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Sia ? 6 years, 6 months ago
b + c, c + a, a + b will be in AP, if
(c + a) - (b + c) = (a + b) - (c + a)
{tex}\Rightarrow{/tex} a - b = b - c {tex}\Rightarrow{/tex} 2b = a + c
i.e., a, b, c are in AP.
Thus, a, b, c are in AP implies b + c, c + a, a + b are also in AP.
Hence proved.
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Sia ? 6 years, 6 months ago
{tex}x - y + 1 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}y = x + 1{/tex}
| -1 | 1 | 0 |
| 0 | 2 | 1 |
{tex}2x + y - 10 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}y = 10 - 2x{/tex}
| 2 | 4 | 5 |
| 6 | 2 | 0 |
Shaded portion is the area bounded by the lines and x-axis.
Area = {tex}\frac{1}{2} A C \times B L{/tex}
= {tex}\frac{1}{2} \times 6 \times 4{/tex}
= {tex}12 sq. units{/tex}

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