Question

The coordinates of a point are (0,1) & the ordinate of
another point in -3. If the distance blw 2 point in 5,
then the abscissa of
another point is​

Answer 1

Answer:

A(0,1) and B(x,-3) let's take these as the points

Step-by-step explanation:

so by applying distance formula =

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

we get dis

$\sqrt{ {(x - 0)}^{2} + {( - 3 - 1)}^{2} } \\ { ({x}^{2} + 16) }^{ \frac{1}{2} } = 5 \\ squaring \: both \: sides \: \\ {x}^{2} + 16 = 25 \\ {x}^{2} = 9 \\ x = 3$

three absicca is 3

Answer 2

Answer:

A(0,1) and B(x,-3) let's take these as the points

Step-by-step explanation:

so by applying distance formula =

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

(x2−x1)

2

+(y2−y1)

2

we get dis

\begin{gathered}\sqrt{ {(x - 0)}^{2} + {( - 3 - 1)}^{2} } \\ { ({x}^{2} + 16) }^{ \frac{1}{2} } = 5 \\ squaring \: both \: sides \: \\ {x}^{2} + 16 = 25 \\ {x}^{2} = 9 \\ x = 3\end{gathered}

(x−0)

2

+(−3−1)

2

(x

2

+16)

2

1

=5

squaringbothsides

x

2

+16=25

x

2

=9

x=3