Question

Which equations represent the line that is perpendicular to the line 5x - 2y = -6 and passes through the point

Answer 1

Answer:  2x+5y+10=0

Step-by-step explanation:

The equation of the given line : $5x - 2y = -6$ which can be written as $2y=5x+6$ or  $y=\dfrac{5}{2}x+3$ . (i)

Linear equation : $y=mx+c$, where m= slope and c = intercept.

Comparing this to (i), we get $m=\dfrac{5}{2}, c=3$

Let $m_1$ be the slope of the line perpendicular to the line 5x - 2y = -6 and passes through the point  (5,-4).

Since, the product of slopes of two perpendicular lines is -1.

So,

$\dfrac{5}{2}\times m_1=-1\\\\\Rightarrow\ m_1=-\dfrac{2}{5}$

Equation of line passes through (a,b) and have slope n is given by :-

$(y-b)=n(x-a)$

So, Equation of line passes through (5,-4) and have slope  $\dfrac{2}{5}$ would be

$(y-(-4))=\dfrac{-2}{5}(x-5)\\\\\Rightarrow\ 5(y+4)=-2(x-5)\\\\\Rightarrow\ 5y+20=-2x+10\\\\\Rightarrow\ 2x+5y+10=0$

Required equation : $2x+5y+10=0$