Section formula :-

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\sf\implies\boxed{\tt{ R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}}$

2. Mid-point formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\sf\implies\boxed{\tt{ R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}}$

3. Centroid of a triangle :-

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\sf\implies\boxed{\tt{ R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}}$

4. Distance Formula :-

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane, the distance PQ is

$\rm \implies\:\boxed{\tt{ PQ \: = \: \sqrt{ {(x_{1} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2} }}}$