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Line l passes through point (6,0) and line p is the graph of 2x-3y=4. If l is perpendicular to line p, what is the equation of l.

Sagot :

RedBeetle

Equation of line p:

[tex]{\sf{2x - 3y = 4}}[/tex]

[tex]{\sf{y = \frac{2x}{3} - \frac{4}{3}}}[/tex]

Slope of line p (m):

[tex]{\sf{\frac{2}{3}}} [/tex]

Since, l is perpendicular to line p, the product of slopes of line l & p should be -1. We assume slope of line l be m2

Hence,

[tex]{\sf{m \times m2 = - 1}}[/tex]

[tex]{\sf{ \frac{2}{3} \times m2 = - 1}}[/tex]

[tex]{\sf{m2 = \frac{ - 3}{2}}}[/tex]

Since, line l passes through points (6, 0).

We apply,

[tex]{\sf{(y - y1) = m2(x - x1)}}[/tex]

[tex]{\sf{y - 0 = \frac{ - 3}{2}(x - 6)}} [/tex]

[tex]{\sf{2y - 0 = - 3x + 18}}[/tex]

[tex]{\sf{3x + 2y - 18 = 0}}[/tex]

The equation of line l:

[tex]{\sf{\red{\boxed{\sf{3x+2y-18=0}}}}}[/tex]

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