Question

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answer this with clear explanation ⤵️
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→The distance of point P(-7,-8) from the y-axis is (in units)
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Answer 1

Solution :

We are given with a point P(-7,-8) and asked to find the distace of the point from Y - axis

Now we have to know the coordinates of the point from Y - axis

In P(-7,-8)

• x = - 7
• y = - 8

On Y - axis at any point the coordinates will be (0,y)

Therefore the coordinates on Y - axis will be (0, - 8)

Let the coordinates on Y - axis be Y(0, - 8)

Y(0, - 8) A(- 7, - 8)

→ x₁ = 0

→ x₂ = - 7

→ y₁ = - 8

→ y₂ = - 8

Since y₁ = y₂, the line is parallel to X - axis

Therefore, Distance AB = |x₂ - x₁|

= |-7 - 0|

= |-7|

= 7 units

Hence, the distance of point P(-7,-8) from the y-axis is 7 units.

Answer 2

Correct Question :----

• we have to Find the perpendicular distance between point (-7,-8) and y -axis.

Concept and formula used :----

• At y -axis we know that, value of x = 0 .

Solution :----

we have to Find the perpendicular distance between P(-7,-8) and y-axis...

we know that now the distance of any point from the y-axis is given by the x - coordinates ..

Here , in Question x - coordinate is given = (-7)

Hence , we can say that the perpendicular distance from p(-7,-8) is $\red{\bold{</strong><strong>(</strong><strong>-</strong><strong>7</strong><strong>)</strong><strong>}}$

(Hope it helps you)