Question

* Find the distance of the line 4x – 3y + 10 = 0 from the origin

Answer 1

Answer: We know, the formula to find distance of the line (ax+by+c=0) and a point (x1,y1) is |ax1+by1+c|/sqrt(a^2+b^2).

So.here distance d= |4×0-3×0+10|/sqrt(4^2+(-3)^2)

=10/sqrt(25)=10/5=5

Thus, the answer is 5 units.

Answer 2

Answer:

Required distance is

$\frac{10}{ \sqrt{21} } \: unit$

Step-by-step explanation:

If one straight line is ax+by+c=0 and a point (p,q)

Then distance between them

$d = \frac{ |ap + bq + c| }{ \sqrt{ {a}^{2} + {b}^{2} } } \: unit$

Now given line here 4x – 3y + 10 = 0 and point is (0.0)

Let distance between them be d

So,

$d = \frac{ |4 \times 0 - 3 \times 0 + 10| }{ \sqrt{ {4}^{2} + {3}^{2} } } = \frac{ |0 - 0 + 10| }{ \sqrt{12 + 9} } = \frac{ |10| }{ \sqrt{21} } \: unit$

By this formula we can find distance between any line from any point.

Required distance is

$\frac{10}{ \sqrt{21} } \: unit$