Straight Lines

The line that connects these two points rises by 6 as it runs along 2 and so rise over run is 6/2 which equals 3. It follows that the gradient of the line is 3.

Hence, this tells us that the line is of the form $y=3x+c$. To then find $c$, plug one of the given points into the equation so far:

$4=3\times 2 +c$ and so c must be -2.

The final equation of the line as $y=3x-2$.

1. Parallel lines have the same gradient and so the line required has gradient 3. Since it intersects the y-axis at (0,4), the y-intercept is 4 and so the equation is given by [math]y=3x+4[/math].
2. The line perpendicular to $y=2x+3$ ($m=2$) has gradient $-\frac{1}{2}$ (taking $-1/m$). The equation so far is $y=-\frac{1}{2}x+c$. We can find $c$ by inserting the coordinates $(2,7)$ giving$7=-\frac{1}{2}\times 2+c$. Hence, $c=8$ and the equation of the line is $y=-\frac{1}{2}x+8$. Multiplying through by two to make all coefficients integer gives $2y=-x+16$. Finally, we have the equation in the required form: $x+2y-8=0$.

Always draw sketches to accompany your working – they help with sense checking.

We start by sketching the information as this will tell us more about the problem. To find the distance $OA$ we require the coordinates of point $A$. To do this, we must first find the equation of line $l_2$. The line $l_1$ can be rewritten as $y=\frac{1}{3}x+\frac{8}{3}$, which has gradient 1/3. The gradient of $l_2$ is thus -3 and if it intersects the $x$-axis at 5, it must intersect the $y$-axis at 15, using rise over run. Hence, $l_2$ has equation $y=-3x+15$. To find the coordinates of $A$, the intersection of the two lines, we solve simultaneous equations: $y=\frac{1}{3}x+\frac{8}{3}$ and [math]y=-3x+15[/math], to get $x=3.7$ and $y=3.9$. See Simultaneous Equations.

The length $OA$ can now be found using Pythagoras: $OA=\sqrt{3.7^2+3.9^2}=5.38$ to two decimal places. The area of triangle $OAB$ can now be found using the height of the triangle (3.9) and its base (5), i.e. $0.5\times 5\times 3.9=9.75$.