The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3 respectively. If its orthocenter is (2, a), <math> <mrow> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo><</mo> <mi>a</mi> <mo><</mo> <mn>2</mn> <mo>,</mo> </mrow> </math> then p is equal to…………

Slope of AH = a + 2 1 slope of BC = − 1 p ∴ p = a + 2 ∴ C ( 1 8 p − 3 0 p + 1 , 1 5 p − 3 3 p + 1 ) slope of HC = 1 6 p − p 2 − 3 1 1 6 p − 3 2 slope of BC × slope of HC = -1 Þ p = 3 or 5hence p = 3 is only possible value.

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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Straight Lines Ncert Solutions Maths class 11th Maths Class 11th

The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3 respectively. If its orthocenter is (2, a), $- \frac{1}{2} < a < 2,$ then p is equal to…………

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Raj Pandey | Contributor-Level 9

2 weeks ago

Slope of AH = $\frac{a + 2}{1}$ slope of BC = $- \frac{1}{p} ∴ p = a + 2$ $∴ C (\frac{18 p - 30}{p + 1}, \frac{15 p - 33}{p + 1})$

slope of HC = $\frac{16 p - p^{2} - 31}{16 p - 32}$

slope of BC × slope of HC = -1 Þ p = 3 or 5

hence p = 3 is only possible value.

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The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3 respectively. If its orthocenter is (2, a), $- \frac{1}{2} < a < 2,$ then p is equal to…………

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The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3 respectively. If its orthocenter is (2, a), − 1 2 < a < 2 , then p is equal to…………

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The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3 respectively. If its orthocenter is (2, a), $- \frac{1}{2} < a < 2,$ then p is equal to…………