直线 $l_1$ 的方程为 $x + 4y = 20$。 该直线交 x 轴于点 A，交 y 轴于点 B。 另一条与 $l_1$ 垂直的直线 $l_2$ 通过原点 $O$ 并在点 $P$ 处与 $l_1$ 相交。

1. Find the coordinates of the x-intercept A and y-intercept B of $l_1$.

For the x-intercept, $y=0$.

$x+4(0)=20$

$x=20$

So $A=(20,0)$.

For the y-intercept, $x=0$.

$0+4y=20$

$y=5$

So $B=(0,5)$.

2. Use the distance formula to calculate $AB$.

$AB=\sqrt{(20-0)^2+(0-5)^2}$

$=\sqrt{400+25}$

$=\sqrt{425}$

3. Simplify the surd length.

$\sqrt{425}=\sqrt{25\times17}=5\sqrt{17}$

1. Translate “equidistant from the coordinate axes” into an equation relating the coordinates of C.

Let $C=(x,y)$.

Since $C$ is equidistant from the coordinate axes, and $C$ lies on the first-quadrant part of $l_1$,
$x=y.$

2. Substitute $x=y$ into the equation of $l_1$ to find the coordinates of C.

Since $C$ lies on $l_1$,
$x+4y=20.$
Substitute $x=y$:
$y+4y=20$
$5y=20$
$y=4.$
Therefore,
$x=4.$

1. Find the gradient of $AB$ from the equation of line $l_1$.

Since $AB$ lies on $l_1$,
$x+4y=20$
$4y=-x+20$
$y=-\frac14x+5$
So the gradient of $AB$ is $-\frac14$.

2. Use the given product of perpendicular gradients to find the gradient of $OP$.

Given
$(\text{gradient of }AB)\times(\text{gradient of }OP)=-1,$
$-\frac14\times(\text{gradient of }OP)=-1.$
Hence,
$\text{gradient of }OP=4.$

3. Form the equation of $OP$ using its gradient and the fact that it passes through the origin.

Line $OP$ passes through the origin, so its $y$-intercept is $0$.

Using $y=mx+c$ with $m=4$ and $c=0$,
$y=4x.$

1. Set up simultaneous equations for the intersection point $P$.

At $P$, the point lies on both $l_1$ and $OP$.

From $l_1$: $x+4y=20$

From part (c): $y=4x$

2. Substitute $y=4x$ into $x+4y=20$ and solve for the coordinates of $P$.

Substitute $y=4x$ into $x+4y=20$:

$x+4(4x)=20$

$17x=20$

$x=\frac{20}{17}$

$y=4\left(\frac{20}{17}\right)=\frac{80}{17}$

So $P=\left(\frac{20}{17},\frac{80}{17}\right)$.

1. Identify a pair of triangles with equal corresponding angles.

Since $OP \perp AB$, both $\triangle AOP$ and $\triangle OPB$ are right-angled at $P$.

Also, the acute angles are equal because $OA \perp OB$ and $OP \perp AB$.

Hence,
$\triangle AOP \sim \triangle OPB.$