What is the equation of a line that passes through the point (10, 5) and is perpendicular to the line whose equation is y=5/4 x-2?

Hey there :)

[tex]y = \frac{5}{4}x - 2 [/tex]
Slope is the coefficient of x ⇒ [tex] \frac{5}{4} [/tex]

If the second line is perpendicular to the given line, then the slope of the 2nd line is the reverse/reciprocal of the 1st line ⇒ [tex] \frac{4}{5} [/tex]

(10, 5)
↑ ↑
x₁ y₁

Point-slope form is ( y - y₁ ) = m ( x - x₁ )

y - 4 = [tex] \frac{4}{5} [/tex] ( x - 10 )
y - 4 = [tex] \frac{4}{5} x[/tex] - 8

y = [tex] \frac{4}{5} x - 4[/tex] ⇔ The equation of the line that passes through the point (10,5) ans perpendicular to the given line

ksalmon1 ksalmon1 · Answer 2 · 2019-01-16T21:00:31+01:00

ksalmon1 ksalmon1

16-01-2019

Answer:

Question with point (10,5)

y=-4/5x+13

Slope of the perpendicular line would be the NEGATIVE reciprical!

Step-by-step explanation:

The answer above is good (except for forgetting it is the negative reciprical) and the person used the point (10,4) not (10,5) from the question...

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