The distance between points $(a + b, b + c)$ and $(a - b, c - b)$ is :

(1)

2 \sqrt{a^{2} + b^{2}}

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

(2)

2 \sqrt{b^{2} + c^{2}}

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

(3)

2 \sqrt{2} b

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

(4)

\sqrt{a^{2} - c^{2}}

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

Open in App

Solution

The correct option is B (3) $2 \sqrt{2} b$
Distance between two points $(x_{1}, y_{1})$
and $(x_{2}, y_{2})$ can be calculated
using the formula $\sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$
Distance
between the points $(a + b, b + c)$ and $(a - b, c - b) = \sqrt{{(a - b - a - b)}^{2} + {(c - b - b - c)}^{2}} = \sqrt{4 b^{2} + 4 b^{2}} = 2 \sqrt{2} b$

Suggest Corrections

Similar questions

The distance between points $(a + b, b + c)$ and $(a - b, c - b)$ is :

a² + b² + c² - ab - bc - ca equals:

\frac{cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2}}{cot A + cot B + cot C} = \frac{{(a + b + c)}^{2}}{a^{2} + b^{2} + c^{2}}

In a triangle ABC, $2 c a s i n \frac{A - B + C}{2}$ is equal to

In a $△ A B C, 2 c a s i n (\frac{A - B + C}{2})$ is equal to

Join BYJU'S Learning Program