Question

WILL GIVE BRANLIEST!!! Pls help! Determine the coordinates of the point on the straight line y=3x+1 that is equidistant from the origin and (−3, 4).

Answer 1

Let , coordinate of points are P( h,k ).

Also , k = 3h + 1

Distance of P from origin :

$d=\sqrt{h^2+k^2}$

Distance of P from ( -3, 4 ) :

$d=\sqrt{(h+3)^2+(k-4)^2}$

Now , these distance are equal :

$h^2+(3h+1)^2=(h+3)^2+(3h+1-4)^2\\\\h^2+(3h+1)^2=(h+3)^2+(3h-3)^2$

Solving above equation , we get :

$P=(\dfrac{16}{21},\dfrac{23}{7})$

Hence , this is the required solution.