Question

Find the ratio in which the v-axis divides the line
the y-axis divides the line segment joining the
pour
1-4) and (5, -6). Also find the coordinates of the point of
intersection.​

Answer 1

$\textbf{Given:}$

$\textsf{Points are(1,-4) and (5,-6)}$

$\textbf{To find:}$

$\textsf{The ratio in which the line segment joining the}$

$\textsf{points (1,-4), and (5,-6) is divided by y-axis}$

$\textbf{Solution:}$

$\textbf{Section formula:}$

$\textbf{The co ordinates of the point which divides the}$

$\textbf{line segment joining (x_1,y_1) and (x_2,y_2) internally in the ratio m:n are}$

$\boxed{\bf\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)}$

$\textsf{By using section formula, the co-ordination of point of division is}$

$\mathsf{\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}$

$\mathsf{\left(\dfrac{m(5)+n(1)}{m+n},\dfrac{m(-6)+n(-4)}{m+n}\right)}$

$\mathsf{\left(\dfrac{5m+n}{m+n},\dfrac{-6m-4n}{m+n}\right)}$

$\textbf{Since the point of division lies on y-axis,}$

$\mathsf{\dfrac{5m+n}{m+n}=0}$

$\implies\mathsf{5m+n=0}$

$\implies\mathsf{5m=-n}$

$\implies\mathsf{\dfrac{m}{n}=\dfrac{-1}{5}}$

$\therefore\textsf{y-axis dvides the line segment joining (1,-4) and (5,-6)}$

$\therefore\textsf{externally in the ratio 1:5}$

$\textsf{The point of division is}$

$\mathsf{\left(\dfrac{mx_2-nx_1}{m-n},\dfrac{my_2-ny_1}{m-n}\right)}$

$\mathsf{\left(\dfrac{5m-n}{m-n},\dfrac{-6m+4n}{m-n}\right)}$

$\mathsf{\left(\dfrac{5(1)-5}{1-5},\dfrac{-6(1)+4(5)}{1-5}\right)}$

$\mathsf{\left(0,\dfrac{14}{-4}\right)}$

$\boxed{\mathsf{\left(0,\dfrac{-7}{2}\right)}}$

Find more:

Find the ratio in which line segment joining (-3,10) and (6,-8) is divided by y axis also find the point of division

https://brainly.in/question/20781027

Answer 2

$\huge \purple{\fbox \blue{\fbox\pink{\fbox\red{Solution }}}}$

$\textbf{Given:}$

$\textsf{Points are(1,-4) and (5,-6)}$

$\textbf{To find:}$

$\textsf{The ratio in which the line segment joining the}$

$\textsf{points (1,-4), and (5,-6) is divided by y-axis}$

$\textbf{Solution:}$

$\textbf{Section formula:}$

$\textbf{The co ordinates of the point which divides the}$

$\textbf{line segment joining (x_1,y_1) and (x_2,y_2) internally in the ratio m:n are}$

$\boxed{\bf\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)}$

$\textsf{By using section formula, the co-ordination of point of division is}$

$\mathsf{\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}$

$\mathsf{\left(\dfrac{m(5)+n(1)}{m+n},\dfrac{m(-6)+n(-4)}{m+n}\right)}$

$\mathsf{\left(\dfrac{5m+n}{m+n},\dfrac{-6m-4n}{m+n}\right)}$

$\textbf{Since the point of division lies on y-axis,}$

$\mathsf{\dfrac{5m+n}{m+n}=0}$

$\implies\mathsf{5m+n=0}$

$\implies\mathsf{5m=-n}$

$\implies\mathsf{\dfrac{m}{n}=\dfrac{-1}{5}}$

$\therefore\textsf{y-axis dvides the line segment joining (1,-4) and (5,-6)}$

$\therefore\textsf{externally in the ratio 1:5}$

$\textsf{The point of division is}$

$\mathsf{\left(\dfrac{mx_2-nx_1}{m-n},\dfrac{my_2-ny_1}{m-n}\right)}$

$\mathsf{\left(\dfrac{5m-n}{m-n},\dfrac{-6m+4n}{m-n}\right)}$

$\mathsf{\left(\dfrac{5(1)-5}{1-5},\dfrac{-6(1)+4(5)}{1-5}\right)}$

$\mathsf{\left(0,\dfrac{14}{-4}\right)}$

$\boxed{\mathsf{\left(0,\dfrac{-7}{2}\right)}}$