Question

coordinate geometry ​

Answer 1

ans... of your first one

Answer 2

Coordinate Geometry :-

If three distinct points $\sf{A(x_1,y_1),B(x_2,y_2)\:and\:C(x_3,y_3)}$ are collinear, then we cannot form a triangle, because for such a triangle there will be no altitude (height). Therefore, three points $\sf{A(x_1,y_1),B(x_2,y_2)\:and\:C(x_3,y_3)}$ will be collinear if the area of ΔABC = 0.

(i) Given that,

• $\sf{A(x_1,y_1)\:=\:(4,-1)}$,
• $\sf{B(x_2,y_2)\:=\:(0,k)\:and\:}$
• $\sf{C(x_3,y_3)\:=\:(-2,-3)}$

We know that,

$\sf⟹{\dfrac{1}{2}}\:[(x_1y_2+x_2y_3+x_3y_1)\:-\:(x_2y_1+x_3y_2+x_1y_3)]\:=\:0$

$\sf⟹{\dfrac{1}{2}}\:[(4k-0+2)\:-\:(0-2k-12)]\:=\:0$

$\sf⟹{\dfrac{1}{2}}\:[4k-0+2-0+2k+12]\:=\:0$

$\sf⟹{\dfrac{1}{2}}\:[6k+14]\:=\:0$

$\sf⟹{6k+14\:=\:0\:×\:{\dfrac{1}{2}}}$

$\sf⟹{6k+14\:=\:0}$

$\sf⟹{6k\:=\:-14}$

$\sf⟹{k\:=\:{\dfrac{-14}{6}}}$

Therefore,$\sf{k\:=\:{\dfrac{-14}{6}}}$

(ii) Let,

• $\sf{P(x_1,y_1)\:and\:Q(x_2,y_2)}$be the points of trisection of line segment.

So, AP = PQ = QB.

• PB = PQ + QB
• PB = k + k = 2k
• Hence, ratio = $\sf{\dfrac{AP}{PB}}\:=\:{\dfrac{k}{2k}}\:=\:{\dfrac{1}{2}}$

Note : [Refer to the attachment]

Solving,

$\sf⟹{x_1\:={\dfrac{1\:×\:(-2)\:+\:2\:×\:4}{1\:+\:2}}}$

$\sf⟹{x_1\:={\dfrac{-2\:+\:8}{3}}}$

$\sf⟹{x_1\:={\dfrac{6}{3}}}$

$\sf⟹{x_1\:=\:2}$

$\sf⟹{y_1\:={\dfrac{1\:×\:(-3)\:+\:2\:×\:(-1)}{1\:+\:2}}}$

$\sf⟹{y_1\:={\dfrac{-3\:+\:-2}{3}}}$

$\sf⟹{y_1\:={\dfrac{-5}{3}}}$

$\sf{∴\:P(x_1,y_1)\:=\:(2,{\dfrac{-5}{3}})}$

$\sf⟹{x_2\:={\dfrac{2\:×\:(-2)\:+\:1\:×\:4}{2\:+\:1}}}$

$\sf⟹{x_2\:={\dfrac{-4\:+\:4}{3}}}$

$\sf⟹{x_2\:={\dfrac{0}{3}}}$

$\sf⟹{x_2\:=\:0}$

$\sf⟹{y_2\:={\dfrac{2\:×\:(-3)\:+\:1\:×\:(-1)}{2\:+\:1}}}$

$\sf⟹{y_2\:={\dfrac{-6\:-1}{3}}}$

$\sf⟹{y_2\:={\dfrac{-7}{3}}}$

$\sf{∴\:Q(x_2,y_2)\:=\:(0,{\dfrac{-7}{3}})}$

Therefore, $\sf{P(x_1,y_1)\:=\:(2,{\dfrac{-5}{3}})}$ and $\sf{Q(x_2,y_2)\:=\:(0,{\dfrac{-7}{3}})}$