Question

Determine the distance between the straight line 4x-7y+15=0 and the point (-1,6)

Answer 1

Answer :

d = 31/√65 units

Solution :

• Note : The perpendicular distance between a straight line ax + by + c = 0 and the point (x1 , y1) is given by the formula :

d = | ax1 + by1 + c | /√(a² + b²)

Here ,

The given line is 4x - 7y + 15 = 0 . Comparing it by the general equation ax + by + c = 0 , we have ;

a = 4

b = -7

c = 15

Also ,

The given point is (-1 , 6) , thus x1 = -1 and y1 = 6 .

Thus ,

The perpendicular distance between the straight line 4x - 7y + 15 = 0 and the point (-1 , 6) will be given as ;

=> d = | ax1 + by1 + c | /√(a² + b²)

=> d = | 4•(-1) + (-7)•6 + 15 | /√[4² + (-7)²]

=> d = | -4 - 42 + 15 | /√(16 + 49)

=> d = | -31 | /√65

=> d = 31/√65

Hence ,

d = 31/√65 units

Answer 2

Step-by-step explanation:

Given :-

line :- 4x - 7y + 15 = 0

point :- (-1, 6)

The distance from a point (m, n) to the line Ax + By + C = 0 is given by:-

∣ Am + Bn + C ∣ /√(A²+B²)

.

.

where A = 4 ; B = -7 ; C = 15 (from line)

where A = 4 ; B = -7 ; C = 15 (from line) and m = (-1) ; n = (6)

| (4x(-1)) + (-7) x (6) + 15 | / √ 4² + (-7)²

| (-4) + (-42) +15 | / √ 16 + 49

31 / √65

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