You visited us 1 times! Enjoying our articles? Unlock Full Access!

Distance Formula

Let the point...

Question

Let the point $(a, b)$ be equidistant from the points $(6, - 1)$ and $(2, 3)$ . If $a = k_{1} b + k_{2}$ , then the value of $k_{1} + k_{2}$ is

Open in App

Solution

Given:
$(a - 6)^{2} + (b + 1)^{2} = (a - 2)^{2} + (b - 3)^{2} \Rightarrow - 12 a + 36 + 2 b + 1 = - 4 a + 4 - 6 b + 9 \Rightarrow - 8 a = - 8 b - 24 \Rightarrow a = b + 3 \Rightarrow k_{1} b + k_{2} = b + 3 \Rightarrow k_{1} = 1, k_{2} = 3 ∴ k_{1} + k_{2} = 4$

Suggest Corrections

Similar questions

Equilibrium constants $K_{1}$ and $K_{2}$ for the following equilibria are

$N O (g) + 0.5 O_{2} (g) \to K_{1} N O_{2} (g)$ and
$2 N O_{2} (g) \to K_{2} 2 N O (g) + O_{2} (g)$

k_{1} a n d k_{2}

(0, a)

(0, - a)

y

y = 3 x + 4

a = \pm \frac{k_{1}}{\sqrt{k_{2}}}

k_{1}, k_{2}

k_{1} + k_{2}

P (a, b)

P

\frac{a}{b}

k_{1}

k_{2},

k_{1} + k_{2}

A \to B (k = k_{1}) B \to C (k = k_{2}) C \to D (k = k_{3})

A \to D

Join BYJU'S Learning Program