Question

what is the perpendicular distance of the point P(4,3) from x axis

Answer 1

$\textbf{Answer}$

Given point is P(4,3)
Lets suppose (x1,y1) = (4,3)
&
Required point is Q(x2,y2)

We know that $\textbf{value of y on x axis is always 0.}$

Since the given point P(4,3) is perpendicular to the point Q(x2,y2)

=> Q(x2,y2) = (4,0)

$\textbf{We know the Distance Formula,}$
Distance between points P(x1,y1) and Q(x2,y2) is calculated as,
PQ = √{(x2 - x1)^2 + (y2 - y1)^2}

$\textbf{P(4,3)}$
$\textbf{Q(4,0)}$

=> PQ = √{(4 - 4)^2 + (0 - 3)^2}

=> PQ = √(0 + 9)

PQ = 3 that satisfies the conditions.

So the perpendicular distance of point P(4,3) from the x-axis is $\textbf{3 Unit.}$

$\textbf{Hope My Answer Helped}$

$\textbf{Thanks}$

Answer 2

$\textbf{Answer :}$

$\textbf{Method (1) -}$

The perpendicular distance between the two points (x₁, y₁) and (x₂, y₂) is

d = √{(x₁ - x₂)² + (y₁ - y₂)²} units

The given point is P (4, 3) and the perpendicular point on x-axis be (4, 0).

Thus, the required perpendicular distance be

= √{(4 _ 4)² + (3 - 0)²} units

= √(0² + 3²) units

= √(3²) units

= $\textbf{3 units}$

$\textbf{Method (2) -}$

The given point is (4, 3) and the line is the x-axis, i.e., y = 0, i.e., 0.x + 1.y = 0

So, the required distance be

= (0.4 + 1.3)/√(0² + 1²) units

= (0 + 3)/√(0 + 1) units

= 3/√1 units

= 3/1 units

= $\textbf{3 units}$

$\textbf{Method (3) -}$

The given point is P (4, 3).

we see that, the ordinate of the P is 3 and this determines the required distance of the point P (4, 3) from the x-axis.

So, $\textbf{3 units}$ is the required perpendicular distance.

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