Find the equation of the straight line.

Question

Find the equation of the straight line that is parallel to $2y-x=7$ and bisects the line joining the points $(3,1)$ and $(1,-5)$.

So, I found the gradient = $\frac{1}{2}$

And I solved the midpoint: $[\frac{3+1}{2}, \frac{1+(-5)}{2}] = (2,-2)$

The equation of straight line is $y=mx+c$

To find $c$, $-2 = \frac{1}{2}(2)+c$, which gives $c=-3$

So my equation is $2y - x + 3=0$

But my book says $2y-x+6=0$

What is my error? Help please :'(

Answer 1

$$y=mx+c=\frac12x-3=\frac{x-6}2\implies 2y=x-6\implies 2y-x+6=0$$

You forgot to multiply $\;C\;$ by two...

Answer 2

As you rightly say, the gradient of the line $2y-x=7$ is $\frac{1}{2}$ because we may rearrange $2y-x=7$ to give $y=\frac{1}{2}x+\frac{7}{2}$. The line you want to find is parallel to this and so also has gradient $\frac{1}{2}$.

Again, you rightly say that the bisector of $(3,1)$ and $(1,-5)$ passes through the midpoint, i.e. $$\left( \frac{3+1}{2},\frac{1-5}{2}\right)=(2,-2)$$ We want the line that passes through $(2,-2)$ and has gradient $\frac{1}{2}$.

We have the line $y=\frac{1}{2}x+c$ which passes through $(2,-2)$. We need to find $c$. Putting $x=2$ and $y=-2$ gives $-2=\frac{1}{2}(2)+c$, i.e. $-2=1+c$, i.e. $c=-3$. The equation is then $$y = \frac{1}{2}x - 3$$

Answer 3

You can use another formula which is $y-y_1 = m(x-x_1)$ Where $y_1 = -24$ and $4x_1 = 2$.

Answer 4

Here's an another method using point slope form of straight line.

• $m = \dfrac12$
• $(x_1,y_1) = (2,-2)$

Equation of straight line is given by,

$(y-y_1) = m(x-x_1)$

$(y-(-2)) = \dfrac12(x-2)$

$ 2(y+2) =x-2 $

$2y + 4 = x-2$

$2y - x + 6 = 0$