Question

Street L is perpendicular to and has the same y-intercept
as the line 5x+6y= 78. Write this street's equation in
standard form.

Answer 1

Answer:

$6x-5y+65=0$

Step-by-step explanation:

Given equation of line ,

$\implies 5x + 6y = 78$

Convert it to Slope Intercept Form which is $y=mx + c$ , we have ,

$\implies 6y = 78 - 5x \\\\\implies y= \dfrac{-5}{6}x + 13$

Therefore

• $m =\dfrac{-5}{6}$
• $c = 13$

As we know that the product of slopes of two perpendicular lines is -1 . Hence the slope of the perpendicular line will be ,

• $m_{perp}= \dfrac{6}{5}$
• $y- intercept = (0,13)$

On using point slope form of the line , we have ,

$\implies y - y_1 = m(x - x_1) \\\\\implies y - 13 = \dfrac{6}{5}( x - 0 ) \\\\\implies 5y - 65 = 6x \\\\\implies \underline{ \boxed{ 6x - 5y + 65 = 0 }}$