The distance between the points $C (- 1, - 2)$ and $D (4, 7)$ is:

\sqrt{34} u n i t s .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{116} u n i t s .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{130} u n i t s .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{106} u n i t s .

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $\sqrt{106} u n i t s .$
$T h e g i v e n p o i n t s a r e C (x_{1}, y_{1}) = (- 1, - 2) & D (x_{2}, y_{2}) = (4, 7) . N o w t h e d i s t a n c e b e t w e e n t w o p o i n t s C (x_{1}, y_{1}) & D (x_{2}, y_{2}) i s d = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} . ∴ T h e l e n g t h o f C D = d = \sqrt{{(- 1 - 4)}^{2} + {(- 2 - 7)}^{2}} u n i t s = \sqrt{25 + 81} u n i t s = \sqrt{106} u n i t s . A n s - O p t i o n D .$

Suggest Corrections

Similar questions

A (- 3, 5, 6), B (1, - 5, 7), C (8, - 3, - 1)

D (4, 7, - 2)

In parallelogram ABCD, A(2,3), B(6,5) and D(4,7). Then the coordinates of C is

x y = c (c > 0)

x^{2} + y^{2} = 1

The distance between the points (-1, 5) and (2, 1) is ___ units.

\sqrt{(a - c)^{2} + (b - d)^{2}}

Join BYJU'S Learning Program