help me sir and ma'am how to solved the equation using directed distance
here in question Question: find the equation of the bisector of angles for the pair of line line one 4x+2y=9 line two 2x-y=8 please sir I need your help now and thanks in advanced.
here in question Question: find the equation of the bisector of angles for the pair of line line one 4x+2y=9 line two 2x-y=8 please sir I need your help now and thanks in advanced.
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There are two angle bisectors. First write both lines in general form.
4x + 2y - 9 = 0
2x - y - 8 = 0
Any point (x, y) on either angle bisector is equidistant from the two lines, so equate those distances.
|4x + 2y - 9|/√(4² + 2²) = |2x - y - 8|/√(2² + 1²)
(4x + 2y - 9)²/(4² + 2²) = (2x - y - 8)²/(2² + 1²)
5(4x + 2y - 9)² = 20(2x - y - 8)²
(4x + 2y - 9)² = 4(2x - y - 8)²
16x² + 16xy + 4y² - 72x - 36y + 81 = 4(4x² - 4xy + y² - 32x + 16y + 64)
16x² + 16xy + 4y² - 72x - 36y + 81 = 16x² - 16xy + 4y² - 128x + 64y + 256
32xy + 56x - 100y - 175 = 0
That is the locus. It may be factored to find the component lines.
(8x - 25)(4y + 7) = 0
The two bisectors:
8x - 25 = 0
4y + 7 = 0
4x + 2y - 9 = 0
2x - y - 8 = 0
Any point (x, y) on either angle bisector is equidistant from the two lines, so equate those distances.
|4x + 2y - 9|/√(4² + 2²) = |2x - y - 8|/√(2² + 1²)
(4x + 2y - 9)²/(4² + 2²) = (2x - y - 8)²/(2² + 1²)
5(4x + 2y - 9)² = 20(2x - y - 8)²
(4x + 2y - 9)² = 4(2x - y - 8)²
16x² + 16xy + 4y² - 72x - 36y + 81 = 4(4x² - 4xy + y² - 32x + 16y + 64)
16x² + 16xy + 4y² - 72x - 36y + 81 = 16x² - 16xy + 4y² - 128x + 64y + 256
32xy + 56x - 100y - 175 = 0
That is the locus. It may be factored to find the component lines.
(8x - 25)(4y + 7) = 0
The two bisectors:
8x - 25 = 0
4y + 7 = 0
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Find the intersection between those two equations. The slope will be the average, so 3. Then plug in the intersection to y=mx+b to find b.
y=3x+b
y=3x+b
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These lines are parallel to each other. They do not intersect.