Question
Prove that the points $A(a, 0), B(0, b)$ and $C(1, 1)$ are collinear, if $\frac{1}{\text{a}{}}+\frac{1}{\text{b}}=1.$
✓
Answer
Consider the point $A(a, 0), B(0, b), $ and $C(1, 1)$
Here, $(x_1 = x, y_1 = y), B (x_2 = -5, y_2 = 7)$ and $(x_3 = -4, y_3 = 5)$ be the given points.
it is given that the point are collinear so,
$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$
$\Rightarrow a(b - 1) + 0(1 - 0) + 1(0 - b) = 0$
$\Rightarrow ab - a - b = 0$
Dividing the equation by ab:
$\Rightarrow1-\frac{1}{\text{b}}-\frac{1}{\text{a}}=0$
$\Rightarrow1-\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)=0$
$\Rightarrow\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)=1$
Therefore, the given point are collinear if $\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)=1.$
Here, $(x_1 = x, y_1 = y), B (x_2 = -5, y_2 = 7)$ and $(x_3 = -4, y_3 = 5)$ be the given points.
it is given that the point are collinear so,
$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$
$\Rightarrow a(b - 1) + 0(1 - 0) + 1(0 - b) = 0$
$\Rightarrow ab - a - b = 0$
Dividing the equation by ab:
$\Rightarrow1-\frac{1}{\text{b}}-\frac{1}{\text{a}}=0$
$\Rightarrow1-\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)=0$
$\Rightarrow\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)=1$
Therefore, the given point are collinear if $\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)=1.$
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