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Prove that perpendiculars drawn on intersecting lines are also intersecting.

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nasirjam | Student, Grade 9 | (Level 2) Honors

Posted January 8, 2013 at 3:59 PM via web

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Prove that perpendiculars drawn on intersecting lines are also intersecting.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted January 8, 2013 at 5:40 PM (Answer #1)

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You need to consider the slope intersection forms of equations of two intersecting lines, such that:

`{(y = m_1*x + n_1),(y = m_2*x + n_2):}` `=>` `{(-m_1*x + y = n_1),(-m_2*x + y = n_2):}`

You should remember that the lines are intercepting each other if the determinant of the system does not have the value 0, such that:

`[(-m_1,1),(-m_2,1)] != 0 => ` -`m_1 + m_2 != 0 => m_2 - m_1 != 0`

You need to remember that the product of the slopes of two perpendicular lines is `-1` , hence, you may evaluate the slopes of perpendicular lines, such that:

`m_1*m'_1 = -1 => m'_1 = -1/(m_1)`

`m_2*m'_2 = -1 => m'_2 = -1/(m_2)`

You need to write the equations of perpendicular lines such that:

`{(y = m'_1*x + n'_1),(y = m'_2*x + n'_2):}`

You need to consider the condition of two intersecting lines, such that:

`[(-m'_1,1),(-m'_2,1)] != 0 `=> `-m'_1 + m'_2 != 0 => m'_2 - m'_1 != 0`

You need to substitute `-1/(m_1)` for `m'_1` and `-1/(m_2)` for `m'_2` such that:

`-1/(m_2) + 1/(m_1) != 0 => (-m_1 + m_2)/(m_1*m_2) != 0`

Since `m_1, m_2 != 0` and `m_2 - m_1 != 0` (condition of intersecting lines) yields that `(-m_1 + m_2)/(m_1*m_2) != 0` , hence, the perpendicular lines are intersecting each other.

Hence, using the slope intercept forms of equations of two lines and the condition of two perpendicular lines yields that the perpendicular lines to two intercepting lines are also intercepting each other.

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