Find the slope of any line perpendicular to the line passing through (3,−2) and (12,19)

hazbijav6
2022-09-01
Answered

Find the slope of any line perpendicular to the line passing through (3,−2) and (12,19)

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Giancarlo Phelps

Answered 2022-09-02
Author has **10** answers

If the two points are $({x}_{1},{y}_{1})$ and $({x}_{2},{y}_{2})$, the slope of the line joining them is defined as

$\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ or $\frac{{y}_{1}-{y}_{2}}{{x}_{1}-{x}_{2}}$

As the points are (3,−2) and (12,19)

the slope of line joining them is $\frac{19-(-2)}{12-3}$ or $\frac{21}{9}$

i.e. $\frac{7}{3}$

Further product of slopes of two lines perpendicular to each other is −1.

Hence slope of line perpendicular to the line passing through (3,−2) and (12,19) will be $-\frac{1}{\frac{7}{3}}$ or $-\frac{3}{7}$

$\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ or $\frac{{y}_{1}-{y}_{2}}{{x}_{1}-{x}_{2}}$

As the points are (3,−2) and (12,19)

the slope of line joining them is $\frac{19-(-2)}{12-3}$ or $\frac{21}{9}$

i.e. $\frac{7}{3}$

Further product of slopes of two lines perpendicular to each other is −1.

Hence slope of line perpendicular to the line passing through (3,−2) and (12,19) will be $-\frac{1}{\frac{7}{3}}$ or $-\frac{3}{7}$

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