36. By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.

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alok kumar singh | Contributor-Level 10

4 months ago

36.

Let the given points be A (3, 0), B (–2, –2) and C (8, 2). Then by two point form we can write equation of line passing point A (3, 0) and B (–2, –2) as

$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})$

$\Rightarrow y - 0 = \frac{- 2 - 0}{- 2 - 3} (x - 3)$

$\Rightarrow y = \frac{- 2}{- 5} (x - 3)$

$\Rightarrow 5 y = 2 (x - 3)$

$\Rightarrow 5 y = 2 x - 6$

$\Rightarrow 2 x - 5 y - 6 = 0 (1)$

If the three points A, B and C are co-linear, C will also lieonm the line formed by AB or satisfies equation (1).

Hence, putting x = 8 and y = 2 we have

L.H.S. = 2 × 8 – 5 × 2 – 6

= 16 – 10 – 6

= 0 = R.H.S.

The given three points are collinear.

36. Let the given points be A (3, 0), B (–2, –2) and C (8, 2). Then by two point form we can write equation of line passing point A (3, 0) and B (–2, –2) as<math><mrow><mi>y</mi><mo>−</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><mfrac><mrow><msub><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>−</mo><msub><mrow><mi>y</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow><mrow><msub><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>−</mo><msub><mrow><mi>x</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></mfrac><mrow><mo> (</mo><mrow><mi>x</mi><mo>−</mo><msub><mrow><mi>x</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow><mo>)</mo></mrow></mrow></math><math><mrow><mo>⇒</mo><mi>y</mi><mo>−</mo><mn>0</mn><mo>=</mo><mfrac><mrow><mo>−</mo><mn>2</mn><mo>−</mo><mn>0</mn></mrow><mrow><mo>−</mo><mn>2</mn><mo>−</mo><mn>3</mn></mrow></mfrac><mrow><mo> (</mo><mrow><mi>x</mi><mo>−</mo><mn>3</mn></mrow><mo>)</mo></mrow></mrow></math><math><mrow><mo>⇒</mo><mi>y</mi><mo>=</mo><mfrac><mrow><mo>−</mo><mn>2</mn></mrow><mrow><mo>−</mo><mn>5</mn></mrow></mfrac><mrow><mo> (</mo><mrow><mi>x</mi><mo>−</mo><mn>3</mn></mrow><mo>)</mo></mrow></mrow></math><math><mrow><mo>⇒</mo><mn>5</mn><mi>y</mi><mo>=</mo><mn>2</mn><mrow><mo> (</mo><mrow><mi>x</mi><mo>−</mo><mn>3</mn></mrow><mo>)</mo></mrow></mrow></math><math><mrow><mo>⇒</mo><mn>5</mn><mi>y</mi><mo>=</mo><mn>2</mn><mi>x</mi><mo>−</mo><mn>6</mn></mrow></math><math><mrow><mo>⇒</mo><mn>2</mn><mi>x</mi><mo>−</mo><mn>5</mn><mi>y</mi><mo>−</mo><mn>6</mn><mo>=</mo><mn>0</mn><mtext> </mtext><mtext> </mtext><mrow><mo> (</mo><mrow><mn>1</mn></mrow><mo>)</mo></mrow></mrow></math>If the three points A, B and C are co-linear, C will also lieonm the line formed by AB or satisfies equation (1).Hence, putting x = 8 and y = 2 we haveL.H.S. = 2 × 8 – 5 × 2 – 6= 16 – 10 – 6= 0 = R.H.S.The given three points are collinear.

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36. By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.

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