Question

Solve the attachment!!!

Answer 1

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Reffer to the attachment!!

Answer 2

ANSWER:

Given:

• Two points A(5,-6) and B(-1,-4)
• y-axis divides the line segment AB.

To Find:

• Ratio in which AB is divided
• Point of intersection

Solution:

$\text{\sf{Let the point dividing AB be C, with coordinates (x,y)}}\\\\\text{\sf{But, it is given that y-axis divides AB. So, coordinates of C = (0,y)}}\\\\\text{\sf{Section Formula}}:\longrightarrow\sf{(x,y) = \left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)} \\\\\text{\sf{Where, x_1,x_2, y_1, y_2 are coordinates of the two points and,}}\\\text{\sf{m:n is the ratio in which the line is divided.}}\\\\\text{\sf{Here, x_1=5,x_2=-1,y_1=-6,y_2=-4 and x = 0. So,}}$

$:\implies\sf{(0,y) = \left(\dfrac{m(-1)+n(5)}{m+n},\dfrac{m(-4)+n(-6)}{m+n}\right)}\\\\:\implies\sf{(0,y) = \left(\dfrac{-m+5n}{m+n},\dfrac{-4m-6n}{m+n}\right)}\\\\\text{\sf{Taking x coordinates,}}\\\\:\implies\sf{0 = \dfrac{-m+5n}{m+n}}\\\\:\implies\sf{5n-m=0}\\\\:\implies \sf{m=5n}\\\\\bf{:\implies \dfrac{m}{n}=5\implies m:n = 5:1}\\\\\text{\sf{Hence,}} \\\\:\implies\sf{y=\dfrac{-4(5)+-6(1)}{5+1}}\\\\:\implies\sf{y=\dfrac{-20-6}{6} \implies \dfrac{-26}{6}=\dfrac{-13}{3}}$

$\text{\sf{Hence, the point of intersection C(0,y) is}} \\\\\bf{:\implies C = (0,y) = \left(0,\dfrac{-13}{3}\right)}$

$\text{\bf{FINAL ANSWER:}}\\\\\bf{Ratio=5:1\:\:\:\&\:\:\:C=\left(0,\dfrac{-13}{3}\right)}$

Formula Used:

$\:\:\:\:\:\:\bullet\:\:\:\text{\sf{Section Formula}}:\longrightarrow\sf{(x,y) = \left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}$