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  • Points A (–6, 10), B (–4, 6) and C (3, –8) are collinear such that AB = 2 by 9AC.59080

Points A (–6, 10), B (–4, 6) and C (3, –8) are collinear such that AB =2/9 AC.

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Answer.      [Ture]

Solution.     If the points A (–6, 10), B(–4, 6) and C(3, –8) are collinear then area of \triangleABC = 0
Area \, of \, \triangle ABC= \frac{1}{2}\left [ -6\left ( 6+8 \right ) +\left ( -4 \right )\left ( -8-10 \right )+3\left ( 10-6 \right )\right ]
                                   = \frac{1}{2}\left [ -84+72+12 \right ]
Area of \triangleABC = 0
Hence, A, B and C are collinear
Distance \, between \, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}

AB= \sqrt{\left ( -4+6 \right )^{2}+\left ( 6-10 \right )^{2}}

AB=\sqrt{4+16}= \sqrt{20}= 2\sqrt{5}
Distance \, between \, AC= \sqrt{\left ( 3+6 \right )^{2}+\left ( -8-10 \right )^{2}}

= \sqrt{81+324}
= \sqrt{405}

AC= 9\sqrt{5}
Hence,
 \frac{AB}{AC}= \frac{2\sqrt{5}}{9\sqrt{5}}
AB= \frac{2}{9}AC

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