Question

The point on x-axis which divides the line segment joining (2, 3) and
16.-9) in the ratio 1:3 is
(A) (4, -3)
(B) (6,0)
(C) (3,0)
(D) (0,3)
​

Answer 1

Correct Question:-

The point on x-axis which divides the line segment joining (2, 3) and

(6,-9) in the ratio 1:3 is

(A) (4, -3)

(B) (6,0)

(C) (3,0)

(D) (0,3)

SoluTion:-

Lets the points that divides the line joining points (2,3) and (6,-9) in the ratio 1:3 be equals to (x,y).

Here we use section Formula to find the coordinates of point -

$\sf x = \dfrac{( m_1 x_2 + m_2 x_1 )}{( m_1 + m_2 )} \\ \\ \sf y = \dfrac{( m_1 y_2 + m_2 y_1 )}{( m_1 + m_2 )}$

Here Given that,

$\;\;\bullet\;\sf m_1 = 1$

$\;\;\bullet\;\sf m_2 = 3$

$\;\;\bullet\;\sf x_1 = 2$

$\;\;\bullet\;\sf x_2 = 6$

$\;\;\bullet\;\sf y_1 = 3$

$\;\;\bullet\;\sf y_2 = -9$

Now Putting values in formula:-

$\sf x = \dfrac{( m_1 x_2 + m_2 x_1 )}{( m_1 + m_2 )} \\ \\ \sf y = \dfrac{( 1 \times 6 + 3 \times 2 )}{( 1 + 3 )} \\ \\ \sf y = \dfrac{6 + 6}{4} \\ \\ \sf y = \cancel{\dfrac{12}{4}} \\ \\ \bf \red{y = 3}$

$\rule{150}{4}$

$\sf y= \dfrac{( m_1 y_2 + m_2 y_1 )}{( m_1 + m_2 )} \\ \\ \sf y = \dfrac{( 1 \times (-9) + 3 \times 3 )}{( 1 + 3 )} \\ \\ \sf y = \dfrac{(-9) + 9}{4} \\ \\ \sf y = \dfrac{0}{4} \\ \\ \bf \red{y = 0}$

$\rule{150}{4}$

We have found the value of x and y which are the coordinates of points on x - axis.

Therefore, the coordinates of the points on x - axis is (3,0).

Hence, Option (C) is correct.

$\rule{150}{4}$

Answer 2

Given ,

A point on x-axis which divides the line segment joining (2, 3) and (6,-9) in the ratio 1 : 3

Let , the point on the x - axis be (x,0)

We know that , the internal section formula is given by

$\large \sf \fbox{x = \frac{m x_{2} + nx_{1} }{m + n} \: , \: y = \frac{m y_{2} + ny_{1} }{m + n} \: }$

Thus ,

$\sf \mapsto x = \frac{1(6) + 3(2)}{1 + 3} \\ \\\sf \mapsto x = \frac{12}{4} \\ \\\sf \mapsto x = 3$

$\therefore \sf \underline{The \: required \: point \: on \: the \: x - axis \: is \: (3,0)}$