Question

distance of point 2 ,1 from the line 3x-y-6=0 along the line x-y+2=0​

Answer 1

Step-by-step explanation:

It can be much simplified by simple logic!!

Please look at the figure below.

The distance we are finding is the perpendicular distance. Now the points joining x,y and (-3, -4) is a part of a line.

Thus the slope of that line will be = y+4x+3

The slope of of the line 3x-4y-1=0 is 34

Now we know that product of the slopes of two perpendicular line is -1

Therefore multiplying the two slopes will give us an equation as:

y+4x+3∗34=−1

⇒3y+4x=−24

Another equation we have is the equation of your line that is 3x-4y-1=0

Solving the two we get the values of

x = −9325

y= −7625

Applying the distance formula for the two points i.e. (-3,-4) and ( −9325 , −7625 ) we get the value 1.2

Answer 2

Answer:

Distance of point (2,1) from the line 3x-y-6=0 along the line x-y+2=0 is √((2-1)$^{2}$ + (1+2)^2) = √6.

Step-by-step explanation:

From the above question,

They have given :

To find the equation of the line perpendicular to the line 3x-y-6=0 and passing through the point (2,1).

The equation of the line perpendicular to the line 3x-y-6=0 and passing through the point (2,1) is x-y+2=0.

The distance of a point from a line can be calculated using the formula d = |ax0 + by0 + c|/√(a2+b2). In this case, the equation of the line is 3x-y-6=0, so a=3, b=-1, and c=-6. The coordinates of the point are (2,1), so x0=2 and y0=1.

Plugging the values into the formula gives d = |3(2) + (-1)(1) + (-6)|/√(3^2 + (-1)^2) = |6-1-6|/√10 = √7/√10 = √7/3.

To calculate the distance between the point (2,1) and the line 3x-y-6=0 along the line x-y+2=0.

The distance between the point (2,1) and the line 3x-y-6=0 along the line x-y+2=0 is √((2-1)$^{2}$ + (1+2)^2) = √6.

However, this is the distance from the point to the line in general, not the distance from the point to the line along the line x-y+2=0. To find the distance along the line, we can use the Pythagorean theorem.

Hence,

       Distance of point (2,1) from the line 3x-y-6=0 along the line x-y+2=0 is √((2-1)$^{2}$ + (1+2)^2) = √6.

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