Question

Find the equation of the line passing through (2,$-$1) and parallel to 3x + 4y = 10.

Answer 1

The given equation of the line is 3x + 4y = 10.

Slope of the line = $\frac{-\mathrm{coefficient}\mathrm{of}x}{\mathrm{coefficient}\mathrm{of}y}=\frac{-3}{4}$
The required line is parallel to the line 3x + 4y = 10 and we know that parallel lines have equal slopes.
Hence, slope of the required line, m = $\frac{-3}{4}$ and point (2, $-$1) lies on the required line.
So, the point slope form of the equation is
y $-$y₁ = m (x $-$ x₁)
⇒ y $-$ ($-$ 1) = $\frac{-3}{4}$(x $-$ 2)
⇒ y + 1 = $\frac{-3}{4}$ (x $-$ 2)
⇒ 4(y + 1) = $-$ 3 (x$-$ 2)
⇒ 4y + 4 = $-$ 3x + 6
∴ 3x + 4y $-$ 2 = 0, which is the required equation.