Find the equation of the line with the following conditions
slope $=$ -3 and y-intercept $=$ -1

3 x + y + 1 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x - 3 y + 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 x + y - 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + 3 y - 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $3 x + y + 1 = 0$
The
equation of any straight line can be written as $y = m x + c$ , where $m$ is its slope and $c$ is its y - intercept.
Hence, the equation of the line withslope $= - 3$ and y - intercept $= - 1$ is $y = - 3 x - 1$ or $3 x + y + 1 = 0$ .

Suggest Corrections

Similar questions

2

\sqrt{3} y^{2} - \sqrt{3} x^{2} - 2 x y = 0

Prove that the lines $\sqrt{3} x + y = 0, \sqrt{3} y + x = 0, \sqrt{3} x + y = 1$ and $\sqrt{3} y + x = 1$ form a rhombus.

x

y

3 x - y = [\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix}]

x - 3 y = [\begin{matrix} 0 & - 1 0 & - 1 \end{matrix}]

x + \sqrt{3} y - 2 = 0, \sqrt{3} x - y + 1 = 0, y = 0

Join BYJU'S Learning Program