Question

33.
Find the ratio in which the y-axis divides the line segment joining the
points (6,- 4) and (-2,- 7). Also find the point of intersection.​

Answer 1

★ Given :-

• A line segment with two end points (6,-4) & (-2 ,-7)

★ To Find :-

• The ratio in which y-axis devides the given points .
• And also we have to find the point of intersection .

★ Solution :-

Let the point be A(6,-4) & B(-2,-7)

• The point which divided the line segment lies on y-axis , hence the value of x on y-axis is 0
• Let the point be P which divides the line segment .
• So Co-ordinate of P be P(0,y)

Now , Using section formula

$\boxed{\sf x=\dfrac{ mx_2+nx_1}{m+n} \ \ ; \ \ y= \dfrac{my_2+ny_1}{m+n}}$

So , Let the Ratio be K:1

• We have ,

$\bullet\sf\ \ x_1= 6 \ \ ; \ \ \bullet\ \ y_1= -4 \\ \\ \bullet\sf \ \ x_2= -2 \ \ ; \ \bullet \ \ y_2= -7 \\ \\ \bullet\sf \ \ m= k \ \ ; \ \ \bullet \ \ n= 1\\ \\ \bullet\sf \ \ x= 0 \ \ ; \ \ \bullet\ \ y= y$

$\dashrightarrow\sf x = \dfrac{mx_2+nx_1}{m+n}\\ \\ \\\dashrightarrow\sf 0 = \dfrac{(k\times -2 ) +( 1\times 6)}{k+1} \ \ \ \ \ \Big[ x=0 (Point \ lies \ on \ y-axis)\Big]\ \\ \\ \\\dashrightarrow\sf 0 = \dfrac{-2k+6}{k+1} \\ \\ \\\dashrightarrow\sf 0(k+1)= -2k+6 \\ \\ \\\dashrightarrow\sf 0= -2k+6 \\ \\ \\\dashrightarrow\sf 2k= 6 \\ \\ \\\dashrightarrow\sf k = \cancel{\dfrac{6}{2}}\\ \\ \\\dashrightarrow\sf {\boxed{\tt{\red{ k= 3 }}}} - - - - - eq.1$

$\underline{\boxed{\red{\sf Hence ,\ Ratio \ will \ be \ 3:1 }}}$

• Now we have to find y also

$\dashrightarrow\sf y = \dfrac{my_2+ny_1}{m+n}\\ \\ \\ \dashrightarrow\sf y= \dfrac{(k\times -7)+(1\times -4)}{k+1}\\ \\ \\ \dashrightarrow\sf y= \dfrac{-7k-4}{k+1} \\ \\ \\ \dashrightarrow\sf y= \dfrac{-7(3)-4}{3+1)}\ \ \ \ \ \therefore \Big[(k= 3)\ from \ eq.1\Big]\\ \\ \\ \dashrightarrow\sf y= \dfrac{-21 -4}{4}\\ \\ \\ \dashrightarrow\sf y= \dfrac{-25}{4}$

$\boxed{\tt{\red{\ \ y= \dfrac{-25}{4}}}}$

$\underline{\bigstar{\sf { Hence ,\ the \ Coordinate \ of \ Point \ be\ P(0, -25/4 )}}}$