Question

find the distance of the point (3,5) from the line 2x+3y=14 measured parallel to the line x-2y=1

Answer 1

Hi,

Here is your answer,

The line is parallel to x-2y=1 passing through the point (3,5)(x,y)

Now, we need to put the line equation and we need to put the values according the sum.

3-2×5=c 
c=-7

So, the equation is x-2y-=-7

Now we need to  find point of intersection of (x-2y=-7) and (2x+3y-14=0). 
we need to solve equation (2)

x - 2y + 7 => 0   -->(1)

2x + 3y -14 => 0   -->(2)

2x - 4y +14 => 0   -->(3)( now we need to multiply (2) with equation (1)

Now, by Subtracting eq.3 from eq.2

7y -28 =>0

--> y = 4

Now, we need to put y in eq (1)

x--> 1

Now,we need to find the distances between (3,5) and (1,4)

Now, by using Distance formula √(X2 - X1)² + (Y2 - Y1)²


Distance = √(3-1)²+(5-4)² = √2²+1² = √5



Hope it helps you!