Question

What is the equation in slope-intercept form of the line that crosses the x-axis at 36
and is perpendicular to the line represented by y = -4/9x + 5?

Answer 1

SOLUTION

TO DETERMINE

The equation in slope-intercept form of the line that crosses the x-axis at ( 36, 0) and is perpendicular to the line represented by

$\displaystyle \sf{y = - \frac{4}{9}x + 5 }$

EVALUATION

Here the given equation of the line is

$\displaystyle \sf{y = - \frac{4}{9}x + 5 } \: \: \: \: .......(1)$

So the slope of the given line is

$\displaystyle \sf{m = - \frac{4}{9} }$

So the slope of the line perpendicular to the given line is

$\displaystyle \sf{m_1 = \frac{9}{4} }$

Hence the equation of the line perpendicular to the line given by Equation (1) is

$\displaystyle \sf{y = \frac{9}{4}x + k } \: \: \: \: .......(2)$

Now the line given by Equation (2) passes through the point ( 36,0)

$\displaystyle \sf{0 = \frac{9}{4} \times 36 + k } \: \: \: \:$

$\displaystyle \sf{ \implies \: k + 81 = 0 } \: \: \: \:$

$\displaystyle \sf{ \implies \: k = - 81 } \: \: \: \:$

Hence the required equation of the line in Slope intercept form is

$\displaystyle \sf{y = \frac{9}{4}x - 81 } \: \: \: \:$

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