Question

Find the ratio in which the line 2x + y-5=0 divides the line segment joining A( 2, -3) and B( 3, 7).

Answer 1

Answer:

The ratio in which the line 2x+y-5=0 divides the line segment joining points A(2,-3) and B(3,7) is 1:2.

Step-by-step explanation:

w.k.t., Equation of line joining two points (x₁,y₁) and (x₂,y₂) is $(y-y_{1}) = (\frac{y_{2}-y_{1}}{x_{2}-x_{1}})(x-x_{1})$.

⇒Equation of line joining the points A and B is $(y-(-3)) = (\frac{7-(-3)}{3-2})(x-2)$

⇒ $(y+3) = (\frac{10}{1})(x-2)$

⇒ 10x-y-23 = 0

On solving 2x+y-5 = 0 and the above equation, we get the point of intersection.

⇒ Point of intersection of two lines, P ≡ ($\frac{7}{3} , \frac{1}{3}$).

Let point P divides the line segment in the ratio α:1.

w.k.t, the co-ordinates of point which divide a line segment joining two points (x₁,y₁) and (x₂,y₂) in the ratio m:n are ≡ ($\frac{mx_{2}+nx_{1}}{m+n} , \frac{my_{2}+ny_{1}}{m+n}$).

⇒Coordinates of point P = ($\frac{\alpha.3+1.2}{\alpha+1} , \frac{\alpha.7+1.(-3)}{\alpha+1}$) = $\frac{3\alpha+2}{\alpha+1} , \frac{7\alpha-3}{\alpha+1}$

On compairing the coordinates of point P, we get

$\frac{3\alpha+2}{\alpha+1} = \frac{7}{3}$ and $\frac{7\alpha-3}{\alpha+1} = \frac{1}{3}$

On solving above two, we get

$\alpha = \frac{1}{2}$

⇒ α:1 = 1:2

⇒The line 2x+y-5 = 0 divides the line segment AB in the ratio 1:2.

Please mark it as Brainliest.