The shortest distance of the points $(a, b, c)$ from the x-axis is

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

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

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

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

\sqrt{(c^{2} + a^{2})}

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

\sqrt{(a^{2} + b^{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 $\sqrt{(b^{2} + c^{2})}$
The point is $(a, b, c)$
the perpendicular distance from $x$ axis will be the shortest distance
$∴$ distance is $\sqrt{b^{2} + c^{2}}$

Suggest Corrections

Similar questions

|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

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

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

a + b + c = 0

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

Join BYJU'S Learning Program