Question

find the distance froma point (1,2) to the line 5x+12y-3=0​

Answer 1

Step-by-step explanation:

This is essentially asking in maths speak to find the shortest distance from (2,1) to the line 5x - 12y + 3 = 0

To answer you need to consider the perpendicular line (at 90^\circ) that goes through (2,1) and where it crosses the original line. Then find how far apart these two points are.

The following is a Desmos plot showing you this situation and it is quite close.

To do this algebraically, rearrange the equation in the form y = mx + c

5x - 12y + 3 = 0

12y = 5x + 3

y = 512x + 312

The gradient is 512. Therefore the perpendicular gradient is -125.

So now use the point (2,1) to find the intercept (constant).

1 = -125 x 2 + C

55 + 245 = C = 295

So to find the crossing point, we look at the crossing point as simultaneous equations as shown in the Desmos plot.

Algebraically we use both y equivalences:

y = 512x + 312 with y = -125x + 295

512x + 312 = -125x + 295

112(5x + 3) = 15(-12x + 29)

5(5x + 3) = 12(-12x + 29)

25x + 15 = -144x + 348

169x = 333

x = 333169

y = 181169 without showing the calculations.

The last step is to find the differences between the x-values and the y-values. Then use Pythagoras’ Theorem to calculate the distance.

So the horizontal (x) difference is 5169

and the vertical difference is 12169.

Apply Pythagoras it looks like the answer will be 13169 knowing it is a 5, 12, 13 Pythagorean triple. This means the final answer is 113.

Answer 2

Answer:

Hence the distance between the line and point is 2 units.

Given:

Given equation of line = 5x + 12y -3 =0

Given point , from where the distance is to be calculated = (1,2)

To Find:

find the distance of a point from the line.

solution:

This line is represented by Ax + By + C = 0. The distance of point from a line, 'd' is the length of the perpendicular drawn from N to l. The x and y-intercepts are −C/A and −C/B respectively. NM = d = |Ax₁ + By₁ + C| / (A² + B²)½.

Here,( x₁,y₁) = (1,2)

given equation = 5x+12y-3=0​

A = 5 , B = 12 , C = -3

So the perpendicular distance of point from the line, d= |Ax₁ + By₁ + C| / (A² + B²)½.

d = 5x1 + 12x 2 -3/√ 5² + 12²

= 5+ 24 - 3/√13²

=26/13

=2

Hence the distance between the line and point is 2 units.