meebuigenzg

2022-10-02

Find the equation of the line perpendicular to $y=-\frac{5}{9}x$ that passes through (−7,3)

Brendan Bradley

Beginner2022-10-03Added 11 answers

One of the forms of the equation of a straight line is y = mx + c where m represents the gradient and c , the y-intercept.

the line $y=-\frac{5}{9}x$

is in this form with c = 0 and m = $-\frac{5}{9}$

When 2 lines are perpendicular then the product of their gradients :

${m}_{1}{m}_{2}=-1$

The gradient of the perpendicular line is : $-\frac{5}{9}\times {m}_{2}=-1$

$\Rightarrow {m}_{2}=-\frac{1}{-\frac{5}{9}}=\frac{9}{5}$

equation : y - b = m(x - a ) , m = $\frac{9}{5},(a,b)=(-7,3)$

$\Rightarrow y-3=\frac{9}{5}(x-7)$

multiply both sides by 5 to eliminate fraction : $5y-15=9x-63$

equation of perpendicular line is 5y - 9x + 48 = 0

the line $y=-\frac{5}{9}x$

is in this form with c = 0 and m = $-\frac{5}{9}$

When 2 lines are perpendicular then the product of their gradients :

${m}_{1}{m}_{2}=-1$

The gradient of the perpendicular line is : $-\frac{5}{9}\times {m}_{2}=-1$

$\Rightarrow {m}_{2}=-\frac{1}{-\frac{5}{9}}=\frac{9}{5}$

equation : y - b = m(x - a ) , m = $\frac{9}{5},(a,b)=(-7,3)$

$\Rightarrow y-3=\frac{9}{5}(x-7)$

multiply both sides by 5 to eliminate fraction : $5y-15=9x-63$

equation of perpendicular line is 5y - 9x + 48 = 0

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