Question

A
standing at a junction (crossing) of two straight paths represented by the
Equations 2x - 3y + 4 = 0 and 3x + 4y - 5 = 0 seek to reach the path whose equation
2X-7Y + 8 = 0 in the least time. Find the equation of the path that he should follow.​

Answer 1

Answer

119x+34y-37=0

The meeting point will be solved by system equation

2x-3y+4=0, 3x+4y-5=0.

The solution is (x,y)=($-\frac{1}{17}$, $\frac{22}{17}$).

Therefore, the meeting point will be ($-\frac{1}{17}$, $\frac{22}{17}$).

If he uses least time to reach the path, then it means he uses shortest route.

Our assumption is :

The shortest route will be a perpendicular line from point down the path.

Now,

if we express 2x-7y+8=0 in a slope-intercept form, we have $y=\frac{2x}{7} +\frac{8}{7}$.

The slope of a perpendicular line will be $-\frac{7}{2}$.

Now, we use point-slope form of linear equation.

We have two conditions : slope is $-\frac{7}{2}$, the line passes ($-\frac{1}{17}$, $\frac{22}{17}$)

We have $y-\frac{22}{17} =-\frac{7}{2}(x+\frac{1}{17} )$.

On simplifying, we get $y -\frac{22}{17} = -\frac{7x}{2} - \frac{7}{34}$.

In a standard form, it is $119x+34y-37=0$.

The answer is $119x+34y-37=0$.

Answer 2

Answer:

0 is answer dear friend