Question

9. Find points on the x-axis, each of which is at a distance of 10 units from
the point A(11,-8).
​

Answer 1

Answer:

Answer:

The points are (5,0) and (17,0)

Step-by-step explanation:

At the x axis, the y coordinate is zero. Hence, any point on the x axis has a coordinate (x,0).

Now, this point is at a distance of 10 units from the point a(11,-8)

The distance formula is given by

\begin{lgathered}d=\sqrt{(x_2-x_1)^2-(y_2-y_1)^2}\\\\10=\sqrt{(11-x)^2+(-8-0)^2}\end{lgathered}

d=

(x

2

−x

1

)

2

−(y

2

−y

1

)

2

10=

(11−x)

2

+(−8−0)

2

Squaring both sides, we get

\begin{lgathered}100=(11-x)^2+(-8-0)^2\\100=121+x^2-22x+64\\x^2-22x+85=0\\x^2-17x-5x+85=0\\x(x-17)-5(x-17)=0\\(x-17)(x-5)=0\\x=5,17\end{lgathered}

100=(11−x)

2

+(−8−0)

2

100=121+x

2

−22x+64

x

2

−22x+85=0

x

2

−17x−5x+85=0

x(x−17)−5(x−17)=0

(x−17)(x−5)=0

x=5,17

Therefore, the points are (5,0) and (17,0)

Answer 2

Answer:

The points are (5,0) and (17,0)

Step-by-step explanation:

At the x axis, the y coordinate is zero. Hence, any point on the x axis has a coordinate (x,0).

Now, this point is at a distance of 10 units from the point a(11,-8)

The distance formula is given by

d=\sqrt{(x_2-x_1)^2-(y_2-y_1)^2}\\\\10=\sqrt{(11-x)^2+(-8-0)^2}

Squaring both sides, we get

100=(11-x)^2+(-8-0)^2\\100=121+x^2-22x+64\\x^2-22x+85=0\\x^2-17x-5x+85=0\\x(x-17)-5(x-17)=0\\(x-17)(x-5)=0\\x=5,17

Therefore, the points are (5,0) and (17,0)