Question

Find the equation of a straight line passing through (-2,1) and is perpendicular to the line
2y = 5x + 2.​

Answer 1

Let the line which is required be $\sf L_{2}$.

And let the line $\sf 2y = 5x + 2$ be $\sf L_{1}$.

$\sf 2y = 5x +2 \\ \sf \longrightarrow y = \dfrac{5x + 2}{2}\\\longrightarrow \sf y = \dfrac{5}{2}x + 1$

It is now in the form of $\sf y = mx +c$, which is the slope intercept form of equation of straight line, where m is the gradient/slope, and c is the y intercept. On comparing it with $\sf y = \dfrac{5}{2}x + 1$, we find that $\sf slope\:of\:L_1 = m_1 = \dfrac{5}{2}$ .

It is given that $\sf L_{1}$ is perpendicular to $\sf L_{2}$.

We know that the product of slopes of two perpendicular lines is -1.

Let the slope of $\sf L_2$ be $\sf m_2$.

So we can say that,

$\sf m_1 \times m_2 = -1\\\longrightarrow \sf \dfrac{5}{2} \times m_2 = -1\\\longrightarrow \sf m_2 = \dfrac{-2}{5}$

Now, we have to use the point slope form of straight line.

$\boxed{\sf y - y_2 = m(x - x_2)}$

Here, m is the slope of the line and $\sf (x_2, y_2)$ is a point on the line.

We have already found that $\sf slope\; of\; L_2 = m_2 = \dfrac{-2}{5}$. And it is given that $\sf (-2,1)$ is a point on it. Hence, $\sf x_2 = -2, y_2 = 1$.

So,

$\sf y - y_2 = m(x - x_2)\\\longrightarrow \sf y - 1= \dfrac{-2}{5}\Big(x - (-2)\Big)\\\longrightarrow \sf y - 1= \dfrac{-2}{5}(x + 2)\\\longrightarrow \sf 5(y - 1) = -2(x + 2)\\\longrightarrow \sf 5y - 5 = -2x - 4\\\longrightarrow \sf 5y + 2x = 1$

Hence, the answer is,

$\longrightarrow \underline{\underline{\sf{\green{5y + 2x = 1}}}}$