nearest point on line 2x-3y = 5 from origin is
Answers
Step-by-step explanation:
step1:- find the per..lr length. ie 5/√13.
step2:- apply the parametric form to write the coordinate of points at distance r from a given point (x1, y1).
ie, x1 +_ r cos( theta), y1 +_ rsin(theta).here theta should be slope angle of the line where the point (x1,y1) exist.
slope of given line is -2/3. so slope of perpendicular line would be 3/2.
tan theta = 3/2, so, sin( theta)= 3/√13 and cos( theta) = 2/√13.
Thus possible points are ( 0+_ 5/√13 × 2/√13), ( 0+_ 5/√13 × 3/√13.
ie, ( +_10/13, +_15/ 13).
step 3:- put both the points in the equation of the line to check ,which is the required point between two possible points.
here equation of given line is 2x+3y =5. so we, can easily observe point ( 10/13,15/13) is satisfying the given equation.
Hence, foot of perpendicular is= ( 10/13, 15/13).
this would be the closest point as well.
Hope it helps:
Answer:
Simplifying 5xy + -2xy = 0 Combine like terms: 5xy + -2xy = 3xy 3xy = 0 Solving 3xy = 0 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Divide each side by '3'. xy = 0 Simplifying xy = 0 The solution to this equation could not be determined.