Question

what is the distance of the point P (5,7) from x axis.

Answer 1

Answer:

Distance between point P(5,7) and (0,y) is $\sqrt{(y^2-14y+74)}$ units

Step-by-step explanation:

Given points,

• P(5,7)
• in x-axis (0,y)

let us take,

• $x_1$ = 5 --------------- (i)
• $y_1$ = 7 ---------------(ii)
• $x_2$ = 0 ---------------(iii)
• $y_2$ = y ---------------(iv)

Now,

we have to find the distance between two given points

the formula used for distance is given below-

Distance = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ -----------------(v)

Substitute the equation (i),(ii),(iii),(iv) in equation (v), it becomes-

Distance = $\sqrt{(0-5)^2+(y-7)^2}$

Distance = $\sqrt{25+y^2+49-14y}$

Distance = $\sqrt{y^2-14y+74}$ units

Answer 2

Answer:

1) As we have 4 quadrants, first we need to locate the given point.

2) The given point P(x,y) is P(5,7). Both x and y values are positive.   So, the point is in the first quadrant.

3) As shown in the figure at the attachment, the point is located at  points from the y-axis and 7 points from the x-axis.

4) So, the distance of the point P (5,7) from x-axis is 7.

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