A question on co-ordinates of intersecting lines...Given in picture below

Question

Please do also MENTION how you got the solution.........

Answer 1

Let $A(-3,5),B(6,-7)$. Also let $C(p,q),D(s,t)$ be the points you want.

Then, they satisfy the following :

$$\vec{AC}=(1/3)\vec{AB}\iff (p-(-3),q-5)=(1/3)(6-(-3),-7-5),$$ $$\vec{AD}=(2/3)\vec{AB}\iff (s-(-3),t-5)=(2/3)(6-(-3),-7-5).$$ Then, you can find $p,q,r,s$.

Answer 2

A=(-3,5) B=(6,-7)

Delta X = 6-(-3) = 9 and so the trisection points are 9/3 = +3 Delta Y = -7-5 = -12 and so the trisection points are -12/3 = -4.

so just add these to the A co-ordinate to get the points

First trisection point = -3 + 3,5+(-4) = 0,1 Second trisection point is got from the first one by adding the deltas 0,1 + 3,-4 = 3,-3

so the answers are (0,1) and (3,-3) which is answer A.

Answer 3

Okay, since I see coordinates which are of form $0,y$ or $x,0$ I would prefer to take out the equation of the line and put the values.

$9(y-5)=-12(x+3)$ So the equation is $12x+9y-9=0$ or $4x+3y-3=0$

Putting x=0, y=1. So (0,1) satisfies the line. Also, (3,-3) satisfies it too. (A) option is correct.