Question

the distance of point p (-3,-4) from the x-axis (in units) is​

Answer 1

Hey mate here is your answer:

Given:-

p=(-3,-4),let point on x-axis be q(x,0)

To Find:-

Distance (d)=?

Solution:-

we know that,

$d = \sqrt{(x2 - x1) {}^{2} + (y2 - y1)} {}^{2}$

Put the values we get:

$d = \sqrt{( - x + 3) {}^{2} + ( 0 + 4) {}^{2} }$

We know that x-axis on origin has coordinates (0,0)

Now put x as 0

$d = \sqrt{9 + 16}$

$d = \sqrt{25}$

$d = 5 \: units$

Answer 2

The distance of point (-3,-4) from the x-axis (in units) is 4 unit.

Step-by-step explanation:

To find : The distance of point (-3,-4) from the x-axis (in units) is ​?

Solution :

The distance of a point P(a,b) from x-axis is given by the y-coordinate of the point.

i.e. d=|b|

Here, the point is P(a,b)=(-3,-4)

The y-coordinate is -4.

The distance of point (-3,-4) from x axis is d=|-4|= 4 unit.

Therefore, the distance of point (-3,-4) from the x-axis (in units) is 4 unit.

#Learn more

Find the point on the x axis which is equal distance from ( 2,-5) (-2,9)

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